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Technical Mathematics
Technical Mathematics
Author: Todd Simeone (Technology Faculty, Heartland Community College)
Volume I
Updated November 18, 2016
Technical Mathematics
Table of Contents
Chapter 1 – Prerequisite material................................................................................................................. 5
1.1 – Arithmetic with whole numbers ...................................................................................................... 5
1.2 – Fractions ........................................................................................................................................... 7
1.3 – Mixed numbers and Decimals.......................................................................................................... 8
1.4 – Order of Operations ....................................................................................................................... 10
1.5 – Solving Linear Equations ................................................................................................................ 12
1.6 – Ratio and Proportion ...................................................................................................................... 13
1.7 – Scientific Notation .......................................................................................................................... 14
Chapter 1 Homework problems.............................................................................................................. 17
Chapter 2 – Measurement Systems ............................................................................................................ 21
2.1 – Standard Systems ........................................................................................................................... 21
2.2 – Areas and Volumes ........................................................................................................................ 25
2.3 – Temperature .................................................................................................................................. 27
2.4 – Converting between systems ......................................................................................................... 29
Chapter 2 Homework problems.............................................................................................................. 30
Chapter 3 – Percentages, precision, error, and tolerance intervals ........................................................... 32
3.1 – Percentages: basic conversions ..................................................................................................... 32
3.2 – Percentages: base, part, and rate .................................................................................................. 34
3.3 – Precision and error ......................................................................................................................... 36
3.4 – Tolerance intervals ......................................................................................................................... 38
Chapter 3 Homework problems.............................................................................................................. 40
Chapter 4 – Number systems...................................................................................................................... 42
4.1 – Decimal number system ................................................................................................................ 42
4.2 – Electronic machines ....................................................................................................................... 43
4.3 – Base-2 (Binary) ............................................................................................................................... 44
4.4 – Base – 16 (Hexadecimal) ................................................................................................................ 46
Chapter 4 Homework problems.............................................................................................................. 51
Chapter 5 – Formulas and data substitution .............................................................................................. 53
5.1 – Substituting data into formulas ..................................................................................................... 53
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5.2 – Simple algebraic manipulation of formulas ................................................................................... 55
Chapter 5 Homework problems.............................................................................................................. 57
Chapter 6 – Basic Geometry ....................................................................................................................... 58
6.1 – Angles ............................................................................................................................................. 58
6.2 – Triangles ......................................................................................................................................... 60
The Pythagorean Theorem ................................................................................................................. 61
Area ..................................................................................................................................................... 64
6.3 – Quadrilaterals................................................................................................................................. 67
6.4 – Circles ............................................................................................................................................. 69
Circle terminology and basic formulas................................................................................................ 69
Arcs and associate angles ................................................................................................................... 70
Chapter 6 Homework problems.............................................................................................................. 74
Chapter 7 – Basic Trigonometry.................................................................................................................. 80
7.1 – The Trigonometric ratios ................................................................................................................ 80
7.2 – Finding trigonometric values on a calculator ................................................................................. 83
Using the calculator to determine trigonometric values of angles. ................................................... 84
Using the calculator to determine angles given their trigonometric values. ..................................... 85
7.3 – Using trigonometry to find unknown components of a triangle ................................................... 87
Chapter 7 Homework problems.............................................................................................................. 89
Chapter 8 – Interpreting Charts and Graphs............................................................................................... 93
8.1 – Charts and Graphs .......................................................................................................................... 93
Bar charts ............................................................................................................................................ 93
Pie charts............................................................................................................................................. 96
Coordinate system graphs .................................................................................................................. 97
Chapter 8 Homework problems............................................................................................................ 100
Chapter 9 – Basic Statistics ....................................................................................................................... 105
9.1 – Data presentation ........................................................................................................................ 105
9.2 – Measures of central tendency ..................................................................................................... 109
9.3 – Measures of dispersion ................................................................................................................ 112
9.4 – Normally distributed data sets..................................................................................................... 116
9.5 – Controlled and capable processes ............................................................................................... 119
Chapter 9 Homework problems............................................................................................................ 121
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Technical Mathematics
Chapter 10 – Basic Logic ........................................................................................................................... 125
10.1 – Logical expressions..................................................................................................................... 125
The ‘And’ logical operator................................................................................................................. 126
The ‘Or’ logical operator ................................................................................................................... 127
The ‘Not’ logical operator ................................................................................................................. 127
10.2 – Logic truth tables ....................................................................................................................... 128
10.3 – Schematics ................................................................................................................................. 131
Chapter 10 Homework problems.......................................................................................................... 133
Supplemental Chapter – Systems of Linear Equations ............................................................................. 135
S.1 – Solving systems of 2 linear equations using a graphing tool ....................................................... 135
S.2 – Solving systems of 2 equations using the Addition-Subtraction method .................................... 138
S.3 – Solving systems of 2 equations using the Substitution method .................................................. 142
S.4 – Applications associated with systems of equations ..................................................................... 146
S.5 – Systems of 3 linear equations ...................................................................................................... 148
Chapter S Homework problems ............................................................................................................ 151
Answers to Homework Problems ............................................................................................................. 155
Chapter 1 Answers ................................................................................................................................ 155
Chapter 2 Answers ................................................................................................................................ 158
Chapter 3 Answers ................................................................................................................................ 160
Chapter 4 Answers ................................................................................................................................ 162
Chapter 5 Answers ................................................................................................................................ 163
Chapter 6 Answers ................................................................................................................................ 164
Chapter 7 Answers ................................................................................................................................ 166
Chapter 8 Answers ................................................................................................................................ 167
Chapter 9 Answers ................................................................................................................................ 169
Chapter 10 Answers .............................................................................................................................. 172
Supplemental Chapter Answers............................................................................................................ 174
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Technical Mathematics
Chapter 1 – Prerequisite material
It is assumed that readers of this text have been exposed to the majority of the Chapter 1 content (if not
all of it). As such much of the content is provided with brevity.
1.1 – Arithmetic with whole numbers
It is assumed that the reader is comfortable with the content of section 1.1 more than any other section
in this chapter. A small sampling of examples is provided with no formal discussion.
Adding whole numbers:
Example 1
3  14  17
3  (14)  11
 3  14  11
(3)  (14)  17
Subtracting whole numbers:
Example 2
3  14  11
3  (14)  17
 3  14  17
(3)  (14)  11
Multiplying whole numbers:
Example 3
3  14  42
3  (14)  42
 3  14  42
(3)  (14)  42
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Technical Mathematics
Dividing whole numbers:
Example 4
15  3  5
15  (3)  5
(15)  3  5
(15)  (3)  5
Exponentiation:
Example 5
112  1111  121
Example 6
23  2  2  2  8
Example 7
86  8  8  8  8  8  8  262,144
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Technical Mathematics
1.2 – Fractions
Adding and subtracting fractions
When adding fractions, the denominator must be the same. This is referred to as a ‘common
denominator’. As in the first section, only a limited discussion is provided for this review material.
Example 1
4 1 5
 
6 6 6
Since the two fractions had 6 as the common denominator, the two numerators
are added and placed over the common denominator
A similar process is used to subtract fractions:
Example 2
5 4 1
 
8 8 8
When adding fractions that do not have the same denominator, a common denominator must be found.
Example 3
4 1

7 3
21 can be used as the common denominator since 7 and 3 are both
divisors of 21
4 3 12
 
7 3 21
1 7 7
 
3 7 21
The fractions in the original problem are replaced with equivalent
fractions over that have a common denominator
4 1 12 7 19
 


7 3 21 21 21
The answer to the original problem is
19
.
21
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Technical Mathematics
A similar process is used when subtracting fractions with different denominators.
Example 4
11 2

6 3
2 2 4
 
3 2 6
11 2 11 4 7
 
 
6 3
6 6 6
The answer to the original problem is
7
.
6
1.3 – Mixed numbers and Decimals
Mixed numbers are numbers that contain both an integer and a fraction. They always have an
equivalent fractional representation knows as an improper fraction.
Example 1
Write the improper fraction
23
as a mixed number.
7
Solution: When 23 is divided by 7, the result is 3 with a remainder of two. 3 becomes the integer in the
mixed number, 2 is the numerator of the fraction, and 7 is the denominator as follows:
23
2
3
7
7
Example 2
Write the mixed number 4
11
as an improper fraction.
13
Solution: To determine the numerator, the following calculation is used: (4 13)  11  63 . As a result,
63 becomes the numerator and the denominator is 13 as follows:
4
11 63

13 13
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Technical Mathematics
In addition to improper fractions and mixed numbers, decimal numbers are another way to represent
these values.
Example 3
Write the fraction
3
as a decimal.
8
Solution: Using a calculator, 3  8  0.375 , and so
3
 0.375 .
8
Example 4
Write the mixed number 5
7
as a decimal.
16
Solution: Using a calculator, 5  7 16  5.4375 , and so 5
7
 5.4375 .
16
Example 5
Write the decimal 0.2 as a fraction.
Solution: Since the 2 is in the tenths place, 0.2 
2 1

10 5
Example 6
Write the decimal 3.21 as a mixed number.
Solution: Since the 1 is in the hundredths place, 3.21  3
21
100
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Technical Mathematics
1.4 – Order of Operations
When multiple operations are to be performed, there is an accepted order in which the operations must
take place. Many students remember the acronym PEMDAS to help keep track of this order. PEMDAS
stands for the following:
1. P = Parenthesis
Any operations within parentheses are done first. Parentheses are used to override the normal
order of operations.
2. E = Exponents
3. MD = Multiplication and division
In step 3, all multiplication and division operations are done. If there is more than one, they are
done in the order they appear from left to right.
4. AS = Addition and subtraction
In step 4, all addition and subtraction operations are done. If there are more than one, they are
done in the order they appear from left to right.
Example 1
Calculate: 3  3  3
Solution:
3  3 3
3 3 is calculated first since multiplication is done before addition
 39
3  9 is calculated since it is the lone remaining calculation to perform
 12
The resulting answer is 12
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Technical Mathematics
Example 2
Calculate: 10  2  7 2  42  15  (3  4)
Solution:
10  2  7 2  42  15  (3  4)
(3  4) is calculated first since it is within parentheses
 10  2  72  42  15  7
7 2 is calculated next since it is an exponent
 10  2  49  42 15  7
10 2 is calculated next since it is the left-most
multiplication
 20  49  42 15  7
15 7 is calculated next since it is multiplication
 20  49  42 105
20  49 is calculated next since all remaining
operations have the same precedence and it is the
left-most operation remaining
 29  42 105
 29  42 is calculated next since all remaining
operations have the same precedence and it is the
left-most operation remaining
 13  105
13  105 is the lone remaining calculation to perform
 92
The resulting answer is  92
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Technical Mathematics
1.5 – Solving Linear Equations
Linear equations are equations that have only one variable that has an exponent of 1. For example, the
following is a linear equation:
x  5  14
The goal when solving a linear equation is to determine all of the numbers that can be substituted in
place of the variable which result in a true statement. 9 is a solution to the above equation because
when 9 is substituted in place of x, a true statement results:
Example 1a
x  5  14
9  5  14
14  14
Following a similar line of thinking, 10 is NOT a solution because when 10 is substituted in place of x, a
false statement results:
Example 1b
x  5  14
10  5  14
15  14
Additional values can be substituted in for x, but only 9 will result in a true statement (and hence only 9
is a solution to this equation).
As equations become more complex, there are a few rules that can be applied to help determine the
solutions:
1.
2.
Any value can be added to or subtracted from both sides of an equation.
Both sides of the equation can be multiplied or divided by any non-zero value.
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Technical Mathematics
Example 2
Find the solution of the equation: 3x  17  2
Solution:
3x  17  2
3x  17  17  2  17
17 is added to both sides of the equation
3x  15
3 x 15

3
3
Both sides of the equation are divided by 3
x5
x has been isolated, and the solution is 5
1.6 – Ratio and Proportion
Ratios are ways to compare one measurement to another and are represented as fractions. For
example, if the wall in a room is 30m, and the opposite wall is 100m, the values can be represented as:
30
3
which can be simplified to .
100
10
2 ratios are in proportion if they are equal to each other. For example,
because
7
21
and
are in proportion
11
33
7
21
.

11 33
Sometimes it is necessary to determine a specific value which will put two ratios in proportion. Crossmultiplication can be used to accomplish this task as follows:
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Technical Mathematics
Example 1
Find the value for x which solves the following proportion:
x
5

217 31
Solution:
x
5

217 31
31x  217  5
Cross-multiplication is performed on the proportion
31x  1085
31x 1085

31
31
Both sides of the equation are divided by 31
x  35
x has been isolated, and the solution is 35
1.7 – Scientific Notation
Scientific notation provides another way to represent numbers. This notation is often used in a technical
setting to provide a short-hand notation for numbers that might otherwise be very long. Scientific
notation depends on powers of 10 to represent numbers. If you consider the following calculation:
3.6 1013
the calculated value is:
36, 000, 000, 000, 000 .
Since these are just two different representations of the same value, either can be used. However, since
the first one is shorter and more concise it is often used in technical settings, and is in fact known as the
scientific notation representation. The second representation is in what is known as decimal notation.
It can be seen that the second representation was generated from the first number by simply moving
the decimal point 13 places to the right. Scientific notation representations always work this way. The
only rules to remember when converting from scientific notation are:
1. If the exponent is positive, move the decimal point to the right
2. If the exponent is negative, move the decimal point to the left
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Technical Mathematics
The following table demonstrates this process:
Scientific notation
4.7 107
1.26 1010
7.3 1011
2.514 105
Rule to be applied
Positive exponent – move decimal to right
Positive exponent – move decimal to right
Negative exponent – move decimal to left
Negative exponent – move decimal to left
Converted to decimal notation
47, 000, 000
12, 600, 000, 000
0.000000000073
0.00002514
To convert from decimal notation to scientific notation, the process is reversed. Furthermore, the
decimal point is always moved so that the remaining number only has one digit to the left of the decimal
point. The rules to convert from decimal notation to scientific notation are as follows:
1. Move the decimal point so the number has exactly one digit to the left of the decimal
2. The exponent is the number of places the decimal point was moved
a. If the decimal was moved to the left, the exponent is positive
b. If the decimal was moved to the right, the exponent is negative
The following table demonstrates this process:
Decimal notation
Rule to be applied
231, 000, 000
4,127, 000, 000, 000, 000
Move decimal to left – positive exponent
Move decimal to left – positive exponent
Move decimal to right – negative exponent
Move decimal to right – negative exponent
0.000061
0.0003
Converted to scientific
notation
2.31108
4.127 1015
6.1105
3 104
When performing calculations with numbers in scientific notation, the most straightforward way to
handle it is to perform the calculations on the numbers separately from the calculations on the powers
of 10 as in the following example:
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Technical Mathematics
Example 1
Calculate the following:
(3.2 108 )  (2.11105 )
Solution:
(3.2 108 )  (2.11105 )
 (3.2  2.11)  (108 105 )
 6.752 1013
Example 2
Calculate the following:
3
(6.21  10 )  (4.11  10 )
11
9.1  10
4
Solution:
3
(6.21  10 )  (4.11  10 )
11
9.1  10
4
6.21  4.11 10  10

4
9.1
10
11

 2.8  10
3
4
As demonstrated by the next example, care must be taken to ensure the answer is in scientific notation.
Example 2
Calculate the following:
(9.2  105 )  (8.5  103 )
Solution:
(9.2 105 )  (8.5  103 )
 (9.2  8.5)  (105 103 )
 78.2 108
 7.82 109
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Technical Mathematics
Chapter 1 Homework problems
Section 1.2 Homework
Calculate the following.
1.
3 5

7 7
2.

3.
3 13

14 7
4.

5.
2 5

3 9
6.
1 4

2 7
7.
9  3
  
13  11 
8.
2 4

3 17
9.

10. 
4 12

11 11
41 5

6 6
12 1

13 2
33
  11
3
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Section 1.3 Homework
Fill in the following chart.
Mixed number
7
Improper Fraction
Decimal
4
7
11
13
41
24
9
1582
14
–6.62
74.6
Section 1.4 Homework
Calculate the following.
1.
7  2 11  8  2
2.
(7  2) 11  8  2
3.
(7  2)  (11  8)  2
4.
(82  33 ) 2
2
4
5. 9  (3  7)2  14  2  4
14 

6.  0.3  17  0.012  
11 

4
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Technical Mathematics
Section 1.5 Homework
In each of the equations, solve for the variable.
1.
x 11  28
2.
2x  7  71
3.
5x  4  3x  16
4.
3
7 1
2
x  x
2
3 6
3
5.
0.007 x  0.314  2.16
Section 1.6 Homework
In each of the following, find the value of x which solves the proportion.
1.
x 22

7 14
2.
x
4

3
5
3.
9
x

27 33
4.
23 x 1035

4833 1611
5.
0.023x 1.07

4.69
0.943
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Technical Mathematics
Section 1.7 Homework
Fill in the following chart:
Scientific Notation
9.21108
8.7 105
5 1014
Decimal Notation
509, 000, 000, 000
0.00007
0.0000000314
Calculate the following and express the answer in scientific notation.
1.
(2  10 )  (4  10 )
2.
(2.01  10 )  (3.72  10 )
3.
(3  10 )  (3  10 )
4.
(5.1  10 )  (3.2  10 )
7
5
3
7
17
53
9
4
8.21  10
14
5.
9.1  10
5
3
(7.21  10 )  (3.94  10 )  (1.7  10 )
11
6.
5
6
(9.1  10 )  (4  10 )
4
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Technical Mathematics
Chapter 2 – Measurement Systems
2.1 – Standard Systems
While the U.S. typically uses the Imperial System for measurement, the rest of the world uses the Metric
System. For this reason, the metric system is also known as the international system, which in other
languages is referenced as the System International, or SI system. The Metric System is easier to use
because everything is based on powers of 10. The following table provides the base values in the metric
system:
Prefix
Symbol
Power of 10
Decimal Equivalent
Tera
Giga
Mega
Kilo
Hecto
Deka
Standard Unit
Deci
Centi
Milli
Micro
Nano
Pico
T
G
M
k
h
da
12
9
6
3
2
1
1,000,000,000,000
1,000,000,000
1,000,000
1,000
100
10
d
c
m
u
n
p
–1
–2
–3
–6
–9
–12
0.1
0.01
0.001
0.000001
0.000000001
0.000000000001
The standard unit is dictated by the type of measurement that is being done according to the following
table:
Measurement
Length
Mass
Volume (liquid)
Current
Power
Resistance
Time
Temperature
Unit
Meter
Gram
Liter
Ampere (amp)
Watt
Ohm
Second
Celsius
Symbol
m
g
L
A
W
Ω
s
C
Using these charts, measurements can be converted from one prefix to another as in the following
examples.
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Technical Mathematics
Example 1
Convert 10 km to dm.
Solution:

Looking at the charts, it can be determined that km is the abbreviation for kilometers (k = kilo
and m = meters). Similarly, dm is the abbreviation for decimeters

The exponent for km is 3, and the exponent for dm is –1

The exponents are subtracted to find how many places the decimal point needs to be moved:
3 – (-1) = 4

The decimal point is moved 4 places. Since decimeters are smaller than kilometers, it is moved
to the right
10 km = 100,000 dm
Example 2
Convert 0.0032 hm to m.
Solution:

Looking at the charts, it can be determined that hm is the abbreviation for hectometers and m is
the abbreviation for meters (since it has no prefix, it is just the standard unit)

The exponent for hm is 2, and the exponent for m is 0 (the standard unit always has an
exponent of 0)

The exponents are subtracted to find how many places the decimal point needs to be moved:
2–0=2

The decimal point is moved 2 places. Since meters are smaller than hectometers, it is moved to
the right
0.0032 hm = 0.32 m
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Technical Mathematics
Example 3
Convert 432,000 cm to dam.
Solution:

Looking at the charts, it can be determined that cm is the abbreviation for centimeters and dam
is the abbreviation for dekameters

The exponent for dam is 1, and the exponent for cm is –2

The exponents are subtracted to find how many places the decimal point needs to be moved:
1 – (-2) = 3

The decimal point is moved 3 places. Since dekameters are larger than centimeters, it is moved
to the left
432,000 cm = 432 dam
One of the powerful aspects of the metric system is that the standard unit does not affect the way the
conversions are done. In the following examples, notice how the standard unit changes, but the process
is the same as for the above examples in meters (and would be the same for watts, liters, seconds, etc.).
Example 4
Convert 0.0013 A to mA.
Solution:

Looking at the charts, it can be determined that A is the abbreviation for Amp and mA is the
abbreviation for milliamps

The exponent for A is 0, and the exponent for mA is –3

The exponents are subtracted to find how many places the decimal point needs to be moved:
0 – (-3) = 3

The decimal point is moved 3 places. Since mA are smaller than A, it is moved to the right
0.0013 A = 1.3 mA
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Technical Mathematics
Example 5
Convert 1,000,000,000,000 ug to kg.
Solution:

Looking at the charts, it can be determined that ug is the abbreviation for microgram and kg is
the abbreviation for kilogram

The exponent for ug is –6, and the exponent for k is 3

The exponents are subtracted to find how many places the decimal point needs to be moved:
3 – (-6) = 9

The decimal point is moved 9 places. Since kg are larger than ug, it is moved to the left
1,000,000,000,000 ug = 1,000 kg
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Technical Mathematics
2.2 – Areas and Volumes
Care must be taken when areas or volumes are involved. The following example demonstrates how to
convert square cm (cm2) to square dam (dam2). Again, the standard unit itself (i.e. m2) is inconsequential
to the process.
Example 1
Convert 25,000 cm2 to dam2.
Solution:

Looking at the charts, it can be determined that cm2 is the abbreviation for square centimeters,
and dam2 is the abbreviation for square dekameters

The exponent for cm is –2, and the exponent for dam is 1

The exponents are subtracted and then multiplied by 2 because this is an area measurement to
find how many places the decimal point needs to be moved:
1 – (-2) = 3, and then 3 x 2 = 6

The decimal point is moved 6 places. Since dam2 are larger than cm2, it is moved to the left
25,000 cm2 = 0.025 dam2
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Technical Mathematics
Similarly, volumes will use a factor of 3. The following demonstrates a volume conversion.
Example 2
Convert 3.8 Mm3 to m3.
Solution:

Looking at the charts, it can be determined that Mm3 is the abbreviation for cubic megameters,
and m3 is the abbreviation for cubic meters

The exponent for Mm is 6, and the exponent for m is 0

The exponents are subtracted and then multiplied by 3 because this is a volume measurement
to find how many places the decimal point needs to be moved:
6 – (0) = 6, and then 6 x 3 = 18

The decimal point is moved 18 places. Since m3 are smaller than Mm3, it is moved to the right
3.8 Mm3 = 3,800,000,000,000,000,000 m3 or 3.8 x 1018 m3
26
Technical Mathematics
2.3 – Temperature
In the metric system, temperatures are handled in a completely different way than other
measurements. The Celsius scale provides the unit of measurement (you may hear about the Kelvin
scale which is also used in the metric system but it is used primarily for more scientific and engineering
applications). In the English system, the Fahrenheit system is used. The following table represents 3
important temperatures on the scales:
Celsius temperature
100°
0°
–273°
Fahrenheit temperature
212°
32°
–459.4°
Significance of measurement
Boiling point of water
Freezing point of water
Absolute zero
The following formulas provide the means to convert between these two temperature systems:
C  95 ( F  32 )
F  95 C  32
Example 1
Convert 22.3° C to F.
Solution:

Since we are converting to Fahrenheit, we will use the second formula listed above
F  95 C  32
9
F  (22.3)  32
5
F  40.14  32
F  72.14

So 22.3° C = 72.14° F
27
Technical Mathematics
Example 2
Convert –13.67° F to C.
Solution:

Since we are converting to Celsius, we will use the first formula listed above
C  95 ( F  32)
C  95 (13.67  32)
C  95 (45.67)
C  25.37

So –13.67° F = –25.37° C
28
Technical Mathematics
2.4 – Converting between systems
It is often necessary to convert from one number system to the other. Because the systems are so
different and have so many units to deal with, many times some research must be done to determine
how one unit relates to another.
Example 1
Chicago, IL and New York City, NY are approximately 800 miles apart. How far is this distance in km?
Solution:

A quick internet search reveals that 1 mi = 1.609 km. In order to convert mi to km, this
conversion factor must be used as follows:
800mi  800(1.609
km
)  1287.2km
mi
Example 2
What is the weight capacity in kg of a 0.5 ton truck?
Solution:

A quick internet search reveals that 1 ton = 907.185 kg. In order to convert tons to kg, this
conversion factor must be used as follows:
0.5tons  0.5(907.185
kg
)  453.593kg
ton
29
Technical Mathematics
Chapter 2 Homework problems
Section 2.1 Homework
1. Convert 46,120 cm to:
a. mm
b. m
c. dam
d. km
2. Convert 17 MW to:
a. mW
b. W
c. kW
d. GW
3. Convert 0.0005 L to:
a. dL
b. cL
c. kL
d. daL
Section 2.2 Homework
1. Convert 1,000,000 mm2 to:
a. um2
b. m2
c. km2
2. Convert 3.211013 km3 to:
a. cm3
b. m3
c. Tm3
30
Technical Mathematics
Section 2.3 Homework
1. Fill in the following chart:
Degrees Fahrenheit
–48°
111°
Degrees Celsius
3.2°
27°
2. A third temperature scale that is often used in Science and Engineering applications is the Kelvin
scale. The units on the Kelvin scale are the same as those on Celsius, except the scale is shifted
so that zero on the Kelvin scale is the coldest possible temperature (known as absolute zero).
The following equation relates Kelvin and Celsius: K  C  273 .
Use this formula to verify the values of absolute zero on the Celsius and Fahrenheit scales that
were given earlier in the text.
Section 2.4 Homework (an internet search may be required to obtain the necessary conversion
factors).
1. Convert 7.2 gallons to liters.
2. Convert 10,000 feet to meters.
3. Convert 727 grams to pounds.
4. Convert 10 hectares to acres.
5. Convert 2800 cm3 to in3.
31
Technical Mathematics
Chapter 3 – Percentages, precision, error, and tolerance intervals
3.1 – Percentages: basic conversions
Percentages are another way to numbers. If one half of a quantity is being discussed, that quantity can
be represented as a fraction or a decimal as follows:
1
 0.5
2
Using a percentage, the decimal representation is multiplied by 100. So a third way to represent this
quantity is illustrated as follows:
1
 0.5  50%
2
In general, conversions are done according to the rules below:



To convert a decimal to a percentage, move the decimal point 2 places to the right
To convert a percentage to a decimal, move the decimal point 2 places to the left
To convert a fraction to a percentage, first convert it to a decimal and then move the decimal
point 2 places to the right
The following table provides examples of these relationships:
Fractional representation
2
5
3
10
13
500
4
3000
27
2
Decimal representation
Percentage representation
0.4
40%
0.3
30%
0.026
2.6
0.0013
0.13%
13.5
1350%
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Technical Mathematics
Example 1
Convert 0.31 to a percentage
Solution:

To convert a decimal to a percentage, the decimal point is moved 2 places to the right:
0.31 = 31%
Example 2
Convert 43.21% to a decimal
Solution:

To convert a percentage to a decimal, the decimal point is moved 2 places to the left:
43.21% = 0.4321
Example 3
Convert
4
to a percentage
11
Solution:

To convert a fraction to a percent, the fraction is converted to a decimal and then the decimal is
converted to a percent:
4
 0.3636
11
0.3636  36.36%

So
4
 36.36%
11
33
Technical Mathematics
3.2 – Percentages: base, part, and rate
There are times when what percentage one number is of another needs to be determined. For example,
700 is 50% of 1400.
In this case, we have the following:
1400 is called the ‘base’ (B in the formulas below)
700 is called the ‘part’ (P in the formulas below)
50% is called the ‘rate’ (R in the formulas below)
Note: the decimal version of R is used in the formulas below
The formula that relates these components together are:
P  BR
used when the ‘part’ needs to be found
B
P
R
used when the ‘base’ needs to be found
R
P
B
used when the ‘rate’ needs to be found
Example 1
What is 38% of 654?
Solution:

Here 38% is the rate (0.38 is used in the formula), and 654 is the base. Since we need to find the
part, the problem is solved as follows:
P  BR
P  654  0.38
P  248.52

So 248.52 is 38% of 654
34
Technical Mathematics
Example 2
29 is what percent of 514?
Solution:

Here 29 is the part and 514 is the base. Since we need to find the rate, the problem is solved as
follows:
P
B
29
R
514
R  0.0564
R

So 29 is 5.64% of 514
Example 3
63 is 0.03% of what number?
Solution:

Here 63 is the part and 0.03% is the rate. Since we need to find the base, the problem is solved
as follows:
B
P
R
63
0.0003
B  210, 000
B

So 63 is 0.03% of 210,000
35
Technical Mathematics
3.3 – Precision and error
The precision of a measurement is defined to be the place of the digit that the measurement was
rounded to. The following table represents the precisions of sample measurements:
Measurement
Precision
170 m
10 m
420,000 mi
10,000 mi
0.023 A
0.001 A
0.10 ft
0.01 ft
4250 cm
1 cm
Explanation
Measurement was rounded to
the tens place
Measurement was rounded to
the ten thousands places
Measurement was rounded to
the thousandths place
Measurement was rounded to
the hundredths place (as
indicated by the last zero)
Measurement was rounded to
the ones place (as indicated by
the tagged zero)
The precision of a measurement allows for different analytical values to be calculated related to error.
The following definitions are associated with this concept:



Greatest possible error – one half of the precision
Relative error – greatest possible error divided by the actual measurement
Percent of error – relative error expressed as a percent
Example 1
Find the precision, greatest possible error, relative error, and percent of error in the measurement
7200 m.
Solution:

The precision is 100 m since the measurement was rounded to the hundreds place

The greatest possible error is 50 m since that is one half of the precision

The relative error is 0.00694 since

The percent of error is 0.694% since that is the percentage representation of the relative error
50
 0.00694
7200
36
Technical Mathematics
Example 2
Find the precision, greatest possible error, relative error, and percent of error in the measurement
0.0012 A.
Solution:

The precision is 0.0001 A since the measurement was rounded to the ten-thousandths place

The greatest possible error is 0.00005 A since that is one half of the precision

The relative error is 0.0417 since

The percent of error is 4.17% since that is the percentage representation of the relative error
0.00005
 0.0417
0.0012
37
Technical Mathematics
3.4 – Tolerance intervals
When manufacturing items that are to meet a specific measurement, no process can be perfect due to
what is called common cause variation. As a result, manufacturers must be presented with tolerance
intervals that indicate how much a particular measurement can differ from the specifications. The
following definitions are associated with this concept:




Tolerance – acceptable amount a measurement may vary from the specification
Upper tolerance limit – the largest measurement that is still within tolerance
Lower tolerance limit – the smallest measurement that is still within tolerance
Tolerance interval – twice the tolerance
The following table represents sample tolerance values:
Specification
Tolerance
Upper limit
Lower limit
Tolerance Interval
0.001 m
60 g
110 lb
0.0002 m
0.05 g
2 lb
0.0012 m
60.05 g
112 lb
0.0008 m
59.95 g
108 lb
0.0004 m
0.1 g
4 lb
Example 1
If an order is placed for a widget with a length specification of 2.12 mm and a tolerance of 0.0001
mm, find the upper limit, lower limit, and tolerance interval.
Solution:

The upper limit is 2.12 + 0.0001 = 2.1201 mm

The lower limit is 2.12 – 0.0001 = 2.1199 mm

The tolerance interval is 2 x 0.0001 = 0.0002
38
Technical Mathematics
Example 2
If an order is placed for a widget with a weight of 170 kg and a tolerance interval of 15, find the
tolerance, the upper limit, and the lower limit.
Solution:

The tolerance is 15/2 = 7.5 kg

The upper limit is 170 + 7.5 = 177.5 kg

The lower limit is 170 – 7.5 = 162.5 kg
39
Technical Mathematics
Chapter 3 Homework problems
Section 3.1 Homework
Fill in the blank spaces in the following table:
Fractional representation
Decimal representation
Percentage representation
1
10
13
74
3
21
84
0.05
0.0023
17.44
71%
17.33%
0.0056%
Section 3.2 Homework
1. What is 14% of 150?
2. What is 11.35% of 720?
3. What is 0.001% of 4410?
4. 15 is what percent of 22?
40
Technical Mathematics
5. 400 is what percent of 100?
6. 0.62 is what percent of 96?
7. 29 is 32% of what number?
8. 1.35 is 0.37% of what number?
9. 18 is 300% of what number?
10. If Julie received a 7% raise and her new salary is $33,560, what was her original salary?
Section 3.3 Homework
Fill in the blank spaces in the following table:
Measurement
Precision
Greatest Possible
Error
Relative Error
Percent of Error
30 V
21.5 m
0.0003 g
105,000,000 mi
Section 3.4 Homework
Fill in the blank spaces in the following table:
Specification
17 m
60.01 mL
Tolerance
1m
0.005 mL
0.5 oz
0.0002 W
Upper limit
Lower limit
Tolerance Interval
1.5 oz.
59.9999
2.5 cm
4.5 m
0.0003 cm
1m
41
Technical Mathematics
Chapter 4 – Number systems
4.1 – Decimal number system
The typical system that is used in everyday mathematics is the decimal number system, also known as
the base-10 system. The system is structured based on powers of 10 (hence the name of base-10). For
example, the number 50,413 can be written in expanded form as follows:
50, 413  5(10000)  0(1000)  4(100)  1(10)  3(1)
or more specifically:
50, 413  5(104 )  0(103 )  4(102 )  1(101 )  3(100 )
The place of every digit is important, as it indicates the power of ten that number is associated with.
Another aspect of the base-10 number system is that there are 10 digits necessary to represent
numbers in the system. These are the familiar digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Every number in base-10
can be represented by using these 10 digits and the concept regarding their place in the number.
Furthermore, all of the familiar operations (+, –, x, ÷) can be performed on these numbers.
Since electronic systems (such as computers) only store data in terms of ‘on’ and ‘off’, they are unable
to directly store a 10-state system. This concept will be discussed in the following sections.
42
Technical Mathematics
4.2 – Electronic machines
Computers, calculators, and other electronic devices are sometimes known as finite state machines. This
references the fact that these devices have memory which can be in any of a number of different states.
More importantly to this discussion, because these machines are electronic they ultimately can only
store on and off sequences. Because there are only two different options, machines do not have the
ability to store a 10-digit sequence like the numbers in a base-10 system. To simulate this system, we
need to use a binary number system (or base-2) in place of the more familiar base-10 system. However,
different number systems are all structured the same way. The only thing that changes is the base that is
used for each place where the digit resides.
One example of how this manifests itself in technology is how a computer stores a symbol such as a
letter ‘A’. Ultimately, this letter must be stored as a sequence of ons and offs so each letter (and in fact
all letters, digits, symbols, etc.) is given a numeric code. Most personal computers use the Unicode
system to encode these characters. In Unicode, the letter ‘A’ has a code of 65. However, even the
number 65 is one step removed from what the machine can store since 65 is in the decimal (10-state)
base. 65 is converted to binary: 65 = 01000001 in base 2 (groupings on computers are in multiples of 8).
Hence, the number 65 actually looks like this sequence:
off, on, off, off, off, off, off, on
Numbers can be written in expanded form using the same technique as that which was used for base-10
as follows:
010000012  0(2 )  1(2 )  0(2 )  1(2 )  0(2 )  1(2 )  1(2 )  0(2 )
7
6
5
4
3
2
1
0
Note the subscript on the number 010000012 indicates that it is in base-2 as opposed to base-10.
Furthermore, if you add the values on the right side of the equation, you get the original value of 65.
While any positive integer can be used as a base, the bases relevant to a technological discussion are the
following:
Base
2
8
10
16
Name
Binary
Octal
Decimal
Hexadecimal (Hex)
Valid digits
0 and 1
0, 1, 2, 3, 4, 5, 6, and 7
0, 1, 2, 3, 4, 5, 6, 7, 8, and 9
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F
Binary numbers are important because they provide a means of simulating the on/off aspect of
electronic devices. Hexadecimal numbers are used in different technological applications such as
networking and design (among others). Since octal numbers are no longer as prevalent as they used to
be, they are not covered in this text.
43
Technical Mathematics
4.3 – Base-2 (Binary)
As mentioned above, the binary number system has only 2 valid digits: 0 and 1. Furthermore, the place
of each digit in a number indicates the associated power of 2.
Any number in any non-decimal number base can be converted to base 10 by writing the number in
expanded form and then calculating the products and sums on the right-hand side.
Example 1
Convert 101102 to decimal
Solution:
101102 is a binary number as indicated by the subscript. Writing this number in expanded form
yields the following:
101102  1(24 )  0(23 )  1(22 )  1(21 )  0(20 )
1(24 )  0(23 )  1(22 )  1(21 )  0(20 )  22
So the binary number 101102 is equivalent to the decimal number 22
101102 = 22
Example 2
Convert 11100102 to decimal
Solution:
11100102  1(26 )  1(25 )  1(24 )  1(21 )
(Note the zeroes were ignored)
1(26 )  1(25 )  1(24 )  1(21 )  114
So 11100102 = 114
44
Technical Mathematics
When converting numbers from base-10 to base-2, a different approach is used as follows:
Example 3
Convert 152 to binary
Solution:
2 152
2 76
2 38
2 19
29
24
22
21
20
0 76 represents the quotient and 0 represents the remainder when 153 is divided by 2
0
0
1
1
0
0
1 The process stops when a 0 is obtained in the bracket
Reading the remainders from the bottom to the top gives the result.
152 = 100110002
Example 4
Convert 37 to binary
Solution:
2 37
2 18 1
29
0
24
1
22
0
21
0
20
1
37 = 1001012
45
Technical Mathematics
4.4 – Base – 16 (Hexadecimal)
Because the hexadecimal number system (often abbreviated to ‘hex’) is base-16, it needs 16 valid digits.
This is problematic because there are only 10 standard digits (0 through 9). Letters are commonly used
for the remaining 6 digits according to the following chart:
Digit in Hex
0–9
A
B
C
D
E
F
Base-10 value
Standard meaning
10
11
12
13
14
15
As in the other bases, the place of each digit is important. In hex, the placement indicates the associated
power of 16. Converting base-16 numbers to decimal can be done by putting the number in expanded
form and then calculating the products and sums on the right-hand side.
Example 1
Convert 3CD216 to decimal
Solution:
3CD216 is a hexadecimal number as indicated by the subscript. Writing this number in expanded
form yields the following:
3CD216  3(16 )  12(16 )  13(16 )  2(16 )  15,570
3
2
1
0
So 3CD216 = 15,570
46
Technical Mathematics
Example 2
Convert A141E216 to decimal
Solution:
A141E216 is a hexadecimal number as indicated by the subscript. Writing this number in expanded
form yields the following:
A141E 216  10(16 )  1(16 )  4(16 )  1(16 )  14(16 )  2(16 )  10,568,162
5
4
3
2
1
0
So A141E216 = 10,568,162
When converting numbers from base-10 to base-16, the same division/remainder approach is used as
that which was used for binary:
Example 3
Convert 5038 to hexadecimal
Solution:
16 5038
16 314
16 19
16 1
16 0
14 (which is E in base-16)
10 (which is A in base-16)
3
1
5038 = 13AE16
47
Technical Mathematics
Example 4
Convert 13020 to hexadecimal
Solution:
16 13020
16 813 12 (which is C in base-16)
16 50
13 (which is D in base-16)
16 3
2
16 0
3
13020 = 32DC16
Base-16 numbers can be converted to base-2 numbers using the following process:
1. Convert the base-16 number to base-10
2. Convert the base-10 number to base-2
However, base-16 and base-2 numbers have a unique relationship whereby every 4 digits in a base-2
number equates to 1 digit in a base-16 number. This can be useful to provide a shortcut to converting
numbers between these two bases.
Example 5
Convert 1011000101102 to hexadecimal.
Solution:
The digits in 101100010110 is broken up into groupings of 4:
1011 0001 0110
The groupings are now associated with their base-16 number using expanded form:
10112 = 1(23) + 1(22) + 1(20) = 13 = D16
00012 = 1(20) = 116
01102 = 1(22) + 1(21) = 616
Placing the values in their respective locations gives 1011000101102 = D1616
48
Technical Mathematics
Example 6
Convert 11000110101012 to hexadecimal.
Solution:
The digits in 101100010110 is broken up into groupings of 4 and replaced with their hex equivalents:
1
1000
1
8
1101
D
0101
5
So 11000110101012 = 18D516
The reverse process is used to convert base-16 to base-2.
Example 7
Convert A316 to binary.
Solution:
The digits in A3 are associated with their corresponding groupings of 4:
A
3
1010
0011
So A316 = 101000112
Example 8
Convert 1CC16 to binary.
Solution:
The digits in 1CC are associated with their corresponding groupings of 4:
1
C
C
0001
1100
1100
So 1CC16 = 1110011002
49
Technical Mathematics
Number Bases Chart
The integers 0 – 32 in Base 10, 2, 8, and 16
Decimal (Base 10)
Binary (Base 2)
Octal (Base 8)
Hexadecimal (Base 16)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
0
1
10
11
100
101
110
111
1000
1001
1010
1011
1100
1101
1110
1111
10000
10001
10010
10011
10100
10101
10110
10111
11000
11001
11010
11011
11100
11101
11110
11111
100000
0
1
2
3
4
5
6
7
10
11
12
13
14
15
16
17
20
21
22
23
24
25
26
27
30
31
32
33
34
35
36
37
40
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
10
11
12
13
14
15
16
17
18
19
1A
1B
1C
1D
1E
1F
20
50
Technical Mathematics
Chapter 4 Homework problems
Section 4.2 Homework
1. Write 6192 in expanded form (the absence of a subscript indicates this is a base 10 number).
2. Write 1012 in expanded form
3. Write 1010110012 in expanded form
4. Write 72916 in expanded form
5. Write AE316 in expanded form
6. Write 13C2D016 in expanded form
Section 4.3 Homework
1. Convert 1012 to Base 10
2. Convert 1010110012 to Base 10
3. Convert 1011011 to Base 10
4. Convert 41 to Base 2
5. Convert 547 to binary
6. Convert 1024 to binary
Section 4.4 Homework
1. Convert AE316 to Base 10
2. Convert 13C2D016 to Base 10
3. Convert 712 to Base 16
4. Convert 501,322 to Base 16
51
Technical Mathematics
5. Convert 10010112 to Base 16
6. Convert 110011010010112 to Base 16
7. Convert A316 to Base 2
8. Convert 19CD16 to Base 2
52
Technical Mathematics
Chapter 5 – Formulas and data substitution
5.1 – Substituting data into formulas
Many formulas are encountered in technical settings to show relationships among the values in a given
situation. A common simple formula in electronics is Ohm’s law which states:
E  IR
where E is the voltage (measured in volts), I is the current (measured in amps) and R is the resistance
(measured in ohms). Hence, the voltage in a circuit is equal to the current multiplied by the resistance.
Example 1a
Find the voltage in a circuit if the current is 0.0009A and the resistance is 10,000Ω.
Solution:
E  IR
E  0.0009 A 10,000
E  9V
So the voltage in this circuit is 9V.
Example 1b
It is not uncommon for current and resistance values to be given in uA (microamps) and kΩ
(kiloohms or k-ohms). As a result, it would be common to see the problem in Example 1a phrased
this way:
Find the voltage in a circuit if the current is 900uA and the resistance is 10kΩ.
Solution:
E  IR
E  0.0009 A 10, 000
E  9V
(since 900uA=0.0009A and 10k  10, 000)
Using different (but equivalent) units does not change the situation, and hence the voltage is still 9V.
53
Technical Mathematics
There are many formulas which will be encountered in technical settings. A small subset is provided
here.
Formula
f  pa
P  IE
P  I 2R
Q
I
t
E  0.7854b2 SN
9
F  C  32
5
Description
force = (pressure)(area)
power = (current)(voltage)
power = (square of the current)(resistance)
Electric charge
current=
time
Engine displacement = 0.7854(square of the bore)(Stroke)(number of cylinders)
Fahrenheit temp = 9/5(Celsius temp) + 32
Substituting data into formulas always uses essentially the same process. The values are substituted in
and the calculations are performed to determine the unknown quantity.
Example 2
Find the engine displacement given that the bore is 3.63 in, the stroke is 3.65 in, and the engine has
8 cylinders.
Solution:
E  0.7854(3.632 )(3.65)(8)
E  302in3
Example 3
Water boils at 100°C. What is the equivalent Fahrenheit temperature?
Solution:
9
F  C  32
5
9
F  (100)  32
5
F  180  32
F  212
So water boils at 212°F.
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5.2 – Simple algebraic manipulation of formulas
Problems encountered in technology are often not always in terms of the isolated variable. In one of the
situations above, the formula was written in terms of V and the situation required that the value of V be
found. However, it is quite possible that V would be a known quantity and one of the other two values
would need to be found. In this case, minor algebraic manipulation would have to be performed to
isolate the required variable.
Example 1
A light bulb in a circuit has a voltage of 12V and current of 24mA. What is its resistance?
Solution:
E  IR
E
R
I
E
R
I
12V
R
0.024 A
R  500
Both sides of the equation are divided by I
The equation is flipped to place R on the left
The known values are substitued in for E and I
The value of R is calculated (and left in  since k is typically
used for values over 1000)
So the resistance is 500Ω.
Example 2
What is the wattage of the bulb in example 1?
Solution:
P  IE
P  (12V )(0.024 A)
P  0.288W
P  288mW
The known values are substitued in for I and E
The value for P is calculated and converted to mW
So the power is 288mW.
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Example 3
As of the end of 2013, the highest recorded temperature in Normal, IL was 107.6°F. What is this
temperature in Celsius?
Solution:
9
F  C  32
5
9
F  32  C
5
5
( F  32)  C
9
5
C  ( F  32)
9
5
C  (107.6  32)
9
C  42
32 is subtracted from both sides of the equation
both sides of the equation are multiplied by
5
9
the equation is flipped to place C on the left
the known value is substituted for F
the value of C is calculated
So as of the end of 2013, the highest recorded temperature in Normal, IL was 42°C.
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Chapter 5 Homework problems
Section 5.1 Homework
1. The pressure between a boy and the ground he stands on is 1.875 N/cm2. If the boy’s feet cover
an area of 240 cm2, what is his weight (in N)?
2. If the current in a circuit is 4A, and the voltage is 12V, what is the power?
3. If the current in a circuit is 2000mA, and the voltage is 9V, what is the power?
4. If the current in a circuit is 3A, and the resistance is 2 Ω, what is the power?
5. Find the engine displacement of a twin cylinder engine in cm3 if the bore is 5.4 cm and the
stroke is 4.0 cm.
Section 5.2 Homework
1. If the Power in a circuit is 50W, and the voltage is 12V, what is the current?
2. If the Power in a circuit is 50W, and the current is 9A, what is the voltage?
3. If the Power in a circuit is 50W, and the current is 9A, what is the resistance?
4. If the Power in a circuit is 50W, and the resistance is 5 Ω, what is the current?
5. In the late 1960s, Toyota produced a small 1988 cm3 6-cylinder engine. If the bore of the engine
was 75 mm, what was the stroke?
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Chapter 6 – Basic Geometry
6.1 – Angles
Most shapes are in some way related to line segments and angles. There are many different
relationships inherent in geometry, so it is important to first understand what angles are. An angle is
formed when two line segments meet in a point. This point is known as the vertex of the angle. In the
following image, O is the angle vertex:
Angles can be measured using a few different approaches. One measurement systems uses degrees. Any
angle that is formed by two perpendicular segments as in the image below is known as a right angle and
has a measurement of 90°.
The question can reasonably be asked “What is a degree?” If a circle is ‘cut’ into 360 equal pie-shaped
wedges, each of these wedges is exactly 1°, regardless of the size of the circle:
Angles can now be classified as follows:





An ‘acute’ angle measures less than 90°
A ‘right’ angle measures exactly 90°
An obtuse angle measures more than 90°
Two angles are complements if the sum of their measures is 90°
Two angles are supplements if the sum of their measures is 180°
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There are many angular relationships that exist when angles are formed by intersecting lines as follows.
In the following image, angles A and B are called vertical angles and will always have the same
measurement. Angles C and D are also vertical angles.
A somewhat more complicated example occurs when two parallel lines are intersected by a third line
(called a transversal). The image and table below show the various relationships that result in this
scenario (the table does not show all possible relationships).
Angle pair
A and D
D and E
A and C
C and E
E and G
A and E
A and G
Relationship
Equal
Equal
Supplements
Supplements
Supplements
Equal
Supplements
Reason
Vertical angles
Alternate-interior
Form a straight line
E = A and A and C are supplements
Form a straight line
A=D=E
A = E and E and G are supplements
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6.2 – Triangles
Polygons are closed figures that are made up of line segments. They are categorized by the number of
sides. Triangles are the 3-sided polygons where the sum of the interior angles is always 180°. There are
two different ways to categorize triangles as follows:
Categorization 1 – based on the interior angles



A triangle is ‘acute’ if all three angles measure less than 90°
A triangle is ‘right’ if one of its angles measures 90° (the other two angles will always be less
than 90° and will be complements of each other). The longest side of a right triangle is called the
hypotenuse, and the two shorter sides are called the legs.
A triangle is obtuse if one of the angles measures more than 90° (the other two angles will
always be less than 90°)
Categorization 2 – based on the lengths of the sides



A triangle is ‘scalene’ if all three sides have different lengths
A triangle is ‘isosceles’ if two of the sides have the same length
A triangle is ‘equilateral’ if all three sides have the same length
Example 1
What is the measure of angle A in the image below?
Solution
Since A is a right triangle (indicated by the small square in the lower-left corner), the other two
angles are complements of each other. The measure of the angle A can be found by subtracting the
other angle from 90°:
A = 90° - 37° = 53°.
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Example 2
An isosceles triangle has a perimeter of 22m and the shorter side is 4m. What is the length of the
two equal sides?
Solution
The perimeter of a figure is the distance around the figure. In the case of a triangle, it is the sum of
the 3 sides. Using ‘L’ as a variable for the length of the equal sides gives the following:
4 + L + L = 22
4 + 2L = 22
2L = 18
L=9
So the length of the two equal sides is 9m.
The Pythagorean Theorem
When the lengths of two sides of a right triangle are known, the third side can always be calculated
using the Pythagorean Theorem. This theorem is demonstrated as follows:
In the image below, c 2  a 2  b 2
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Example 3
Find the length of the unknown side in the following figure:
Solution
This is a right triangle so the Pythagorean Theorem is used. Since x represents the hypotenuse, the
solution is calculated as follows:
x 2  27 2  162
x  27 2  162
x  729  256
x  985
x  31.4
So the length of x is 31.4 cm.
Care needs to be taken because the variable c here is always indicating the hypotenuse. The following
problem indicates how to solve for one of the legs.
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Example 4
Find the length of the unknown side in the following figure:
Solution
This is a right triangle so the Pythagorean Theorem is used. Since x represents one of the legs, the
solution is calculated as follows:
1002  x 2  37 2
1002  37 2  x 2
x 2  1002  37 2
x  1002  37 2
x  10000  1369
x  8631
x  92.9
So the length of x is 92.9 mm.
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Area
The formula for the area of a triangle is given by
A  12 bh
where b is the base and h is the height. The
following example applies this formula.
Example 5
Find the area of the triangle.
Solution
The height of this triangle is 19 in and the base is 42 in.
A  12 bh
A  12 (42in)(19in)
A  12 (798in 2 )
A  399in 2
So the area of the triangle is 399 in2.
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In a right triangle, the two legs represent the base and the height.
Example 6
Find the area of the triangle.
Solution
The height of this triangle is 14.3 m and the base is 7.1 m.
A  12 bh
A  12 (14.3m)(7.1m)
A  12 (101.53m2 )
A  50.765m 2
So the area of the triangle is 50.765 m2.
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Example 7
Find the area of the triangle.
Solution
Since only the length of one of the legs is known, the other must be found using the Pythagorean
Theorem.
27.52  x 2  20.32
27.52  20.32  x 2
x 2  27.52  20.32
x  27.52  20.32
x  756.25  412.09
x  344.16
x  18.55 cm
Now that the 2 legs are known to be 20.3 cm and 18.55 cm, the area can be found.
A  12 bh
A  12 (20.3cm)(18.55cm)
A  12 (376.565cm2 )
A  188.3cm2
So the area of the triangle is 188.3cm2.
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6.3 – Quadrilaterals
Quadrilaterals are the four-sided polygons. Two of the common types found in a technical setting are
parallelograms and trapezoids.

Parallelograms have two pairs of parallel sides. Examples are rectangle and squares, although
not all parallelograms have 4 right angles.

Trapezoids have one pair of parallel sides.
Name
Area formula
General parallelogram
A  bh
Rectangle
A  lw
Trapezoid
A  12 (a  b)h
Example
Example 1
Find the area of the parallelogram.
Solution
The base of the parallelogram is 9.5 in and the height is 7 in.
A  bh
A  (7 in)(9.5in)
A  66.5in 2
So the area of the parallelogram is 66.5 in2.
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Example 2
Find the area of the trapezoid.
Solution
The height of the trapezoid is 5.1 ft and the two bases are 13.2 and 16.8 ft.
A  12 (a  b)h
A  12 (13.2  16.8)(5.1)
A  12 (30)(5.1)
A  76.5 ft 2
So the area of the trapezoid is 76.5 ft2.
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6.4 – Circles
Circle terminology and basic formulas
Important terminology associated with circles is discussed in the following table. The most important of
these are the radius and diameter.
Line / Line segment
Chord
Diameter
Radius
Secant
Tangent
Description
Line segment that begins on one point of the circle and ends on the other
Chord that goes through the center of the circle
Line segment starting at the center of the circle and going to a point on the circle
Line that goes through the circle, intersecting with it in two points
Line that intersects with the circle in one point
Two important formulas associated with circles given below. For this text, the value of 3.14159 will be
used for the constant π.
Formula
C  2 r
A   r2
Description
Circumference (distance around the circle)
Area
Example 1
Find the circumference and area of the circle.
Solution
The radius of this circle is 6 cm.
C  2 r
C  2(3.14159)(6cm)
C  37.7cm
A   r2
A  (3.14159)(6cm) 2
A  113.1cm 2
So the circumference of the circle is 37.7 cm and the area is 113.1 cm2.
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Example 2
Find the diameter of a circle that has an area of 48.2m2.
Solution
First the area formula is algebraically manipulated to isolate the radius, and then the values are
substituted. Once the radius is found, it is doubled to find the diameter.
A   r2
A
 r2

A

r
r
A

48.2m 2
3.14159
r  3.9m
r
Since the radius of the circle is 3.9m, the diameter is 7.8m.
Arcs and associate angles
Arcs represent a portion of the circle and are measured in degrees using the same measurement system
as for angles (i.e. 1/360th of the way around a circle represents 1°).
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For example, in the following image, the red arc labeled ‘s’ measures 90° since it is ¼ of the circle:
Angles that originate from the center of a circle are called central angles. These angles have the same
measure as the corresponding arc on the circle as demonstrated by the following image:
Angles that originate from a point on a circle are called inscribed angles. These angles also have a
relationship to the corresponding arc on the circle. The measure of an inscribed angle is always half the
measurement of the corresponding arc as demonstrated by the following image:
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A relationship involving triangles and circles is that a triangle inscribed in circle, and using the diameter
as one of the sides of the triangle, is guaranteed to be a right triangle. Furthermore, the diameter of the
circle will always represent the hypotenuse of the right triangle. Hence the red triangle in the image
below (where C marks the center of the circle) must have a 90° angle as marked in the image.
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Example 3
If C is the center of the circle, find length of the unknown side ‘x’ in the triangle, and the measure of
the arc ‘s’.
Solution
The red triangle must be a right angle (since it is inscribed in a circle and the diameter represents
one of its sides). This information will be used to find the values.
First the length of ‘x’ can be found using the Pythagorean theorem.
x  4.7 2  2.42
x  16.33
x  4.0m
Next, the measure of the arc ‘s’ can be found using the following two pieces of information:
1. The angle opposite the hypotenuse is 90°
2. The sum of the angles in a triangle is 180°
3. The unknown angle is the inscribed angle associated with the arc ‘s’
The third angle in the triangle is calculated as follows:
Inscribe angle measure = 180 – 90 – 31 = 59°
And the corresponding arc ‘s’ is calculated as follows:
S = 59 x 2 = 118°
Combining what was calculated above, the values are:
x = 4.0 m
s = 118°
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Technical Mathematics
Chapter 6 Homework problems
Section 6.1 Homework
1. If an angle measures 38°, then:
a. what is the angle’s complement (if it exists)?
b. what is the angle’s supplement (if it exists)?
2. If an angle measures 114°, then:
a. what is the angle’s complement (if it exists)?
b. what is the angle’s supplement (if it exists)?
3. In the image below, what are the two pairs of equal angles?
4. In the image below, find 3 pairs of equal angles and 3 pairs of supplementary angles.
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Technical Mathematics
Section 6.2 Homework
1. Fill in the chart for each triangle:
Triangle
Acute, right, or obtuse
Scalene, Isosceles, or
Equilateral
2. In the image below, find the measure of angle A:
3. In the image below, find the value of x and the perimeter of the triangle.
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4. In the image below, find the value of x and the perimeter of the triangle.
5. Find the area of the triangle in question 3.
6. Find the area of the triangle in question 4.
7. Find the area of the following triangle:
8. If an isosceles triangle has a perimeter of 176 ft, and the shorter side is 10 feet less than the
measure of the two equal sides, what is the length of the three sides of the triangle?
9. Find the area of the triangle in question 8.
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10. When the lengths of the three sides of a triangle are known, but the height is not, Heron’s
formula can be used to calculate the area. Heron’s formula states that if a, b, and c
represent the three sides, then the area is calculated as follows:
A  s( s  a)( s  b)( s  c) where s 
abc
2
Use this formula to find the area of the following triangle:
Section 6.3 Homework
1. Find the area and perimeter of a rectangle with a length of 30m and a width of 42m.
2. Find the length of a rectangle with an area of 700cm2 and a width of 68cm.
3. Find the area of the trapezoid in the image below:
4. Find the area and perimeter of the trapezoid in the image below:
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5. A building has a length of 40 feet, a width of 60 feet, and a height of 80 feet. Find the
amount of paint needed (in gallons) and the cost of the paint if:
a. all 4 sides of the building needed to be painted
b. a gallon of paint covers 420 square feet
c. a gallon of paint costs $11
Section 6.4 Homework
1. Fill in the values in the table below.
Circle radius
3.4 in
17 ft
Circle diameter
Circle circumference
Circle area
3 x 106 km
104 cm
712 mm3
2. Find the measure of the arc s:
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3. Find the measure of the arc s:
4. Find the measure of the arc s:
5. Find the measure of the arc s, and the diameter of the circle:
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Chapter 7 – Basic Trigonometry
7.1 – The Trigonometric ratios
The six trigonometric ratios are relationships between the lengths of the sides in a right triangle. They
are defined in the following chart (in this chart, ‘A’ represents an angle).
Full ratio name
Standard abbreviation
Ratio
sine(A)
sin(A)
side opposite A
O
commonly referred to as
hypotenuse
H
cosine(A)
cos(A)
side adjacent to A
A
commonly referred to as
hypotenuse
H
tangent(A)
tan(A)
side opposite A
O
commonly referred to as
side adjacent to A
A
secant(A)
sec(A)
hypotenuse
H
commonly referred to as
side adjacent to A
A
cosecant(A)
csc(A)
hypotenuse
H
commonly referred to as
side opposite A
O
cotangent(A)
cot(A)
side adjacent to A
A
commonly referred to as
side opposite A
O
Secant, cosecant, and cotangent can all be found if the other ratios are known. As a result, they do not
appear on scientific calculators, they are not often used in technical classes, and they will not be covered
in this text.
The acronym SOHCAHTOA (pronounced soh-ca-toe-a) is often used to remember the ratios of the three
O
A
O
major trigonometric functions, since sin A  , cos A  , and tan A  .
H
H
A
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Technical Mathematics
Example 1
Find sinA, cosA, and tanA in the following triangle.
Solution
sin A 
O 13

 0.6190
H 21
cos A 
A 17

 0.8095
H 21
tan A 
O 13
  0.7647
A 17
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Example 2
Find sinB, cosB, and tanB in the following triangle.
Solution
Since the hypotenuse is needed to find sin and cos, that is calculated first using the Pythagorean
Theorem:
h  4.72  8.22  9.5m
The three trigonometric ratios can now be calculated:
sin B 
O 8.2

 0.8632
H 9.5
cos B 
A 4.7

 0.4947
H 9.5
tan B 
O 8.2

 1.745
A 4.7
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7.2 – Finding trigonometric values on a calculator
All scientific calculators can calculate any of the trigonometric values of a particular angle (and can also
calculate an angle given the trigonometric value). Angles can be measured using different measurement
systems. To be consistent with the previous angular discussions, the degree measurement system will
be used. When using a scientific calculator, care must be taken to ensure that the degree measurement
system is used (as opposed to radians or gradians), as this will affect the answers.
In this text, the following two assumptions are made:
1. The calculator used is the standard Windows OS calculator in ‘Scientific’ mode (other calculators
can be used, but they all have their own processes)
2. The degree measurement system is used
This image shows what the calculator looks like. Note the ‘Degrees’ radio button is selected. The ‘sin’,
‘cos’, and ‘tan’ buttons are also visible in this mode.
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Using the calculator to determine trigonometric values of angles.
Example 1
Find cos(58°).
Solution
The question is asking for the cosine of a 58° angle to be found. This can be done by entering 58 into
the calculator and pressing the ‘cos’ button. The calculator displays the value 0.5299 (trig values are
often rounded to four significant digits), and so:
cos(58°) = 0.5299.
Example 2
Find the sin, cos, and tan of 71.3°.
Solution
Using the calculator, the following values are obtained:
sin(71.3°) = 0.9472
cos(71.3°) = 0.3206
tan(71.3°) = 2.9544
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Using the calculator to determine angles given their trigonometric values.
Example 3
If cosA = 0.3141, find the measure of angle A.
Solution
The question is the opposite of what is being asked in example 1 above. Rather than being given the
angle and being asked for the trig value, this question gives the trig value and is asking for the angle.
Scientific calculators can calculate this, but the process is slightly different. To calculate this value,
0.3141 is entered in the calculator, and the ‘Inv’ button is pressed (this is often marked as the 2nd
key on Scientific calculators). The image below shows the calculator at this point (the ‘Inv’ button is
in yellow):
The ‘cos-1’ button replaces the ‘cos’ button on the calculator. Pressing this button yields the value
71.69, and so:
If cosA = 0.3141, then the measure of angle A = 71.69°.
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Example 4
If sinA = 0.2178, find the measure of angle A.
Solution
Using the calculator, the following value is obtained:
If sinA = 0.2178, then the measure of angle A = 12.58°
Example 5
If tanA = 2.668, find sinA and cosA.
Solution
To find sinA and cosA, first the measure of A must be found. Using the calculator (as in examples 3
and 4), the following is obtained:
If tanA = 2.668, then the measure of angle A = 69.45°
Now, cosA and sinA can be found (as in examples 1 and 2):
sin(69.45°) = 0.9364
cos(69.45°) = 0.3510
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7.3 – Using trigonometry to find unknown components of a triangle
The concepts in sections 7.1 and 7.2 can be used to find the lengths of sides and measures of angles in
right triangles if enough information is provided.
Example 1
Find x in the image below.
Solution:
In this example, the 61° angle and its opposite side are provided. The adjacent side must be
found. Since the sides in question are the opposite and adjacent sides, the appropriate
O
trigonometric function to use is tangent (since tan A  ).
A
The calculation is performed as follows:
tan 61 
14.1
x
x(tan 61)  14.1
x
14.1
tan 61
x  7.816
And so x = 7.816 m.
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There are many applications of right triangles in technical courses. The following provides one such
example.
Example 2
The relationship between Resistance (R), Inductive reactance (XL), and Impedance (Z) (all three
of these are measured in Ohms), and the phase angle in a circuit is given by the image below.
If R = 11Ω and Z = 15Ω, find XL and the phase angle (  ).
Solution:
XL can be found by using the Pythagorean Theorm:
X L  152  112  10.2
 can be found using R, Z, and cos(  ). Cos is used because R and Z represent the adjacent
angle and the hypotenuse.
cos  
R 11
  0.7333
Z 15
Since cos   0.7333, using cos -1 on the calculator yields the following:
  42.84
So: XL = 10.2Ω and  = 42.84°.
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Technical Mathematics
Chapter 7 Homework problems
Section 7.1 Homework
1. Find sinA, cosA, and tanA.
2. Find sinA, cosA, and tanA.
3. Find sinB, cosB, and tanB.
Section 7.2 Homework
1. If A = 67°, find:
a. sinA
b. cosA
c. tanA
2. if B = 23°, find:
a. sinA
b. cosA
c. tanA
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3. Was there any relationship between the sin and cos values in question 1 and 2? If so, explain
why this happened.
4. If sinA = 0.2513, find A.
5. If cosA = 0.5731, find A.
6. If tanA = 3.614, find A.
7. Explain what happens (and why) if you try to find A if cosA = 2.143.
8. If sinA = 0.3141, find cosA and tanA.
Section 7.3 Homework
1. Find x and y in the image below.
2. Find x and y in the image below.
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3. Find the measure of angle A in the image below.
4. Find the measure of angle A in the image below.
5. Find the measure of angle A in the image below.
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For questions 6, 7, and 8, use the image below that represents the relationships between Impedance,
Reactance, Resistance and the phase angle  in a circuit:
6. If R and XL are both 14Ω, find Z and the phase angle.
7. If Z = 20Ω and  = 38°, find R and XL.
8. If R = 8.5Ω and Z = 12Ω, find  and XL.
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Chapter 8 – Interpreting Charts and Graphs
8.1 – Charts and Graphs
Charts and graphs provide a visual representation of data. The data can be representative of many
different situations, from quality control processes to informational trends over time to the power
output of a device.
While charts and graphs present the data in a concise and more easily-read format, it takes practice to
become comfortable reading and analyzing these charts. Furthermore, there are many different kinds of
charts that can be used. Many software products provide ways to enter data and build charts and
graphs. All of the images in this section were either built from Excel, or by using the TI-83 graphing
calculator. Three samples are given below, and these three types are discussed in more detail on the
following pages.
Chart/Graph type
Sample Image
10000
5000
Bar chart
0
1
2
3
Series2
1, 23%
3, 47%
Pie chart
2, 30%
1
2
3
Coordinate system graph
Bar charts
As the name indicates, bar charts use bars to present information. The bars in the chart can be
presented either horizontally or vertically.
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Technical Mathematics
Example 1
The following chart shows the number of defective parts produced during a manufacturing process.
Quality control made a change to the process in September to reduce the number of defective parts.
Quality control report - Defective Parts in Process
900
800
Defective parts
700
600
500
400
300
200
100
0
Jan & Feb
Mar & Apr
May & Jun
Jul & Aug
Sep & Oct
Nov & Dec
Timeframe
Chart analysis
1) Question: How many parts were defective in January and February?
Answer: Looking at the first blue bar, it appears that approximately 720 parts were defective in
these months.
2) Question: How many parts were defective in the months January through August?
Answer: These months comprise the first 4 bars. Adding them together, it appears that there
were approximately 720 + 705 + 800 + 690 = 2915 defective parts in this timeframe.
3) Question: Did the change made to the process in September seem to have a positive affect?
Answer: Additional analysis needs to be done to answer this question. A good approach would
be to calculate the average number of defective parts per month.
720  705  800  690
For Jan – Aug, the average is:
 364 defective parts per month.
8
505  550
For Sep – Dec, the average is:
 264 defective parts per month.
4
While more data and analysis should be done to verify the findings, it appears that the process did
have a positive effect since the average number of defective parts per month was lowered by 100.
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Technical Mathematics
Example 2
The following chart is a clustered (or stacked) horizontal bar chart showing information regarding
employee salaries and taxes.
Payroll Information
Employee
Lori Williams
Joe Black
John James
Lisa Johnson
Jake Smith
0
20,000
40,000
60,000
80,000
100,000
120,000
Dollar Value
Net pay
Taxes
Gross salary
Chart analysis
1) Question: Who paid the most in taxes?
Answer: Looking at the legend at the bottom of the chart show that the orange bars represent
taxes. Lori Williams has the longest orange bar, paying approximately $16,000 in taxes.
2) Question: What is the average employee net salary?
Answer: The green bars indicate the net salary. The average is calculated as follows:
88, 000  37, 000  62, 000  45, 000  43, 000
 55,500 .
5
3) Question: What is the total company payroll?
Answer: To find the total payroll, the gross salaries needed to be added.
Payroll = 110, 00  41, 000  72, 000  52, 000  50, 000  325, 000 .
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Pie charts
As the name indicates, pie charts use circles with wedges. The wedges are reminiscent of pie-shaped
pieces giving them their name.
Example 3
The following chart shows the same information as in Example 1, only this time a pie-chart is used.
Since this is simply a different way of presenting the data, all of the same questions can be
answered.
Defective parts (total 4004 defective parts)
Nov & Dec
14%
Jan & Feb
18%
Sep & Oct
13%
Mar & Apr
18%
Jul & Aug
17%
Jan & Feb
Mar & Apr
May & Jun
May & Jun
20%
Jul & Aug
Sep & Oct
Nov & Dec
Chart analysis
1) Question: How many parts were defective in January and February?
Answer: Looking at the blue area, 4004*.18 = 720 parts were defective in this timeframe.
2) Question: How many parts were defective in the months January through August?
Answer: Looking at the blue, red, green, and purple areas, 4004 * .73 = 2923 parts were
defective in this timeframe.
3) Question: Did the change made to the process in September seem to have a positive affect?
Answer: For Jan – Aug, the average percent is
For Sep – Dec, the average is:
18  18  20  17
 9.1%
8
13  14
 6.8% , so again, it seemed to have a positive effect.
4
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Technical Mathematics
Example 4
The following chart is a version of a pie chart that focuses on a specific part of the chart and
provides a secondary pie chart with only that data.
Defective parts (total 4004 defective parts)
Jul & Aug
17%
Nov & Dec
14%
May & Jun
20%
Other
27%
Mar & Apr
18%
Jan & Feb
Sep & Oct
13%
Jan & Feb
18%
Mar & Apr
May & Jun
Jul & Aug
Sep & Oct
Nov & Dec
Chart analysis
This chart not only shows the defective part totals broken out in their months, but also shows the
timeframe associated with the manufacturing process change. This allows for more detailed analysis
and easier readability so that appropriate decisions can be made regarding the effectiveness of the
change, and if that approach should be continued.
Coordinate system graphs
Coordinate system graphs use a standard x and y axis approach. The x-axis is the horizontal axis and the
y-axis runs vertically. Sometimes these axes will indicate other values (such as time, power, dollar
values, or any value that can be measured).
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Technical Mathematics
Example 5
The following graph displays the relationship from Ohm’s law discussed earlier in the text. Recall
that Ohm’s law states E  IR . For a set value of R = 5Ω, the graph appears in the image below.
Note that E is the vertical axis and I is the horizontal axis.
Chart analysis
In interpreting this chart, it can be seen that there is a smooth linear relationship between E and I.
For every unit I goes up, E goes up 5 units. This can be seen in the two images below:
When I = 1, E = 5:
When I = 2, E = 10:
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Technical Mathematics
Example 6
The following graph displays the relationship Power relationship P  I 2 R (given earlier in the text).
For a set value of R = 5Ω, the graph appears in the image below. Note that P is the vertical axis and I
is the horizontal axis.
Chart analysis
In interpreting this chart, it can be seen that the relationship is no longer linear. Unlike in the first
example, this time as the value of I increases the value of P increases dramatically. This can be seen
in the two images below:
When I = 1A, P = 5W:
When I = 2A, P jumps all the way to 20W (note that the point is off the graph below). This increase
will get more and more dramatic as I gets larger due to the exponential nature of the formula.
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Technical Mathematics
Chapter 8 Homework problems
Section 8.1 Homework
1. Use the chart below to answer the questions that follow.
Chicago Cubs win totals per season
120
100
Wins
80
60
40
20
0
2009
2008
2007
2006
2005
2004
2003
2002
2001
2000
Season
a. In what year did the Cubs win the most games?
b. Approximate the total wins for all even numbered years.
c. If a season consists of 162 games, approximate the number of losses in the year 2000.
d. How many games have the Cubs won since the 2005 season (include 2005 in your
total)?
e. How many games have the Cubs lost since the 2005 season (include 2005 in your total)?
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Technical Mathematics
2. A small company reported $70,000 of revenue during the calendar year. The chart below
indicates the percentage of that revenue that occurred during each quarter of the year. Use the
chart to answer the questions that follow.
Revenue
21%
33%
26%
20%
Q1
Q2
Q3
Q4
a. What percent of the revenue was generated during the first half of the year?
b. How much combined revenue (in dollars) was generated in the months of July, August,
and September?
c. How much combined revenue (in dollars) was generated in the months of April, May,
June, October, November, and December?
d. A fiscal report provided by the company claims that their sales are greatest in the
months leading up to the holiday season. Explain whether or not the chart supports this
claim.
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3. The chart below uses what is called a line graph to show the number of bolts produced by a
company every 5 years. Use the chart to answer the questions that follow.
Analysing bolt production
102,000
100,000
98,000
96,000
94,000
92,000
90,000
88,000
86,000
1985
1990
1995
Bolts meeting specs
2000
2005
2010
2015
Total bolts produced
a. Approximately how many bolts were produced in 1990?
b. Approximately how many bolts produced in 1990 did not meet specifications?
c. Approximately what percent of bolts produced in 1995 did not meet specifications?
d. A report was published by the company for its stock holders in 1994. The report stated
that a new process would be put into production in 1995 and the expectation was that
while there would be some inefficiencies when the process was first implemented, over
time the process should allow for more bolts to be produced. While the chart only
supplies a small subset of the data, does it corroborate the claims made by the
company?
e. How many bolts were produced in 1985?
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4. The following two graphs show the relationship between Power, Current, and Resistance in 10
situations using the formula P  I 2 R .
The first graph uses a standard scale:
P=I2R
60000
50000
40000
30000
20000
10000
0
1
2
3
4
Resistance
5
6
Current (A)
7
8
9
10
Power (W)
Because there is an exponential relationship, a standard scale makes it difficult to see the
smaller values of I and R (because P becomes large very quickly). In this case, a logarithmic scale
is often used as in the following chart:
P=I2R
100000
10000
1000
100
10
1
1
2
3
4
Resistance
5
6
Current (A)
7
8
9
10
Power (W)
Use these charts to answer the question below:
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Technical Mathematics
a. In second chart, what is the Resistance in all 10 cases?
b. In the first chart, can the Resistance be identified? If not, why?
c. How does the current change from one scenario to the next? Which chart helps to
answer this question?
d. If the current increases from 10A to 20A, what is the change in the Power?
e. If the current increases from 10A to 100A, what is the change in the Power?
5. Use the graph below to answer the questions that follow.
a. For each value of x, find the corresponding value of y:
i. X = –2
ii. X = 0
iii. X = 3
iv. X = 6
b. For each value of y, find all corresponding values of x:
i. Y = 0
ii. Y = –2
iii. Y = –4
iv. Y = 3
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Technical Mathematics
Chapter 9 – Basic Statistics
9.1 – Data presentation
Statistics in technical settings often deals with collecting and analyzing data. This process can manifest
itself in many ways including but not limited to the following:

Collecting historical data over time to allow for future predictions that can lead to better
business or processing decision-making

Collecting data to identify trends in industry

Collecting data about a specific process that can lead to better quality control

Collecting data about a specific process that can lead to the determination of whether the
process can meet order specifications in terms of a customer’s tolerance
While there are different mathematical calculations and techniques involved in this process, first and
foremost the data must be presented in a readable format. Consider the following chart of raw data:
Measurement of widgets produced (target measurement: 2.00 mm)
Widgets above target
2.03
2.01
2.01
2.02
2.01
2.01
2.05
2.01
2.01
2.03
2.02
2.01
2.01
2.02
2.05
Widgets below target
1.97
1.99
1.99
1.98
1.99
1.99
1.95
1.99
1.99
1.97
1.98
1.99
1.99
1.98
1.97
Looking at this data, it is hard to determine any kind of categorization or trend as it is not grouped
together or in any kind of natural ordering. One common approach is to create what is known as a
frequency table that will group and order the data in a more meaningful way.
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Technical Mathematics
Measurement of widgets produced (target measurement: 2.00 mm)
Frequency Table
Measurement interval
1.93 - 1.95
1.96 - 1.98
1.99 - 2.01
2.02 - 2.04
2.05 - 2.07
Frequency
1
6
16
5
2
In the frequency table above, the representation helps to form a clearer picture of what is happening. It
can been seen that most of the widgets in this set are close to the target measurement and as the
measurement moves away from the target, fewer and fewer of the widgets fall into that category.
Another way to present data in a more graphical form is to use what is known as a histogram.
Histograms are bar charts that represent the frequency distribution in a way that appeals to many
readers. While histograms can be created by hand, there are many tools that can help to create wellformatted and visually appealing charts. The chart below was created in Microsoft Excel.
Analysing widget production
18
16
Frequency
14
12
10
8
6
4
2
0
1.93 - 1.95
1.96 - 1.98
1.99 - 2.01
2.02 - 2.04
2.05 - 2.07
Measurement of widget in mm
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Technical Mathematics
Example 1
For the following data set, build the frequency table and the histogram using six 1-unit intervals.
Solution:
The data is tallied and categorized to form the frequency table:
The data is now graphed to form the following histogram:
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Technical Mathematics
Analysing student experiment results
9
8
Frequency
7
6
5
4
3
2
1
0
6-7
7-8
8-9
9 - 10
10 - 11
11 - 12
Measurement of acceleration due to gravity
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Technical Mathematics
9.2 – Measures of central tendency
When analyzing data sets, there is often a need to have a single number that indicates a ‘middle’ value.
This can give some indication of the overall tendency of the data set. Common examples are the average
score of a standardized test, the average salary of a given population, or the average measurement of a
widget in a production process.
These measurements are referred to as measurements of central tendency. There are three important
such measurements as follows:
Measure of central tendency
Notation (if applicable)
Description/formula
The sum of all of the data points divided
by the number of data points
mean
x
x
x1  x2  x3  ...  xn
n
median
Middle number of an ordered data set, or
the mean of the two middle numbers if
there are an even number of data points
mode
Most frequently occurring data point
(there may be none or more than 1
mode)
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Technical Mathematics
Example 1
Find the mean, median, and mode of the following set of ACT scores:
17, 17, 17, 17, 18, 18, 19, 19, 20, 20, 21, 21, 22, 23, 25, 25, 26
Solution:
Finding the mean:
x
17  17  17  17  18  18  19  19  20  20  21  21  22  23  25  25  26
17
x  20.29
Finding the median:
Since the data set is an ordered data set with 17 data points, the 9th element is the median
Median = 20
Finding the mode:
Since 17 occurs 4 times, and this is more than any other element, 17 is the mode
Mode = 17
Analyzing these results it can be seen that all 3 of the measurements give a reasonable
representation of the central tendency of this set.
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Technical Mathematics
Example 2
Find the mean, median, and mode of the following house prices:
70,000
72,000
81,000
90,000
94,000
114,000
117,000
118,000
120,000
8,500,000
Solution:
Finding the mean:
x
70,000+114,000+72,000+117,000+81,000+118,000+90,000+120,000+94,000+8,500,000
10
x  937, 600
Finding the median:
Since the data set is an ordered data set with 10 data points, the mean of the 5th and 6th
elements is the median
Median =
94, 000  114, 000
 104, 000
2
Finding the mode:
Since all of the numbers occur exactly once (i.e. no number occurs more than any other
number), this set does not have a mode
Analyzing these results, the following can be seen:
1.
Although the mean was calculated correctly, it is too high to be meaningful in this case.
This is caused by the one house that had a value dramatically higher than all of the other
houses. Data points like this are often called outliers and are sometimes left of
calculations because of the skewing affect they have.
2.
Since there is no mode, this calculation is not helpful at all.
3.
The median is the only meaningful measure of central tendency for this set of data.
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Technical Mathematics
9.3 – Measures of dispersion
Whereas measures of central tendency attempt to provide the single number that indicates a ‘middle’
value, sometimes it is important to know how dispersed or ‘spread out’ the data set is. This value will
often be used in many of the same circumstances as above to provide more insight into the data set.
These measurements are referred to as measurements of dispersion. There are three important such
measurements as follows:
Measure of dispersion
Notation (if applicable)*
Description/formula
The largest number in the set minus the smallest
number in the set
range
n
variance
s2
 (x  x )
2
i
i 1
n 1
n
standard deviation
s
 (x  x )
2
i
i 1
n 1
* Sometimes notations other than ‘s’ are used, but for simplicity this text will use ‘s’.
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Technical Mathematics
Example 1
Find the range, variance, and standard deviation of the set of ACT scores:
17, 17, 17, 17, 18, 18, 19, 19, 20, 20, 21, 21, 22, 23, 25, 25, 26
Solution:
Finding the range:
range  26  17  9
Finding the variance:
n
s2 
 (x  x )
2
i
i 1

n 1
(20.29  17) 2  (20.29  17) 2  ...  (20.29  26) 2
17  1
s 2  9.10
Finding the standard deviation:
n
s
 (x  x )
2
i
i 1
n 1
(20.29  17) 2  (20.29  17) 2  ...  (20.29  26) 2
s
17  1
s  9.10
s  3.02
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The Windows OS calculator (as well as many other statistical calculators) has built in statistical
functionality which can make these calculations much easier. The image below shows what the
calculator looks like in statistical mode. The important keys for this discussion are the mean and
standard deviation keys (circled in red).
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Technical Mathematics
Example 2
Use the calculator to find the standard deviation of the following house prices:
70,000
72,000
81,000
90,000
94,000
114,000
117,000
118,000
120,000
8,500,000
Solution:
The data is added to the statistical calculator and the standard deviation is found.
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Technical Mathematics
9.4 – Normally distributed data sets
Some data sets will have their data naturally categorize into intervals according to a very standard
system. The data sets are called ‘normally distributed data sets’, and occur frequently in both nature
and artificial processes. Normally distributed data sets must meet the following structure (a small
amount of error is allowed in artificial processes):





50% of the data is above the mean, and 50% is below
The data is symmetric when graphed
68.2% of the data falls within one standard deviation of the mean
95.4% of the data falls within two standard deviations of the mean
99.7% of the data falls within three standard deviations of the mean
The following graph encapsulates this structure:
Standardized test scores are typically normally distributed, and analysis can be done based on this fact
as in the following example.
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Technical Mathematics
Example 1
A standardized exam has the following characteristics:
mean: 57
standard deviation: 9
If 250 students took this exam, how many scored between 48 and 66?
Solution:
48 is one standard deviation below the mean (57 – 9 = 48)
66 is one standard deviation above the mean (57 + 9 = 66)
This question amounts to asking how many scores were within one standard deviation of the
mean. Since this is a standardized exam (meaning it is normally distributed), it is known that
68.2% of the scores will be within this range.
The following calculation generates the number:
250 (.682) = 170.5 (rounded to 171)
Hence, 171 students scored between 48 and 66.
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Technical Mathematics
Example 2
9000 widgets are produced in a manufacturing process with a weight that follows a normal
distribution according to the following characteristics:
mean: 3.71 g
standard deviation: 0.003 g
a) How many widgets weigh between 3.704 g and 3.716 g?
b) How many widgets weigh over 3.719 g?
Solution:
a)
3.704 = 3.71 – 2(0.003) = x  2s
3.716 = 3.71 + 2(0.003) = x  2s
This question amounts to asking how many widgets’ weights are within two standard
deviations of the mean. Since this is a normally distributed set, it is known that 95.4% of
the weights will be within this range.
The following calculation generates the number:
9000 (.954) = 8586
Hence, 8586 of the widgets weigh between 3.704 g and 3.716 g.
b)
3.719 = 3.71 + 3(0.003) = x  3s
This question amounts to asking how many widgets’ weights are more than three
standard deviations above the mean. Since this is a normally distributed set, it is known
that 99.7% of the weights will be within three standard deviations, meaning only 0.3%
will not. Of these 0.3%, half (or 0.15%) will be higher than x  3s .
The following calculation generates the number:
9000 (.0015) = 13.5 (rounded up to 14)
Hence, 14 of the widgets weigh more than 3.719 g.
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9.5 – Controlled and capable processes
In manufacturing, it is important to have a process work in such a way that all widgets are produced
according to some specific control limits. While common cause variation prevents widgets from all being
produced according to exact specifications, processes can be refined so that the widgets fall within
certain parameters. When a process is refined to this point, it is called a controlled process.
Controlled process – when the measurements of all widgets produced fall within specified control limits
(note this definition is a bit of an oversimplification but is precise enough for the discussion in this text).
When an order is made by a customer, it will often come with tolerance limits (this represents the fact
that the customer understands common cause variation and has identified the limits of measurement
error they can tolerate). If a process is known to have control limits that fall within the tolerance limits,
it is called a capable process.
Capable process – a controlled process where the control limits are within a customer’s specified
tolerance limits.
Example 1
A controlled process produces widgets with the following characteristics (the target
measurement is 7.5 cm):
Upper control limit (UCL): 7.501 cm
Lower control limit (LCL): 7.499 cm
A customer order is entered with the following tolerance limit specification:
7.5 ± 0.003 cm
Is this process capable?
Solution:
The situation can be depicted by the following graphic:
UTL = 7.503:
UCL = 7.501:
---------------------------------------------------------------------------------------------------------
LCL = 7.499:
LTL = 7.497:
---------------------------------------------------------------------------------------------------------
Since the control limits are within the tolerance limits (in other words, all widgets will be within
the customer’s tolerance), this is considered a capable process. The order can be accepted and
filled.
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Technical Mathematics
Example 2
A controlled process produces widgets with the following characteristics (the target
weight is 110 kg):
Control limits: ± 5 kg
A customer order is entered with the following tolerance limit specification:
110 ± 3.5 kg
Is this process capable?
Solution:
The situation can be depicted by the following graphic:
UCL = 115:
----------------------------------------------------UTL = 113.5: ----------------------------------------------------LTL = 106.5:
LCL = 105:
---------------------------------------------------------------------------------------------------------
Since the control limits are not within the tolerance limits, it cannot be guaranteed that all
widgets will be within the customer’s tolerance. Hence, this is not a capable process. The order
cannot be accepted.
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Technical Mathematics
Chapter 9 Homework problems
Section 9.1 Homework
1. In a manufacturing process, a new starter is being installed into a product during assembly.
Quality control personnel are tracking how many of the starters are failing each day over the
course of a month. For the following data set, build the frequency table and the histogram using
four 5-unit intervals. Start the lowest interval with 2 (so the first interval will be 2 – 6).
2. The following test scores were collected over a 5-year period. The grading scale was a straight
90-80-70-60 grading scale to indicate A, B, C, and D grades respectively (anything below a 60 is
considered a failing grade).
Create the frequency table and histogram using 4 intervals – one for each of the passing grades.
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Technical Mathematics
Section 9.2 Homework
For questions 1 – 3, find the mean, median, and mode. After finding those measures, discuss whether
any of them are a better reflection of the central tendency for the data set.
1. 4, 7, 9, 9, 9, 14, 14, 16, 23, 31, 32
2. 1,000
700
1800
5300
2300
500
600
400
3. 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 5, 5, 5, 6
4. Is the order of the data set important when calculating the mean? The median? The mode?
Section 9.3 Homework
1. Manually calculate the range and the standard deviation of the following data set:
3, 5, 7, 7, 9
2. Use a statistical calculator to verify the standard deviation calculated in question 1.
3. Use a statistical calculator to calculate the range and standard deviation of the following data
sets:
a. 17, 18, 18, 22, 44, 45, 45, 45, 65, 71, 73, 78, 80, 80, 91, 92, 93, 99
b. 100, 300, 250, 175, 188, 337, 299, 171, 152, 201
4. Is the order of the data set important when calculating the range? The standard deviation?
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Technical Mathematics
Section 9.4 Homework
1. 100,000 widgets are manufactured with a target measurement of 11.5 cm. After the widgets are
measured and the data is collected, the distribution characteristics are:
x  11.499 cm
s  0.001 cm
a. How many of the widgets measure between 11.498 cm and 11.5 cm?
b. How many of the widgets measure between 11.497 cm and 11.501 cm?
c. How many of the widgets measure between 11.496 and 11.499 cm?
d. How many of the widgets measure between 11.496 and 11.501 cm?
e. How many of the widgets measure at least 11.499 cm?
f.
If an order is placed for 15,000 widgets, but only under the condition that at least 90%
of them are between 11.496 cm and 11.502 cm, can this order be accepted?
2. Professional baseball batting averages that fell between .280 and .289 were tracked over a
season. The distribution of a sampling of 10 players is:
0.280, 0.282, 0.282, 0.284, 0.285, 0.287, 0.287, 0.287, 0.289, 0.289
Assuming this data is normally distributed, use the sampling to determine the pertinent
characteristics of the data set, and then determine how many of the 73 players who had a
percentage between .280 and .289 were either above 0.28832 or below 0.28208.
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Technical Mathematics
Section 9.5 Homework
1. A controlled process produces widgets with the following characteristics (the target
measurement is 33.2 mm):
Upper control limit (UCL): 33.204 cm
Lower control limit (LCL): 33.196. cm
A customer order is entered with the following tolerance limit specification:
33.2 ± 0.01 cm.
Is this process capable?
2. Another customer order for the widgets in question 1 is entered with the following tolerance
limit specification:
33.2 ± 0.001 cm.
Is this process capable?
3. A controlled process produces widgets with the following characteristics (the target weight is
100 kg):
Control limits: ±2.5 kg
A customer order is entered with the following tolerance limit specification:
220.5 ± 3.3 (units are measured in lb)
Is this process capable?
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Chapter 10 – Basic Logic
10.1 – Logical expressions
Logical expressions revolve around situations that have only one of two ways to be resolved. These
situations are often referred to as Boolean scenarios, and are not entirely unrelated to the binary
discussion in the Number systems chapter.
Situations in technical settings often revolve around circuitry or computer programming. They also occur
in many other areas. A small sample of situations in technology are given below.

In circuitry the Boolean values occur in situations such as:
o
o

Determining if the voltage is high or low
Determining whether a given scenario occurs, and taking action based on that scenario
(for example, if a car is waiting in a left turn lane the logic controller will detect that and
give a green turn arrow when the time comes)
In computer programming the Boolean values occur and dictate the logic flow:
o
o
The value may determine what module is executed (for example, if the employee is a
manager one routine might be executed, otherwise some other routine might be
executed)
The value may determine how many times a module is executed (for example, as long as
there is data in a file the records are processed and as soon as there is no more data the
processing ends)
Evaluating logical expressions is very similar to evaluating algebraic expressions. Recall that in algebraic
expressions, we use the operations +, –, x, and  .
Example 1 (Reviewing algebraic expressions)
If x = 3, y = 2, and z = 11, evaluate the algebraic expression: 2 xy  3 y  4 z .
2
Solution:
2 xy  3 y 2  4 z when x  3, y  2, and z  11
 2(3)(2)  3(2)2  4(11)
 12  12  44
 20
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A similar process is used to evaluate logical expressions. However, the operations are no longer the four
familiar arithmetic operations. The logical operators are:
Boolean operator
And
Or
Not
Notation
˄
˅
‘ (the single apostrophe)
Description
Evaluates to true if both operands are true
Evaluates to true if either operand (or both) is true
Inverts the value of the operand
These three operators are more formally defined by their truth tables. A truth table provides all of the
necessary information for a logical operator.
The ‘And’ logical operator
When logical variables are needed, p, q, and r are often used (as opposed to x, y, and z in algebra).
Additional, the two values used are typically 0 and 1, and these stand for the two states in the system
(the states might be 1 = high voltage and 0 = low voltage in an electronics context, 1 = true and 0 = false
in a programming context, etc.)
The ‘And’ logical operator is defined by the following truth table (where p and q are Boolean variables):
p
0
0
1
1
q
0
1
0
1
p^q
0
0
0
1
This allows for expressions to be evaluated.
Example 2
If p = 1 and q = 0, evaluate the logical expression: ( p  q )  p .
Solution:
In a similar fashion to algebraic expressions, logical expression are evaluated by substituting in
the appropriate data, performing the operations, and obtaining the final value.
( p  q )  p when p  1 and q  0
 (1  0)  1
 0 1
0
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The ‘Or’ logical operator
The ‘Or’ logical operator is defined by the following truth table:
p
0
0
1
1
q
0
1
0
1
p˅q
0
1
1
1
Example 3
If p = 1 and q = 0, evaluate the logical expression: ( p  q )  p .
Solution:
In a similar fashion to algebraic expressions, logical expression are evaluated by substituting in
the appropriate data, performing the operations, and obtaining the final value.
( p  q )  p when p  1 and q  0
 (1  0)  1
 1 1
1
The ‘Not’ logical operator
The ‘Not’ logical operator is defined by the following truth table (since the ‘Not’ operator simply inverts
the operand, only one variable is needed):
p
0
1
p’
1
0
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10.2 – Logic truth tables
Now that the three fundamental operations have been defined, larger truth tables that encompass
more complicated situations can be formed. Again, there are many scenarios in technology where this
process is needed. A couple of examples would be:

In the stop light example above, the controller will have to sense many possible situations. For
example, if there is a car in one left turn lane and a car in the opposite turn lane, then provide
both of them with left turn arrow, if there is only a car in one turn lane then provide the cars on
that side of the street with both a green arrow and a green light, if there is not a car in either
turn lane, then just provide a green light, etc.

In the programming example above, it may be a case where one module is executed if the
employee is a manager and has a salary above a certain value or if they are an administrator,
another module might be executed if they are a programmer, analyst, or DBA, and yet a third
module is executed in all other cases.
In these situations, the logic device will have to evaluate the particular expression and execute the
appropriate commands.
Truth tables allow designers to lay out all of the possibilities so all scenarios can be expressed in one
structure. This allows for analysis and problem-solving of the given situation.
Truth tables for logical expressions will be tables with a certain number of rows and columns. They are
created by following this algorithm (an algorithm is a step-by-step step set of instructions for
accomplishing some task).
1.
2.
3.
4.
5.
Create columns for each of the variables in the expression, in alphabetical order
Use additional columns to build up to the expression, one operation at a time
Create 2n rows, where n is the number of unique variables in the expression
Enter all possible combinations of initial conditions in the columns for the variables
For all of the remaining columns, fill them in one at the time applying the appropriate rules for
the current operation
The following example shows every step in this process:
Example 4
Create a truth table for the logical expression: ( p  q) ' p ' .
Solution:
Steps 1, 2, and 3 form the structure of the truth table
p
q
(p ˅ q)
(p ˅ q)’
p’
(p ˅ q)’ ˄ p’
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Step 4 provides all combinations of initial conditions (these are the first 2 columns)
p
0
0
1
1
q
0
1
0
1
(p ˅ q)
(p ˅ q)’
p’
(p ˅ q)’ ˄ p’
Step 5 is done one column at a time, focusing on the current operands and operator.
Column 3:
p
0
0
1
1
q
0
1
0
1
(p ˅ q)
0
1
1
1
(p ˅ q)’
p’
(p ˅ q)’ ˄ p’
q
0
1
0
1
(p ˅ q)
0
1
1
1
(p ˅ q)’
1
0
0
0
p’
(p ˅ q)’ ˄ p’
q
0
1
0
1
(p ˅ q)
0
1
1
1
(p ˅ q)’
1
0
0
0
p’
1
1
0
0
(p ˅ q)’ ˄ p’
(p ˅ q)’
p’
(p ˅ q)’ ˄ p’
Column 4:
p
0
0
1
1
Column 5:
p
0
0
1
1
And finally column 6 to complete the table:
p
q
(p ˅ q)
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0
0
1
1
0
1
0
1
0
1
1
1
1
0
0
0
1
1
0
0
1
0
0
0
Analyzing the results:
While all of the columns are an important part of the process, the most important column is the
last one. This provides the ultimate result for this expression given a set of initial conditions. For
example if p = 0 and q = 1, then the expression evaluates to 0. This can be seen by identifying
row 2 as the row associated with the initial conditions p = 0 and q = 1, and then looking to the
far right column to see the result of 0.
Typically the table is not replicated for each new column. The steps above are followed, but only one
table is needed as demonstrated in the next example.
Example 5
Create a truth table for the logical expression: ( p ' r )  (q  p ') .
Solution:
p
0
0
0
0
1
1
1
1
q
0
0
1
1
0
0
1
1
r
0
1
0
1
0
1
0
1
p’
1
1
1
1
0
0
0
0
(p’ ˄ r)
0
1
0
1
0
0
0
0
(q ˅ p’)
1
1
1
1
0
0
1
1
(p’ ˄ r) ˅ (q ˅ p’)
1
1
1
1
0
0
1
1
Analyzing the results:
Since there are 3 variables, 23 = 8 rows are needed to handle all possible combinations of initial
conditions. Other than that, the table works in the same way as any other table.
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10.3 – Schematics
Logical expressions can be expressed as in sections 10.1 and 10.2, or then can be expressed using
technical schematics. When schematics are used, the three fundamental operations are defined and
notated as in the following table:
Operator
Image
Image name
And gate
And
Two inputs, one output
Or gate
OR
Two inputs, one output
Inverter
Not
One input, one output
Building the actual schematic follows a similar process as choosing the columns in the corresponding
truth table. The following 2-step algorithm indicates what must be done:
1. Write each of the variables in the expression vertically along the left-hand side of the schematic
2. Build the schematic one operation at a time, thinking through and indicating the inputs to each
gate/inverter
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Example 1
Create the logic schematic for the logical expression from the example above:
( p  q) ' p ' .
Solution:
Following the two steps above, the following schematic is created:
p
q
Schematics can also be used to generate a logical expression. It is important to read the schematics from
left to right to obtain the correct expression.
Example 2
Determine the logical expression associated with the following schematic:
Solution:
Reading the schematic from left to right, the following expression is determined:
( p  q) ' ((q  r )  r)
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Chapter 10 Homework problems
Section 10.1 Homework
1. If x, y, and z all have the value 1, evaluate the logical expression: x  ( y  z ') .
2. If x is 1 and y is 0, evaluate the expression: ( x  y ')  ( y  ( x ' y ))
Section 10.2 Homework
3. Construct the truth table for the expression in question 1.
4. Construct the truth table for the expression in question 2.
Section 10.3 Homework
5. Construct the logic schematic for the expression in question 1.
6. Construct the logic schematic for the expression in question 2.
Section 10.4 Homework
7. Determine the logical expression associated with the following schematic:
p
q
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8. Determine the logical expression associated with the following schematic:
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Supplemental Chapter – Systems of Linear Equations
S.1 – Solving systems of 2 linear equations using a graphing tool
There are often times in technological settings when two linear equations (which would graph as lines)
are part of a system. When this happens, certain applications require knowing when the two lines
intersect, as this point of intersection would provide the (often unique) solution to the system. For
example, consider the following two lines:
y  3x  2
y  12 x  4
When graphed on the TI-83, these two lines look as follows:
If the point of intersection needs to be determined, it can be done on the calculator using the intersect
functionality resulting in the following image:
The intersection point calculated by the TI-83 is (2.4, 5.2). What this means is that while there are an
infinite number of points that satisfy the line y  3x  2 , and an infinite number of points that satisfy
the line y  12 x  4 , there is ONE AND ONLY ONE point that satisfies both of them – specifically the point
(2.4, 5.2) referenced above.
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The following example illustrates the use of a TI-83 to solve a system in more detail.
Example 1 – Solve the following system of equations using a TI-83
y  1.5 x  1
y  2 x  3.2
Solution:
1. Enter the equations into the calculator (press the ‘Y=’ button to do this)
2. Graph the equations (press the ‘GRAPH’ button to do this)
3. Use the TI-83 to calculate the point of intersection by doing the following:
Press ‘2nd' and then ‘TRACE’
Press ‘5’ to choose ‘intersect’
Press ‘ENTER’ to choose the first graph
Press ‘ENTER’ to choose the second graph
Move the cursor close to the point of intersection and press ‘ENTER’
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4. The calculator will now display the point of intersection:
Hence, these two lines intersect at the point (1.2, –0.8)
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S.2 – Solving systems of 2 equations using the Addition-Subtraction method
When a graphing tool is not available, systems of equations can be solved algebraically. Two techniques
for doing this are discussed here:
1. Algebraically solving the system using the Addition-Subtraction method
2. Algebraically solving the system using the Substitution method
The Addition-Subtraction method is discussed in this section. The following algorithm is used when
implementing this method:
1. If necessary, algebraically manipulate both equations so they are in the form ax  by  c
2. If necessary, multiply one or both of the equations so that either the coefficients on the x terms
or the coefficients on the y terms are the same (without regard to the sign).
3. Add or subtract (as appropriate) one equation from the other in order to remove one of the
variables from the equations
4. Solve the resulting linear equation (which will only have 1 variable) to find the first necessary
value
5. Substitute the solution from step 4 into either of the original equations to find the second
necessary value
6. Check the solution by substituting the values into both of the original equations
This algorithm is demonstrated in the following example which uses the equations graphed previously:
Example 1 – Solve the following system of equations using the Addition-Subtraction method:
y  3x  2
y  12 x  4
Solution:
1. Algebraically manipulate the equations (step 1 from algorithm)
y  3x  2
y  12 x  4
2. Multiply the second equation by 6 so the coefficients on x are the same (step 2 from algorithm)
(note that this step is technically not necessary for this problem since the y variables already
have a common coefficient, but is done to demonstrate the process)
y  3x  2
6 y  3x  24
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3. Subtract equation 2 from equation 1 to eliminate the terms with x (step 3 from algorithm)
y  3x  2
6 y  3x  24
---------------------5 y  26
4. Solve 5 y  26 for y (step 4 from the algorithm)
5 y  26
y  5.2
5. Substitute the value 5.2 in for y in either of the original equations (step 5 from algorithm)
y  3x  2
5.2  3 x  2
7.2  3 x
x  2.4
6. Check (step 6 from algorithm)
y  3x  2
5.2  3(2.4)  2
5.2  7.2  2
5.2  5.2
y  12 x  4
5.2  12 (2.4)  4
5.2  1.2  4
5.2  5.2
The solutions check out, and hence the solution is: x  2.4 and y  5.2 which is another way to
notate the point (2.4, 5.2). Note that this is the same solution previously found using the graphing
technique.
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In the next example, both equations must be multiplied by a value in step 2:
Example 2 – Solve the following system of equations using the Addition-Subtraction method:
2 x  7 y  10
3x  4 y  12
Solution:
1. Algebraically manipulate the equations – this step is unnecessary as the equations are already in
the appropriate form.
2. Multiply the first equation by 3 and the second equation by 2 so the coefficients on x are the
same
6 x  21y  30
6 x  8 y  24
3. Subtract equation 2 from equation 1 to eliminate the terms with x
6 x  21y  30
6 x  8 y  24
---------------------13 y  6
4. Solve 13 y  6 for y (step 4 from the algorithm)
13 y  6
y  136
5. Substitute the value
6
13
in for y in either of the original equations
2 x  7 y  10
2 x  7( 136 )  10
42
2 x  13
 10
88
2 x  13
x
44
13
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6. Check (step 6 from algorithm)
2 x  7 y  10
3 x  4 y  12
44
2( 13
)  7( 136 )  10
44
3( 13
)  4( 136 )  12
88
13
42
 13
 10
130
13
 10
132
13
24
 13
 12
156
13
 12
12  12
10  10
The solutions check out, and hence the solution is: x  1344 and y  136 which is another way to notate
44
6
the point ( 13 , 13 ).
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S.3 – Solving systems of 2 equations using the Substitution method
As discussed above, the two techniques for solving a system of equations are:
1. Algebraically solving the system using the Addition-Subtraction method
2. Algebraically solving the system using the Substitution method
In this section, the Substitution method is discussed. The following algorithm is used when
implementing this method:
1. If necessary, algebraically manipulate one of the equations so that one of the variables is
isolated
2. Substitute the corresponding expression for the isolated variable into the second equation
3. Solve the resulting linear equation (which will only have 1 variable) to find the first necessary
value
4. Substitute the solution from step 3 into either of the original equations to find the second
necessary value
5. Check the solution by substituting the values into both equations
This algorithm is demonstrated in the following example which uses the system of equations solved in
the previous sections:
Example 1 – Solve the following system of equations using the Substitution method:
y  3x  2
y  12 x  4
Solution:
1. Algebraically manipulate one of the equations to isolate a variable (step 1 from algorithm) – this
step is unnecessary since the equations already have the y variable isolated
2. Substitute the corresponding value in for y (step 2 from algorithm)
y  3x  2
1
2
x  4  3x  2
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3. Solve
1
2
x  4  3 x  2 for x (step 3 from the algorithm)
1
2
x  4  3x  2
1
2
x  3 x  2  4
 52 x  6
5 x  12
x  2.4
4. Substitute the value 2.4 in for x in either of the original equations (step 4 from algorithm)
y  3x  2
y  3(2.4)  2
y  7.2  2
y  5.2
5. Check (step 5 from algorithm)
y  3x  2
5.2  3(2.4)  2
5.2  7.2  2
5.2  5.2
y  12 x  4
5.2  12 (2.4)  4
5.2  1.2  4
5.2  5.2
The solutions check out, and hence the solution is: x  2.4 and y  5.2 which is another way to
notate the point (2.4, 5.2). Note that this is the same solution found using the graphing technique
and the addition-subtraction technique used previously.
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In the next example, step 1 is required as neither equation is isolated for a variable. In this case, the
addition-subtraction method is typically the preferred method but any method can be used.
Example 2 – Solve the following system of equations using the Substitution method:
2 x  5 y  27
11x  3 y  41
Solution:
1. Algebraically manipulate one of the equation to isolate a variable
2 x  5 y  27
2 x  5 y  27
x
y  272
5
2
2. Substitute the corresponding value in for x into the other equation (step 2 from algorithm)
11x  3 y  41
11( 52 y  272 )  3 y  41
3. Solve 11( 52 y  272 )  3 y  41 for y
11( 52 y  272 )  3 y  41
55
2
6
82
y  297
2  2 y  2
49
2
y
379
2
49 y  379
y
379
49
4. Substitute the value
379
49
in for y in either of the original equations
2 x  5 y  27
1323
2 x  5( 379
49 )   49
1323
2 x  1895
49   49
2x 
x
572
49
286
49
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5. Check (step 5 from algorithm)
2 x  5 y  27
11x  3 y  41
379
2( 286
49 )  5( 49 )  27
379
11( 286
49 )  3( 49 )  41
3146
49
 1137
49  41
 1323
49  27
2009
49
 41
27  27
41  41
572
49
 1895
49  27
The solutions check out, and hence the solution is: x 
286
49
and y 
379
49
which is another way to
379 .
notate the point ( 286
49 , 49 )
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S.4 – Applications associated with systems of equations
There are many technological situations where a system of equations is needed to find solutions. The
following provide examples of this situation.
Example 1
A chemist needs 50 mL of an 8% acid solution. She only has a 5% and a 13% solution on hand. How
much of each needs to be combined to obtain the desired 8% solution?
Solution:
1. The first step is to determine the necessary relationships in order to obtain the relevant system
of equations. Variables are used to symbolize the needed quantities:
A = the mL of the 5% solution needed
B = the mL of the 13% solution needed
Considering the problem, it can be determined that:
A  B  50
(since the amount of the two solutions must add to 50mL)
0.05 A  0.13B  0.08(50) (since the amount of acid will be constant)
2. Now that the system has been found, it is solved using any technique above. Here the AdditionSubtraction technique is used.
A  B  50
0.05 A  0.13B  0.08(50)
A  B  50
A  2.6 B  80
(this equation was multiplied by 20)
1.6 B  30
(the bottom equation was subtracted from the top equation)
B  18.75
A  18.75  50
(the value of B is substitued in to one of the equations)
A  31.25
Since A represented the amount of the 5% solution, and B represented the amount of the 13%
solution, 31.25 mL of the 5% solution must be mixed with 18.75 mL of the 13% solution.
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Example 2
If a concrete mix must have 5 times as much gravel as cement, and enough product must be mixed
to fill a volume of 15 m3, how much gravel and how much cement is needed to make this mix?
Solution:
1. Variables are used to symbolize the needed quantities:
C = m3 of cement needed
G = m3 of gravel needed
Considering the problem, it can be determined that:
C  G  15
G  5C
(since the amount of the two quantities must add to 15m3 )
(since there is 5 times more gravel)
2. Now that the system has been found, it is solved using any technique. Here the substitution
method is used since the second equation is already solved for G.
C  G  15
C  5C  15
6C  15
C  2.5
2.5  G  15
G  12.5
(5C is substituted in place of G)
(The value of C determined above is substituted into one of the equations)
Since C represented the amount of cement and G represented the amount of gravel, 2.5 m3 of
cement is needed, and 12.5 m3 of gravel is needed.
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S.5 – Systems of 3 linear equations
In certain instances there are three unknown quantities which need to be determined. In this case, three
equations are needed. While the process is more complicated than that for systems of two equations,
the actual steps use similar techniques. The algorithm is as follows:
1. If necessary, algebraically manipulate all equations so they are in the form ax  by  cz  d .
2.
3.
4.
5.
6.
Eliminate one of the variables from any pair of the three equations.
Eliminate the same variables from any other pair of the three equations.
Solve the resulting system of two linear equations using any of the three earlier techniques.
Substitute the two values obtained in step 4 to obtain the last unknown.
Check the solution by substituting the values into all three equations.
Example 1 – Solve the following system of equations:
x  y  z  10
2 x  2 y  5z  4
3x  2 y  5 z  20
Solution:
1. All equations are already in the form ax  by  cz  d (step 1 from algorithm)
2. Eliminate y from the first two equations (step 2 from algorithm)
x  y  z  10
2x  2 y  5z  4
2 x  2 y  2 z  20
2x  2 y  5z  4
4 x  7 z  24
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3. Eliminate y from the first and third equations (step 3 from algorithm)
x  y  z  10
3 x  2 y  5 z  20
2 x  2 y  2 z  20
3 x  2 y  5 z  20
5 x  7 z  40
4. Solve the system of two equations formed by combining the equations that resulted from the
previous two steps (step 4 from algorithm)
4 x  7 z  24
5 x  7 z  40
 x  16
x  16
4(16)  7 z  24
64  7 z  24
7 z  40
z   407
5. Substitute the values to find y (step 5 from algorithm)
x  y  z  10
16  y  407  10
112
7
72
7
 y  407  707
 y  707
y   72
149
Technical Mathematics
6. Check (step 6 from algorithm)
2 x  2 y  5z  4
x  y  z  10
16  
2
7
40
7
 10
16  6  10
10  10
2(16)  2( 72 )  5( 407 )  4
32  74  200
7 4
28
7
4
44
3 x  2 y  5 z  20
3(16)  2( 72 )  5( 407 )  20
48  74  200
7  20
336
7
 74  200
7  20
20  20
The solutions check out, and hence the solution is: x  16, y   72 , and z   407 .
150
Technical Mathematics
Chapter S Homework problems
Section S.1 Homework
Use the Virtual TI-83 (or any graphing calculator) to find the point of intersection of the following lines.
1.
y  4x  5
y  3x  9
2. Note: For this problem you will need to manipulate the equation to isolate y, and you will need
to change the Window to:
Xmin = 70
Xmax = 80
Ymin = –30
Ymax = –40
y  2 x  168
y  2 x  128
3. Note: For this problem you will need to manipulate the equation to isolate y, and you will need
to change the Window back to the default values:
Xmin = –10
Xmax = 10
Ymin = –10
Ymax = 10
4 y  27.8  2 x
24 y  153.3  3x
4. Enter the following two equations as written below, and graph them to try to determine the
point of intersection.
y  3x  5
y  3x  2
Can you explain what happened in this case?
151
Technical Mathematics
Section S.2 Homework
Use the Addition-Subtraction method to solve the following system of equations (note that the first 3
problems are the same as the preceding section and will have the same answers).
1.
y  4x  5
y  3x  9
2.
y  2 x  168
y  2 x  128
3.
4 y  27.8  2 x
24 y  153.3  3x
4.
3a  11b  172
7a  4b  57
Section S.3 Homework
Use the Substitution method to solve the following system of equations (note that the first 3 problems
are the same as the preceding sections and will have the same answers).
1.
y  4x  5
y  3x  9
2.
y  2 x  168
y  2 x  128
3.
4 y  27.8  2 x
24 y  153.3  3x
4.
4m  30n  12
7m  3n  17
152
Technical Mathematics
Section S.4 Homework – use any of the three methods when solving the following problems.
1. E10 is a common gasohol blend, as typically no modifications need to be made to a vehicle’s
engine in order to use this mixture. E10 means the mixture would have 10% ethanol mixed with
90% gasoline.
If 30 liters of E10 are needed, but only E12 and E5 are on hand (these are common mixtures in
other countries), how much of the E12 must be mixed with the E5 in order to obtain the 30 liters
of E10?
Note: E12 would contain 12% ethanol and E5 would contain 5% ethanol.
2. A rectangular yard has been fenced using a total of 344m of fencing. Find the dimensions of the
yard if the length is 14m more than the width.
3. A plane travels 1400 miles with the wind in 5 hours. It makes the return trip in 6 hours.
Assuming the wind speed is constant, find the speed of the wind and the speed of the plane.
4. A circuit has the following relationships between two currents I1 and I 2 (currents are measured
in mA):
17.6 I1  8I 2  11.2
18I1  4 I 2  22
Find I1 and I 2 .
153
Technical Mathematics
Section S.5 Homework
Solve the following systems of equations.
2 x  4 y  6 z  14
1. 4 x  2 y  2 z  8
9 x  6 y  6 z  30
3 x  2 y  20
2.  x  2 y  6 z  24
3 x  4 y  4 z  28
3x  2 y  4 z  14
3. x  10 y  5 z  4
5x  4 y  5z  2
154
Technical Mathematics
Answers to Homework Problems
Chapter 1 Answers
Section 1.2 Homework
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
8/7
8/11
–23/14
–6
10/27
2/7
–27/143
34/12
–24/13
1
Section 1.3 Homework
Fill in the following chart.
Mixed number
Improper Fraction
Decimal
74/7
53/7
7.571
–1113/41
–464/41
–11.317
22/3
24/9
2.667
113
1582/14
113
–662/100
–662/100
–6.62
746/10
373/5
74.6
155
Technical Mathematics
Section 1.4 Homework
Calculate the following.
1.
2.
3.
4.
5.
6.
25
95
13.5
1369
145
6.09
Section 1.5 Homework
In each of the equations, solve for the variable.
1.
2.
3.
4.
5.
x = 39
x = 32
x = 10
x = -5/4
x = 2474/7 = 353.43
Section 1.6 Homework
In each of the following, find the value of x which solves the proportion.
1.
2.
3.
4.
5.
x = 11
x = –12/5
x = 11
x = 135
x = 5018300/21689 = 231.38
156
Technical Mathematics
Section 1.7 Homework
Fill in the following chart:
Scientific Notation
Decimal Notation
9.21 x 108
921,000,000
8.7 x 10-5
0.000087
5 x 1014
500,000,000,000,000
5.09 x 1011
509,000,000,000
7.0 x 10-5
0.00007
3.14 x 10-8
0.0000000314
Calculate the following and express the answer in scientific notation.
1.
2.
3.
4.
5.
6.
8 x 1012
7.48 x 104
9 x 1070
1.632 x 10-4
9.02 x 108
1.33 x 1015
157
Technical Mathematics
Chapter 2 Answers
Section 2.1 Homework
1.
a.
b.
c.
d.
461,200 mm
461.20 m
46.120 dam
0.46120 km
a.
b.
c.
d.
1.7 x 1010 mW
1.7 x 107 W
17,000 kW
0.017 GW
a.
b.
c.
d.
0.005 dL
0.05 cL
5 x 10-7 kL
0.00005 daL
2.
3.
Section 2.2 Homework
1.
a. 1.0 x 1015 µm2
b. 1.0 m2
c. 1.0x10-6 km2
2.
a. 321 cm3
b. 3.21 x 10-4 m3
c. 3.21 x 10-40 Tm3
158
Technical Mathematics
Section 2.3 Homework
1.
Degrees Fahrenheit
Degrees Celcius
-48º
-44.444º
111º
43.888º
37.76º
3.2º
80.6º
27º
2. -273 ºC and -459.4 ºF are absolute zero.
Section 2.4 Homework
1.
2.
3.
4.
5.
27.255 liters
3048 meters
1.60276 pounds
24.7105 acres
170.8665 in3
159
Technical Mathematics
Chapter 3 Answers
Section 3.1 Homework
Fractional representation
Decimal representation
Percentage representation
1_
10
0.1
10%
13
74
0.17568
17.568%
3.25
325%
0.05
5%
0.0023
0.23%
17.44
1,744%
0.71
71%
0.1733
17.33%
.000056
0.0056%
5_
100
_23_
10,000
1744
100
71
100
1733
10,000
__56__
1,000,000
Section 3.2 Homework
1. 21
2. 81.72
3. 0.0441
4. 68.182%
5. 400%
6. 0.646%
7. 90.625
8. 364.86486
9. 6
10. $31,364.49
160
Technical Mathematics
Section 3.3 Homework
Measurement
Precision
30 V
21.5 m
0.0003 g
105,000,000 mi
10 V
0.1 m
0.0001 g
1,000,000 mi
Greatest
Possible Error
5V
0.05 m
0.00005 g
500,000 mi
Relative Error
Percent of Error
0.16667 V
0.00233 m
0.16667 g
0.00476 mi
16.667%
0.233%
16.667%
0.476%
Tolerance
Interval
2m
0.01 mL
1.0 oz
0.0004 W
.0003 cm
1m
Section 3.4 Homework
Specification
Tolerance
Upper Limit
Lower Limit
17 m
60.01 mL
1.0 oz
60.0001 W
2.5 cm
4m
1m
0.005 mL
.5 oz
0.0002 W
0.00015 cm
0.5 m
18 m
60.015 mL
1.5 oz
60.0003 W
2.50015 cm
4.5 m
16 m
60.005 mL
0.5 oz
59.9999 W
2.49985 cm
3.5 m
161
Technical Mathematics
Chapter 4 Answers
Section 4.2 Homework
1.
2.
3.
4.
5.
6.
6(1000) + 1(100) + 9(10) + 2(1)
1(22) + 0(21) + 1(20)
1(28) + 0(27) + 1(26) + 0(25) + 1(24) + 1(23) + 0(22) + 0(21) + 1(20)
7(162) + 2(161) + 9(160)
10(162) + 14(161) + 3(160)
1(165) + 3(164) + 12(163) + 2(162) + 13(161) + 0(160)
Section 4.3 Homework
1.
2.
3.
4.
5.
6.
5
345
91
1010012
10001000112
100000000002
Section 4.4 Homework
1.
2.
3.
4.
5.
6.
7.
8.
2,787
1,295,056
2C816
7A64A16
4B16
334B16
101000112
11001110011012
162
Technical Mathematics
Chapter 5 Answers
Section 5.1 Homework
1.
2.
3.
4.
5.
450 N
48W
18W
18W
183.2 cm3
Section 5.2 Homework
1.
2.
3.
4.
5.
4.17A
5.56V
0.617Ω
3.16A
7.50 cm
163
Technical Mathematics
Chapter 6 Answers
Section 6.1 Homework
1.
a. The angle’s complement is 52⁰.
b. The angle’s supplement is 142⁰.
2.
a. The angle’s complement does not exist.
b. The angle’s supplement is 66⁰.
3. Two pairs of equal angles are W & Y and X & Z.
4. Three pairs of equal angles are E & B, B & G, and E & G. Three pairs of supplementary angles are
E & F, F & G, and G & H.
Section 6.2 Homework
1.
Acute, Right, or Obtuse
Acute
Obtuse
Right
2.
3.
4.
5.
6.
7.
8.
9.
10.
Scalene, Isosceles, or Equilateral
Equilateral
Isosceles
Scalene
A = 62⁰
x = 20.24846. The perimeter is 48.24846.
x = 20.19332. The perimeter is 47.69332.
The area is 93.5 m.
The area is 99.95693 in.
The area is 37,960.12 m.
The three sides are 83 ft, 83 ft, and 10 ft.
The area is 415.75 ft.
The area is 2.24 cm.
164
Technical Mathematics
Section 6.3 Homework
1.
2.
3.
4.
5.
The area is 1,260 m2 and the perimeter is 144 m.
The width is 10.29412 cm.
The area is 592.5 mm2.
The area is 25,750 m2 and the perimeter is 746.072 m.
39 gallons of paint are needed and the cost is $429.
Section 6.4 Homework
1.
Circle radius
3.4 in
17 ft
1.5 x 106 km
52
π
15.0545 mm
2.
3.
4.
5.
Circle diameter
6.8 in
34 ft
3 x 106 km
104
π
30.1089 mm
Circle circumference
6.8 π
34 π
3 π x 106 km
104 cm
30.1089 π mm
Circle area
11.56 π
289 π
1.125 π x 1012 km
2704
π
712 mm3
137o
92o
186o
Arc s = 76o and the diameter of the circle is 3.56 in.
165
Technical Mathematics
Chapter 7 Answers
Section 7.1 Homework
1. sinA = 0.8, cosA = 0.6, tanA = 1.333
2. sinA = 0.5519, cosA = 0.8340, tanA = 0.6617
3. sinB = 0.7273, cosA = 0.6864, tanB = 1.0596
Section 7.2 Homework
1. If A = 67°, find:
a. sinA
0.9205
b. cosA 0.3907
c. tanA 2.3559
2. If B = 23°, find:
a. sinA
0.3907
b. cosA 0.9205
c. tanA 0.4245
3. Yes there was a relationship between the two. This occurs because if a right triangle has an
angle of 23°, the other must be 67º (the values used in both questions 1 and 2). If you look at
this triangle more in depth, you will quickly see that the cosine of one angle is equal to the sine
of the other and vice versa.
4. 14.55°
5. 55.03°
6. 74.53°
7. The value you get back is ‘not a number.’ This occurs because two possible values for the
adjacent and hypotenuse lengths cannot exist to satisfy a value of 2.143 in a right triangle. Since
the length of the hypotenuse will always be greater than the length of one of the legs, and the
hypotenuse is in the denominator, the cos of an angle will always be less than 1.
8. cosA = 0.9494, tanA = 0.3309
Section 7.3 Homework
1.
2.
3.
4.
5.
6.
7.
8.
x = 12.72 cm, y = 11.87 cm
x = 82.89 ft, y = 137.73 ft
A = 50.91°
A = 61.95°
A = 52.61°
Z = 19.8, phase angle = 45°
R = 15.76, XL = 12.31
phase angle = 44.9°, XL = 8.47
166
Technical Mathematics
Chapter 8 Answers
Section 8.1 Homework
1.
a.
b.
c.
d.
e.
2008
~388
~96
~479
~399
a.
b.
c.
d.
47%
$14,000
$41,300
This does support the claim. In the fourth quarter (the months leading to the holidays)
revenue was approximately $23,100 dollars which is appreciably higher than the other
quarters.
2.
3.
a.
b.
c.
d.
~95,000 bolts
~500 bolts
~4.74%
Yes. While 1995 produced a larger number of bolts not meeting specifications, the
average number of bolts produced, as well as bolts meeting specifications, has been
greater since 2000.
e. The chart does not specify the bolts produced in 1985. Either no data was collected and
recorded at that time, or the chart doesn’t provide the y-axis numbers necessary for
representing the bolts produced in 1985.
4.
a.
b.
c.
d.
e.
~7Ω
No, the y-axis (Power) scale is too large to represent the resistance values.
The current increases by 10 amps in each subsequent situation. The second chart.
From 700W to 2,800W => increases by ~2,100W.
From 700W to 700,000W => increases by ~699,300W.
167
Technical Mathematics
5.
a. For each value of x, find the corresponding value of y:
i.
Y=8
ii.
Y = -1
iii.
Y = -4
iv.
Y = 11
b. For each value of y, find the corresponding value of x:
i.
X = -0.25
ii.
X = 0.25
iii.
X=1
iv.
X = -1
168
Technical Mathematics
Chapter 9 Answers
Section 9.1 Homework
1. Frequency table:
Histogram:
Analyzing new starter product
12
Frequency
10
8
6
4
2
0
2-6
7 - 11
12 - 16
17 - 21
Number of failures
2. Frequency table:
169
Technical Mathematics
Histogram:
Analyzing student grades
900
800
Frequency
700
600
500
400
300
200
100
0
60 - 69
70 - 79
80 - 89
90 - 99
Grades
Section 9.2 Homework
1. Mean: 15.27
Median: 14
Mode: 9
All values seem to provide a reasonable measure of central tendency.
2. Mean: 1950
Median: 850
Mode: none
The median provides the most reasonable measurement of central tendency. The mean is not
terrible, but it is skewed somewhat as indicated by the fact that it is greater than all but 2 of the
data elements. The mode provides nothing meaningful.
3. The mean, median, and mode are all 3. All provide a reasonable measure of central tendency.
4. Order only matters when calculating the median.
170
Technical Mathematics
Section 9.3 Homework
1. Range: 6, Standard deviation: 2.28
2. Range: 6, Standard deviation: 2.28
3.
a. Range: 82, Standard deviation: 28.48
b. Range: 237, Standard deviation: 76.01
4. The order does not matter when calculating the standard deviation, and only effects the range
to the extent that the greatest and least element need to be identified.
Section 9.4 Homework
1.
a.
b.
c.
d.
e.
f.
68,200 widgets
95,400 widgets
49,850 widgets
97,550 widgets
50,000 widgets
Yes, the order can be accepted because the statistics show that 99.7% are within the
requested range
2. Mean: 0.2852, Standard deviation: 0.00312
49 of the players are within the stated range
Section 9.5 Homework
1. Yes because control limits are within tolerance
2. No, because control limits are not within tolerance
3. Yes because control limits are within tolerance
171
Technical Mathematics
Chapter 10 Answers
Section 10.1 Homework
1. 1
2. 1
Section 10.2 Homework
3.
x
y
z
(y ∧ z’)
x ∨(y ∧ z’)
0
0
0
0
0
0
0
1
0
0
0
1
0
1
1
1
0
0
0
1
1
0
1
0
1
1
1
0
1
1
1
1
1
0
1
4.
x
y
(x ∧ y’)
(x’ ∨ y)
(y ∧ (x’ ∨ y))
(x ∧ y’) ∨ (y ∧ (x’ ∨ y))
0
0
0
1
0
0
0
1
0
1
1
1
1
0
1
0
0
1
1
1
0
1
1
1
172
Technical Mathematics
Section 10.3 Homework
5.
x
y
z
6. Construct the logic schematic for the expression in question 2.
x
y
Section 10.4 Homework
7. ((p ∧ q) ∧ q’)’
8. ((p ∧ q) ∧ ((q ∨ r) ∧ r))’
173
Technical Mathematics
Supplemental Chapter Answers
Section S.1 Homework
1.
x  2 and y  3
2.
x  74 and y  20
3.
x  1.5 and y  6.2
4. The two lines are parallel and hence there is no point of intersection
Section S.2 Homework
1.
x  2 and y  3
2.
x  74 and y  20
3.
x  1.5 and y  6.2
4.
a  0.937 and b  15.892
Section S.3 Homework
1.
x  2 and y  3
2.
x  74 and y  20
3.
x  1.5 and y  6.2
4.
m  2.135 and n  0.685
Section S.4 Homework
1. 21.4 L of E12 and 8.6 L of E10 are needed
2. Length = 93 m and width = 79 m
3. Plane speed = 256.7 mph and wind speed = 23.3 mph
4.
I1  3mA and I 2  8mA
174
Technical Mathematics
Section S.5 Homework
1.
x  2, y  1, and z  1
2.
x
3.
139
x   452
23 , y   23 , and z 
28
17
, y
128
17
, and z 
30
17
350
23
175