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Ratios and Proportion
ALGEBRA 1 LESSON 4-1
(For help, go to Skills Handbook pages 724 and 727.)
Write each fraction in simplest form.
1.
49
84
24
2. 42
135
3. 180
Simplify each product.
4. 35
25

40
14
5. 99
144

96
88
6.
4-1
21
81

108
56
Ratios and Proportion
ALGEBRA 1 LESSON 4-1
Solutions
1. 49 = 7 • 7 = 7
84
7 • 12
12
2. 24 = 6 • 4 = 4
3.
4.
5.
42
6•7
7
135
45 • 3
3
=
=
180
45 • 4
4
35
40
5•7
5•8
5•7•5•8
8
=
=
= =4


25
14
5•5
7•2
5•5•7•2
2
99
96
= 9 • 11  8 • 12 = 9 • 11 • 8 • 12 = 9 = 3

144 88
12 • 12
8 • 11
12 • 12 • 8 • 11
12 4
108 =
3•7

81
56
3•3•3•3
6. 21

3•3•3•4
34 • 7 • 4
= 4
7•8
3 •7•8
4-1
= 4 = 1
8
2
Ratios and Proportion
ALGEBRA 1 LESSON 4-1
Another brand of apple juice costs $1.56 for 48 oz. Find the
unit rate.
cost
ounces
$1.56
= $.0325/oz
48 oz
The unit rate is 3.25¢/oz.
4-1
Ratios and Proportion
ALGEBRA 1 LESSON 4-1
The fastest recorded speed for an eastern gray kangaroo is
40 mi per hour. What is the kangaroo’s speed in feet per second?
40 mi
5280 ft
1h
1 min
•
•
•
1h
1 mi
60 min
60 s
Use appropriate conversion factors.
40 mi
5280 ft
1h
1 min
•
•
•
1h
1 mi
60 min
60 s
Divide the common units.
= 58.6 ft/s
Simplify.
The kangaroo’s speed is about 58.7 ft/s.
4-1
Ratios and Proportion
ALGEBRA 1 LESSON 4-1
Solve y = 3 .
3
y
3
•
12
=
• 12
3
4
4
Multiply each side by the least common multiple of 3
and 4, which is 12.
4y = 9
Simplify.
4y
9
=
4
4
Divide each side by 4.
y = 2.25
Simplify.
4-1
Ratios and Proportion
ALGEBRA 1 LESSON 4-1
Use cross products to solve the proportion
w
6
=–
4.5
5
w(5) = (4.5)(–6)
Write cross products.
5w = –27
Simplify.
5w –27
=
5
5
Divide each side by 5.
w = –5.4
Simplify.
4-1
w
6
=– .
4.5
5
Ratios and Proportion
ALGEBRA 1 LESSON 4-1
In 2000, Lance Armstrong completed the 3630-km Tour de
France course in 92.5 hours. Traveling at his average speed, how
long would it take Lance Armstrong to ride 295 km?
Define: Let t = time needed to ride 295 km.
Relate: Tour de France
average speed
3630
Write:
92.5
3630
295
=
92.5
t
3630t = 92.5(295)
t=
92.5(295)
3630
equals
295-km trip
average speed
295
t
=
kilometers
hours
Write cross products.
Divide each side by 3630.
t
7.5
Simplify. Round to the nearest tenth.
Traveling at his average speed, it would take Lance approximately 7.5
hours to cycle 295 km.
4-1
Ratios and Proportion
ALGEBRA 1 LESSON 4-1
Solve the proportion
z+3
z–4
=
.
4
6
z+3
z–4
=
4
6
(z + 3)(6) = 4(z – 4)
Write cross products.
6z + 18 = 4z – 16
Use the Distributive Property.
2z + 18 = –16
Subtract 4z from each side.
2z = –34
Subtract 18 from each side.
2z –34
= 2
2
Divide each side by 2.
z = –17
Simplify.
4-1
Ratios and Proportion
ALGEBRA 1 LESSON 4-1
12. 1.2
24. 6 2
1. $9.50/h
13. 10,800
25. –5.25
2. $.40/lb
14. 7.5
26. 90
3. 131 cars/week
15. 11.25
27. 28
4. 400 cal/h
16. 5
28. 17.6
5. $.24/oz
17. 25.2
6. $.09/oz
18. 7.5
7. A
19. 6
8. A
20. –20
9. B
21. 14.4
10. A
22. 9
11. 480
23. –16.5
pages 185–188 Exercises
3
29. 67.5
30. 700
31. 105.6 km
32. 0.5
33. 8 11
4-1
12
34. 7 1
3
35. –3 1
2
Ratios and Proportion
ALGEBRA 1 LESSON 4-1
36. 8
48. 1 mi/h
60. 59
37. 165
49. about 0.28 mi/h
61. –8.4
38. 12.5
50. 50.4 min
62. about 646 students
39. 18.75
51. 10.5 mm
63. about 750 students
40. 14.60
52. 246.4 ft/s
64. about 1000 students
41. 18.25
53. 3
42. 504
54. 5.3
43. 2520
55. –16
44. 20 mi/h
56. 115.2
65. Answers may vary. Sample:
Multiply the numerator of each side by
the denominator of the other side.
Set the products equal to each other
and solve the equation.
45. 15 mi/h
57. 45
46. 12 mi/h
58. 4.4
47. 1 mi/h
59. –17
7 = x , (7)(15) = 5x, x = 21
5 15
66. $.05/mi
67. 4 people/mi2, 2485 people/mi2,
78 people/mi2
4-1
Ratios and Proportion
ALGEBRA 1 LESSON 4-1
68. Check students’ work.
77. D
69. a.
b.
c.
d.
78. G
7, 14
21
21
x = 7a
70. Bonnie: $56.00, Tim: $32.00
79. C
80. G
81. [2] 12 in. = 20 in. ;
32 rings x rings
1
1
x = 53 3 rings, which is 53 3 years
71. 48 V
72. 9
OR equivalent explanation
[1] incorrect proportion solved correctly
OR correct proportion solved incorrectly
73. –7.5
74. 9
75. –32
76. a. 5.47 min/mi
b. 5.37 min/mi
82.
83.
84.
4-1
Ratios and Proportion
ALGEBRA 1 LESSON 4-1
85.
97. –5
86.
98. –5.5
87. no solution
99. –90
88. t = height (in.), t <
– 72
100. –6 1
2
89. s = students, s >
– 235
90. m = miles, m <
– 344
91. w = weight (lb), w > 20
92.
93.
94.
95. 6
96. 136
4-1
Ratios and Proportion
ALGEBRA 1 LESSON 4-1
Solve.
1. Find the unit rate of a 12-oz bottle of orange juice that sells for $1.29.
10.75¢/oz.
2. If you are driving 65 mi/h, how many feet per second are you driving?
about 95.3 ft/s
Solve each proportion.
3.
c 12
=
6 15
4.8
4. 21 = 7
5.
3+x
4
=
7
8
1
2
6. 2 + x = 25
12
4
y
x–4
4-1
35
–17
Proportions and Similar Figures
ALGEBRA 1 LESSON 4-2
(For help, go to Skills Handbook and Lesson 4-1.)
Simplify
1.
36
42
2.
81
108
3.
26
52
y
8
=
12 45
6.
w
12
=
15 27
9.
n+1
n
= 24
9
Solve each proportion.
4.
x
7
=
12 30
5.
7.
9
81
=
a
10
8. 25 = z
75
30
4-2
Proportions and Similar Figures
ALGEBRA 1 LESSON 4-2
Solutions
1.
2.
3.
4.
36
6•6 6
=
=
42
6•7 7
81
=
108
26
=
52
27 • 3 3
=
27 • 4 4
26 • 1 1
=
26 • 2 2
x
7
=
12
30
30x = 12(7)
30x = 84
84
x = 30
4
x = 25
y
8
9
81
n
n+1
5. 12 = 45
45y = 12(8)
7. a = 10
81a = 9(10)
9. 9 = 24
24n = 9(n + 1)
45y = 96
81a = 90
24n = 9n + 9
96
90
y = 45
a = 81
2
y = 215
1
a = 19
6. w = 12
15
8. 25 = z
27
75
27w = 15(12)
27w = 180
180
w = 27
2
w=63
4-2
30
75z = 25(30)
75z = 750
750
z = 75
z = 10
15n = 9
9
n = 15
3
n= 5
Proportions and Similar Figures
ALGEBRA 1 LESSON 4-2
In the figure below,
Relate:
EF DE
=
BC AB
ABC ~
DEF. Find AB.
Write a proportion comparing the lengths
of the corresponding sides.
Define: Let x = AB.
Write:
6
8
=
9
x
6x = 9(8)
6x 72
= 6
6
x = 12
AB is 12 mm.
Substitute 6 for EF, 9 for BC, 8 for DE, and
x for AB.
Write cross products.
Divide each side by 6.
Simplify.
4-2
Proportions and Similar Figures
ALGEBRA 1 LESSON 4-2
A flagpole casts a shadow 102 feet long. A 6 ft tall man casts
a shadow 17 feet long. How tall is the flagpole?
102
x
=
17
6
Write a proportion.
17x = 102 • 6
Write cross products.
17x = 612
Simplify.
17x 612
= 17
17
Divide each side by 17.
x = 36
Simplify.
The flagpole is 36 ft tall.
4-2
Proportions and Similar Figures
ALGEBRA 1 LESSON 4-2
The scale of a map is 1 inch : 10 miles. The map distance
from Valkaria to Gifford is 2.25 inches. Approximately how far is the
actual distance?
map
actual
1
2.25
=
10
x
map
actual
1 • x = 10 • 2.25
Write a proportion.
Write cross products.
x = 22.5
Simplify.
The actual distance from Valkaria to Gifford is
approximately 22.5 mi.
4-2
Proportions and Similar Figures
ALGEBRA 1 LESSON 4-2
10. 12 in.
pages 192–195 Exercises
1. AB PQ, BC QR, CA RP,
A
P, B
Q, C
R
11. 87.5 mi
2. ED JH, DF HK, FE KJ,
D
H, E
J, F
13. 325.5 mi
3. 3.125 ft
– cm
4. 13.33
5. 80 in.
6. 40 m
7. 20.25 cm
8. 7.2 ft
9. 4.8 ft
K
12. 145.25 mi
14. 350 mi
15. a. Lincoln to San Paulo = 16 mi
Lincoln to Duncanville = 26 mi
San Paulo to Duncanville = 18 mi
b. 26 mi roundtrip
16. 1 cm : 8 km
17. 4 in. by 6 in.
18. 2 2 in. by 4 in.
3
19. 2 in. by 3 in.
4-2
Proportions and Similar Figures
ALGEBRA 1 LESSON 4-2
20. 3.2 in. by 4.8 in.
21. 33.75 in.
22. 22.5 ft by 27 ft
23. a. Answers may vary.
Sample: GK and RQ
are not corresponding sides.
b. GH = HL
PQ
RQ
24. 1 in. : 12 ft
25. 9 ft by 12 ft
26. 3 ft
27. 216 ft2
28. yes; because it is 6 ft wide and 9 ft long
29. 48 cm long by 20 cm wide
4-2
30. a. 6 m
b. 6 m, 18 m
c. Yes, the ratio of the sides
is equal to the ratio of
perimeters in similar figures.
d. 2 m2, 18 m2
e. Answers may vary. Sample:
The area ratio is the square
of the side ratio.
31. Answers may vary. Sample: doll
house to regular house, model
car to real car
32. a. Yes; the sides are proportional.
b. The volume ratio is the cube
of the side ratio.
c. 27 : 1
33. a = 8, b = 6, c = 10
Proportions and Similar Figures
ALGEBRA 1 LESSON 4-2
34. about 1 in. : 30.5 mi
35. 400,400 km
36. a. 8 = 5
8+x
7
b. 3.2
39. [2] smaller area: 6 • 5 = 30; 30 ft2;
larger area: 10 • 12 = 120; 120 ft2;
0.5 = x
30
120
0.5(120) = 30x
60 = 30x
c. 11.2 in.
d. 39.2 in.2
37. B
38. I
2 =x
Two gallons of paint should cover
a 10 ft  20 ft wall.
[1] incorrect calculation for one area and
proportion solved correctly OR correct
area calculations but proportion
set up incorrectly
4-2
Proportions and Similar Figures
ALGEBRA 1 LESSON 4-2
77.5
55
=/
16
12
40. [4] a.
77.5(12) = 16(55)
930 =/ 880
Since the cross products
are not equal, the
proportion is not true. So
the postcard is not
similar to the painting.
b. 77.5 = 55 OR 77.5 = 55
x
12
16
y
The postcard should be
12 cm  16.9 cm OR
11.4 cm  16 cm.
[3] appropriate methods, but with one
computational error OR found only
one possible postcard size
[2] incorrect proportions solved correctly
[1] correct answer with no work shown
41. 4.5
42. 16 2
3
43. –22 6
7
44. 40
45. b < –4
46. x ≥ 7
47. m < –4
48. h > – 3
2
4-2
Proportions and Similar Figures
ALGEBRA 1 LESSON 4-2
1. In the figure below,
ABC ~
DEF. Find DF.
About 19.7 cm
2. A boy who is 5.5 feet tall casts a shadow that is 8.25 feet long.
The tree next to him casts a shadow that is 18 feet long.
How tall is the tree?
12 ft
3. The scale on a map is 1 in.: 20 mi. What is the actual distance
between two towns that are 3.5 inches apart on the map?
70 mi
4-2
Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
(For help, go to skills handbook pages 727 and 728.)
Find each product.
1. 0.6 • 9
2. 3.8 • 6.8
3.
23 20
•
60 46
4.
17
• 5
135 34
Write each fraction as a decimal and as a percent.
7
5. 10
9.
35
40
23
6. 100
10.
7
16
2
7. 5
11.
4-3
4
25
13
8. 20
12.
170
200
Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
Solutions
1.
0.6 • 9 = 5.4
2.
3.8 • 6.8 = 25.84
3.
23 20
23 • 20
1
1
•
=
=
=
60 46
20 • 3 • 23 • 2
3•2
6
17
5
17 • 5
1
1
4. 135 • 34 = 5 • 27 • 17 • 2 = 27 • 2 = 54
5.
7
= 7 ÷ 10 = 0.7; 0.7(100%) = 70%
10
6.
23
= 23 ÷ 10 = 0.23; 0.23(100%) = 23%
100
7.
2
= 2 ÷ 5 = 0.4; 0.4(100%) = 40%
5
4-3
Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
Solutions (continued)
13
8. 20 = 13 ÷ 20 =0.65; 0.65(100%) = 65%
35
9. 40 = 35 ÷ 40 = 0.875; 0.875(100%) = 87.5%
7
10. 16 = 7 ÷ 16 = 0.4375; 0.4375(100%) = 43.75%
4
11. 25 = 4 ÷ 25 = 0.16; 0.16(100%) = 16%
12.
170
= 170 ÷ 200 = 0.85; 0.85(100%) = 85%
200
4-3
Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
What percent of 90 is 27?
percent
n
27
=
100
90
part
whole
90n = 2700
Find the cross products.
n = 30
Divide each side by 90.
30% of 90 is 27.
4-3
Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
Find 25% of 480.
25
n
=
100 480
part
whole
12,000 = 100n
Find the cross products.
120 = n
Divide each side by 100.
25% of 480 is 120.
4-3
Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
Water covers about 361,736,000 km2, or about 70.8% of the
earth’s surface area. Approximately what is the total surface area of
the earth?
Relate: 70.8% of the total surface area is 361,736,000 km2.
Define: Let t the total surface area.
Write:
70.8
=
100
361,736,000
t
70.8t = 361,736,000,000
t = 510,926,553.7
part
whole
Find cross products.
Divide each side by 70.8.
The total surface area of the earth is approximately 510,926,554 km2.
4-3
Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
What percent of 140 is 84?
Relate: What percent
of
140
is
84?
Define: Let p = the decimal form of the percent.
Write:
p
•
140
=
84
140p = 84
p = 0.6
Divide each side by 140.
p = 60%
Write the decimal as a percent.
60% of 140 is 84.
4-3
Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
What percent of 60 is 144?
Relate: What percent
of
60
is
114?
Define: Let n = the decimal form of the percent.
Write:
n
•
60
=
114
60n = 114
n = 1.90
Divide each side by 60.
n = 190%
Write the decimal as a percent.
190% of 60 is 114.
4-3
Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
a. Estimate the number that is 19% of 323.
19%
1
5
323
325
1
1
= 20%. So is a good approximation of 19%.
5
5
325 and 5 are compatible numbers.
1
• 325 = 65
5
65 is approximately 19% of 323.
b. What is 73% of 125? Use fractions to estimate the answer.
73%
3
4
125
124
3
3
= 75%. So is a good approximation of 73%.
4
4
124 and 4 are compatible numbers.
3
• 124 = 93
4
93 is approximately 73% of 125.
4-3
Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
A candidate for mayor sent out surveys to 8056 people in his
city. After two weeks, about 18% of the surveys were returned. How
many surveys were returned?
Relate: What
is
18%
of
8056?
Define: Let n = the unknown number.
Write:
n
=
0.18
•
n = 0.18 • 8056
n = 1450.08
Simplify.
About 1450 surveys were returned.
4-3
8056
Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
pages 200–202 Exercises
1. 50%
2. 25%
3. 33 1 %
3
13.
14.
15.
4. 20%
5. 25%
23. 45 = n • 60; 75%
12. 20
16.
40
100
80
100
15
100
20
100
60
100
30
100
= 20 , 50
x
= 20 ,
x
= 24 ,
x
= 48 ,
x
= 42 ,
x
= 42 ,
x
25
160
240
24. x = 0.05(300); 15
25. x = 0.05(200); 10
26. 21
27. 200%
28. 4
6. 20%
17.
7. 8
18.
8. 16
19. 30 h
31. 0.3%
9. 21
20. 50 = 0.25x; 200
32. 200
10. 28
21. 25 = 0.50x; 50
11. 10
70
29. 300%
140
30. 0.42
22. 96 = n • 150; 64%
4-3
33. 100
34. 22
Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
35. 150
45. 62; 50% is 61 and 51.3% > 50%.
36. 68
46. 20; 25% is 21 and 23.9% < 25%.
37. 32
47. 73; 10% is 74 and 9.79% < 10%.
38. about 960 students
48. 185; 75% is 180 and 76.02% > 75%.
39–40. Proportions or equations may
vary. Samples are given.
39. 75 = 3 , 4
100 x
40. x = 300 , 400%
100
75
41. x = 0.002(900), 1.8
42. 0.02x = 1.8, 90
43. 1000x = 988, 98.8%
44. 1.4(84) = x, 117.6
49. $297.00
50. $32.70
51. $3896.00
52. Answers may vary. Sample: $1.20;
take 10% and 5% of $8 and add them.
53. a. $74.25
b. 3.75%
c. 6 yr
54. $61.20
4-3
55. $1250.00
56. 7%
57. 2 yr
Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
58. Answers may vary. Sample: 33%;
66. 20
67. 13 1 mi
I sleep 8 h a day, and there are
3
24 h in a day, so x = 8 , x = 33 1 .
100 24
3
59. a. 3
50
68. 60 mi
69. 90 mi
b. 200
70. 106 2 mi
3
c. 400%
71. b < –4;
60. Yes; 16.99(1.15)(0.66) < 13.99(1.06),
12.90 < 14.83.
61. 27237
62. 11712
72. x > 7;
–
73. h > –21;
74. p < – 1 ;
3
63. 425
64. 35
65. 667
4-3
Proportions and Percent Equations
ALGEBRA 1 LESSON 4-3
1. What is 35% of 160?
56
2. What percent of 450 is 36?
8%
3. 32 is 80% of what number?
40
4. What is 0.03% of 260,000?
78
5. What percent of 50 is 75?
150%
6. Estimate 62% of 83?
51
4-3
Percent of Change
ALGEBRA 1 LESSON 4-4
(For help, go to Lesson 4-3.)
Write an equation for each problem and solve.
1. What is 20% of 20?
2. 8 is what percent of 20?
3. 18 is 90% of what number?
4. 27 is 90% of what number?
Estimate each answer.
5. 67.3% of 24
6. 65% of 48
4-4
Percent of Change
ALGEBRA 1 LESSON 4-4
Solutions
1. What is 20% of 20?
n = 0.2(20)
n=4
4. 27 is 90% of what number?
27 = 0.9n
n = 27 = 27 ÷ 0.9 = 30
0.9
2. 8 is what percent of 20?
8 = n • 20
8
n=
= 8 ÷ 20 = 0.4 = 40%
20
5. 67.3% of 24
6. 65% of 48
3. 18 is 90% of what number?
18 = 0.9n
n = 18 = 18 ÷ 0.9 = 20
0.9
4-4
2
2
of 24 = • 24 = 16
3
3
2
2
=
• 48 = 32
3
3
Percent of Change
ALGEBRA 1 LESSON 4-4
The price of a skirt decreased from $32.95 to $28.95. Find
the percent of decrease.
percent of decrease =
amount of change
original amount
=
32.95 – 28.95
32.95
Subtract to find the amount of change.
Substitute the original amount.
=
4
32.95
Simplify the numerator.
0.12 or 12%
Write as a decimal and then as a percent.
The price of the skirt decreased by about 12%.
4-4
Percent of Change
ALGEBRA 1 LESSON 4-4
Between 1940 and 1980, the federal budget increased from
$9.5 billion to $725.3 billion. What was the percent of increase in the
federal budget?
percent of increase =
amount of change
original amount
=
725.3 – 9.5
9.5
Substitute.
=
715.8
9.5
Simplify the numerator.
= 75.35 or 7535%
The federal budget increased nearly 7535%.
4-4
Write as a decimal and then
as a percent.
Percent of Change
ALGEBRA 1 LESSON 4-4
You read the bathroom scale as 122 lb. What is your greatest
possible error?
The scale is read to the nearest 1 lb, so the greatest possible
error is one half of 1 lb, or 0.5 lb.
4-4
Percent of Change
ALGEBRA 1 LESSON 4-4
When a garden plot was measured, the dimensions were
156 in.  84 in. Use the greatest possible error to find the minimum
and maximum possible areas.
Both measurements were made to the nearest whole inch, so the greatest
possible error is 0.5 in.
The length could be as little as 155.5 in. or as great as 156.5 in.
The width could be as little as 83.5 in. or as great as 84.5 in. Find the
minimum and maximum areas.
Minimum Area
Maximum Area
155.5 in.  83.5 in. = 12,984.25 in.2
156.5 in.  84.5 in. = 13,224.25 in.2
The minimum area is 12,984.25 in.2, and the maximum area is 13,224.25 in.2.
4-4
Percent of Change
ALGEBRA 1 LESSON 4-4
Suppose you measure a library book and record its width as
17.6 cm. Find the percent of error in your measurement.
Since the measurement is to the nearest 0.1 cm, the greatest possible
error is 0.05 cm.
percent error =
greatest possible error
measurement
0.05
= 17.6
Use the percent error formula.
Substitute.
0.0028409091
Divide.
= 0.3%
Round and write as a percent.
The percent error is about 0.3%.
4-4
Percent of Change
ALGEBRA 1 LESSON 4-4
A small jewelry box measures 7.4 cm by 12.2 cm by 4.2 cm.
Find the percent error in calculating its volume.
The measurements are to the nearest 0.1 cm. The greatest possible error
is 0.05 cm.
as measured
V=
•w•h
= 7.4 • 12.2 • 4.2
= 379.18 cm3
maximum value
V=
minimum value
•w•h
V=
= 7.45 • 12.25 • 4.25
= 387.87
Possible Error: maximum – measured
387.87 – 379.18 = 8.69
4-4
•w•h
= 7.35 • 12.15 • 4.15
= 370.61
measured – minimum
379 – 370.61 = 8.57
Percent of Change
ALGEBRA 1 LESSON 4-4
(continued)
Use the difference that shows the greatest possible error to find the
percent error.
percent error =
=
greatest possible error
measurement
Use the percent error formula.
387.87 – 379.18
379.18
Substitute.
8.69
= 379.18
Simplify the numerator.
= 0.0229178754
Write as a decimal.
= 2%
Round and write as a percent.
The percent error is about 2%.
4-4
Percent of Change
ALGEBRA 1 LESSON 4-4
pages 207–209 Exercises 12. 14.4%; increase
24. 303.75 km2; 340.75 km2
1. 50%; increase
13. 39%
25. 25%
2. 33 1 %; decrease
14. 60%
26. 25%
3. 25%; increase
15. 0.5 ft
27. 12.5%
4. 20%; decrease
16. 0.05 cm
28. 12.5%
5. 33 1 %; increase
17. 0.005 g
29. a. 48 cm3
6. 25%; decrease
18. 0.5 in.
b. 74.375 cm3
7. 25%; increase
19. 19.25 cm2, 29.25 cm2
8. 20%; increase
20. 48.75 mi2, 63.75 mi2
9. 84.4%; increase
21. 46.75 in.2, 61.75 in.2
c. 28.125 cm3
d. 26.375 cm3
e. 55%
3
3
10. 71.1%; increase
11. 60.7%; decrease
22. 51.75
km2,
23. 253.75
68.75
in.2,
4-4
km2
286.75
in.2
30. 23%; decrease
31. 22%; decrease
Percent of Change
ALGEBRA 1 LESSON 4-4
32. 157%; increase
33. 175%; increase
43. no; increases to $70.40
but decreases to $63.36
49. 11%
50. 34%
37. 9%; decrease
44. Answers may vary. Sample: 51.
Joan bought shoes for $10.
Sarah bought the same shoes
3 days later for $7. What was
the percent change?
30% decrease
38. 17%; increase
45. 24.5 cm2, 25.5 cm2
39. 2%
46. 58 mi2, 59.6 mi2
40. 19%
47. 54.1 in.2, 54.3 in.2
34. 4%; increase
35. 3%; decrease
36. 56%; decrease
41. 1 mm
48. a.
b.
42. no; 16% increase
c.
but a 14% decrease
d.
100%
100%
50%
50%
4-4
Answers may vary.
Sample: Use the
greatest possible error
to calculate the
maximum, minimum,
and measured areas.
Find the amounts by
which the maximum
and minimum differ
from the measured
area. Divide the greater
difference by the
measured area.
Percent of Change
ALGEBRA 1 LESSON 4-4
52. Jorge found the change of $5
but divided by the final price
instead of the original price.
53. a. 9%, 3%
b. Answers may vary. Sample:
The larger a measure,
the smaller is the percent error.
56.
57.
58.
59.
54. Yes; 148.3 > 3 (48.7) = 146.1,
and 148.3 – 48.7
205%.
48.7
55. a. 21%
b. 21%
c. 21%; answers may vary.
Sample: 1.1a • 1.1a = 1.21a2,
which is 21% greater than a • a = a2.
Relationship between % increase of
side and area of the square doesn’t
depend on the side length.
4-4
C
I
A
[2] perimeter of softball diamond:
4(60) = 240, perimeter 240 ft,
side of baseball diamond:
1.5(60) = 90, side 90 ft, perimeter
of baseball diamond: 4(90) = 360,
perimeter 360 ft, % of increase
= 360 – 240 = 50%; area of softball
240
diamond: 60(60) = 3600,
area 3600 ft2, area of baseball
diamond: 90(90) = 8100,
area 8100 ft2, percent of increase
= 8100 – 3600 = 125%
3600
OR computation that gives
same results
Percent of Change
ALGEBRA 1 LESSON 4-4
[1] appropriate methods, but with one
computational error OR finds only
one % of increase
60–65. Equations may vary.
60.
61.
62.
63.
64.
65.
66.
x
100
x
100
44
100
x
100
0.2
100
266
100
5
, 7%
67
= 13 , 87%
15
= 79 , 179.5
x
= 96 , 300%
32
= x , 1.7
834
= x , 37.2
14
n< 1
15
=
4-4
67. q > –17
–
68. x >
– –1
Percent of Change
ALGEBRA 1 LESSON 4-4
Find each percent of change. Describe the percent of change as
an increase or decrease.
1. $6 to $9
50% increase
2. 15 cm to 12 cm
20% decrease
Find the greatest possible error.
3. 13.2 m
4. 34.62 g
0.05 m
0.005 g
5. Find the percent error for the measurement 6 cm.
about 8.3%
6. Find the minimum and maximum possible areas for a rectangle
measured as 3 m x 7 m.
min: 16.25 m2; max: 26.25 m2
4-4
Applying Ratios to Probability
ALGEBRA 1 LESSON 4-5
(For help, go to Skills Handbook page 728.)
Rewrite each decimal or fraction as a percent.
1. 0.32
2. 0.09
45
3. 200
4-5
9
4. 50
Applying Ratios to Probability
ALGEBRA 1 LESSON 4-5
Solutions
32
1. 0.32 = 100 = 32%
9
2. 0.09 = 100 = 9%
2 • 22.5
45
22.5
3. 200 = 2 • 100 = 100 = 22.5%
4.
9
50
=
9•2
50 • 2
= 18 = 18%
100
4-5
Applying Ratios to Probability
ALGEBRA 1 LESSON 4-5
A bowl contains 12 slips of paper, each with a different name
of a month on it. Find the theoretical probability that a slip selected at
random from the bowl has the name of a month that ends with “ber.”
number of favorable outcomes
P(event) = number of possible outcomes
4
= 12
1
= 3
There are 4 months out of 12 that end with “ber”:
September, October, November, and December
Simplify.
1
The probability of picking a month that ends with “ber” is 3 .
4-5
Applying Ratios to Probability
ALGEBRA 1 LESSON 4-5
For a number cube, find the probability of not rolling a
number divisible by 3.
number of favorable outcomes
2
1
P(÷ 3) = number of possible outcomes = 6 = 3
P(not ÷ 3) = 1 – P(÷ 3)
1
2
=1– 3 = 3
Use the complement formula.
Simplify.
2
The probability of not rolling a number divisible by 3 is 3 .
4-5
Applying Ratios to Probability
ALGEBRA 1 LESSON 4-5
Quality control inspected 500 belts at random. They found no
defects in 485 belts. What is the probability that a belt selected at
random will pass quality control?
number of times an event occurs
P(no defects) = number of times the experiment is done
485
= 500
Substitute.
= 0.97 = 97%
Simplify. Write as a percent.
The probability that a belt has no defects is 97%.
4-5
Applying Ratios to Probability
ALGEBRA 1 LESSON 4-5
If the belt manufacturer from Additional Example 3 has 6258
belts, predict how many belts are likely to have no defects.
number with no defects = P(no defects) • number of belts
= 0.97 • 6258
Substitute. Use 0.97 for 97%.
= 6070.26
Simplify.
Approximately 6070 belts are likely to have no defects.
4-5
Applying Ratios to Probability
ALGEBRA 1 LESSON 4-5
pages 214–217 Exercises
1. 1
2.
3.
4.
5.
2
1
3
1
6
1
2
2
3
6. 0
7. 1
3
8. 1
9. 1
3
10. 5
6
11. 5
22. a. 40%
b. about 23 families
6
12. 1
2
23.
13. 1
1
6
14. 80%
24. 0
15. 24%
25. 1
16. 43%
17. 15%
18. 85%
19. 39%
20. 67%
3
26. 5
6
27. 1
2
28. 8
9
29.
1
450
30. 0
21. a. about 40%
1
b. about 200 oak trees 31.
30
4-5
Applying Ratios to Probability
ALGEBRA 1 LESSON 4-5
32. 4
9
33. 1%
34. 1.6%
35. a. Answers may vary.
Sample: 20 students,
12 girls and 8 boys: 5%
b. 60%
c. Answers may vary.
Sample: Subtract P
(picking a boy) from 1.
38. Answers may vary. Sample:
For theoretical probability,
all possible outcomes are
equally likely to happen,
but experimental probability
is based on observed outcomes.
39. 3
40.
41.
36. a. 15%
42.
b. 15%
43.
37. a. 3 b. 2
4
5
c. 3
20
44.
16
3
8
7
16
5
8
1
3
5
3
4-5
Applying Ratios to Probability
ALGEBRA 1 LESSON 4-5
c. 1 , 1 , 1 , 1
45. Answers may vary.
Sample: You can add
the numerator and
denominator and
make the sum the
denominator, keeping
the numerator the same.
18 9 9
d. no
e. Answers may vary.
Sample: Yes; the more
you roll, the closer you
get to the theoretical
probability.
46. 1
4
47. 3
10
48. 3
10
36
50. A
51. G
52. D
49. a. Check students’ work.
b.
4-5
Applying Ratios to Probability
ALGEBRA 1 LESSON 4-5
53. [4] a. theoretical P(red) = 1 OR 20%
5
b. experimental P(red) = 55 OR about 19.6%
280
c. For the red beads, the manufacturer’s claim
seems to be true. However, the experimental
probabilities of the other colors are not as close
to 20%, so Rasheeda’s experiment does not
support the manufacturer’s claim.
[3] one computational error with complete explanation
OR correct computation with weak explanation
[2] correct computation but no conclusion
[1] error(s) in computation and no conclusion
54. [2] P(defective stapler) = 18
350
5.1%; production should be
stopped because 5.1% > 4%.
[1] correct calculation with no conclusion OR incorrect
calculation but correct reasoning based on incorrect calculation
4-5
Applying Ratios to Probability
ALGEBRA 1 LESSON 4-5
55. 25%; increase
67. 6.17, 5, 5
56. 50%; increase
68. 17.29, 11, 35
57. 40%; increase
69. 3
4
6
7
479
19
5
1
70. 1
2
4
6
7
246
37
7
8
9
58. 50%; decrease
59. 25%; decrease
60. 12.5%; decrease
61. –3 ≤ t ≤ 4;
62. 5 < b < 7;
63. h < 2 or h > 5;
64. –2 ≤ w < 1;
65. x < 2 or x ≥ 4;
66. 1 ≤ k ≤ 3;
4-5
Applying Ratios to Probability
ALGEBRA 1 LESSON 4-5
Find each probability for the roll of a number cube.
1. P(4) 1
6
2. P(not 4) 5
6
3. P(odd)
4. You harvest 50 cherry tomatoes from your garden. You
randomly inspect 15 tomatoes and find that 2 have bad spots
on them.
a. What is the experimental probability that a tomato has a
bad spot?
about 13%
b. Predict how many of the tomatoes you picked will have bad
spots.
about 7 tomatoes
4-5
1
2
Probability of Compound Events
ALGEBRA 1 LESSON 4-6
(For help, go to Lessons 4-5.)
Find each probability for one roll of a number cube.
1. P(multiple of 3)
2. P(greater than 4)
3. P(greater than 5)
4. P(greater than 6)
Simplify.
2
7
5. 14 • 6
15
12
6. 24 • 30
6
44
7. 55 • 3
4-6
Probability of Compound Events
ALGEBRA 1 LESSON 4-6
Solutions
2
1
1. P(multiple of 3) = P(3 or 6) = 6 = 3
2. P(greater than 4) = P(5 or 6) =
3. P(greater than 5) = P(6) =
2
1
=
6
3
1
6
4. P(greater than 6) = 0
2
7
2•7
1
•
=
=
14 6 2 • 7 • 6
6
15 12
15 • 12
1
1
6. 24 • 30 = 12 • 2 • 15 • 2 = 2 • 2 = 6
6 44
3 • 2 • 4 • 11
2•4
8
3
=
=
=
1
7. 55 • 3 =
5 • 11 • 3
5
5
5
5.
4-6
Probability of Compound Events
ALGEBRA 1 LESSON 4-6
Suppose you roll two number cubes. What is the probability
that you will roll an odd number on the first cube and a multiple of 3
on the second cube?
3
1
2
1
There are 3 odd numbers out of
six numbers.
P(odd) = 6 = 2
There are 2 multiples of 3 out of
6 numbers.
P(odd and multiple of 3) = P(odd) • P(multiple of 3)
P(multiple of 3) = 6 = 3
1
1
= 2• 3
Substitute.
= 1
Simplify.
6
The probability that you will roll an odd number on the first cube and a
multiple of 3 on the second cube is 1 .
6
4-6
Probability of Compound Events
ALGEBRA 1 LESSON 4-6
Suppose you have 3 quarters and 5 dimes in your pocket.
You take out one coin, and then put it back. Then you take out
another coin. What is the probability that you take out a dime and then
a quarter?
Since you replace the first coin, the events
are independent.
5
P(dime) = 8
There are 5 out of 8 coins that are dimes.
3
P(quarter) = 8
There are 3 out of 8 coins that are quarters.
P(dime and quarter) = P(dime) • P(quarter)
5
3
= 8 •8
Multiply.
15
= 64
15
The probability that you take out a dime and then a quarter is 64 .
4-6
Probability of Compound Events
ALGEBRA 1 LESSON 4-6
Suppose you have 3 quarters and 5 dimes in your pocket.
You take out one coin, but you do not put it back. Then you take out
another coin. What is the probability of first taking out a dime and then
a quarter?
5
P(dime) = 8
3
P(quarter after dime) = 7
There are 5 out of 8 coins that are dimes.
There are 3 out of 8 coins that are quarters.
P(dime then quarter) = P(dime) • P(quarter after dime)
5
3
= 8 •7
Multiply.
15
= 56
15
The probability that you take out a dime and then a quarter is 56 .
4-6
Probability of Compound Events
ALGEBRA 1 LESSON 4-6
A teacher must select 2 students for a conference. The
teacher randomly picks names from among 3 freshmen,
2 sophomores, 4 juniors, and 4 seniors. What is the probability that a
junior and then a senior are chosen?
4
P(junior) = 13
There are 4 juniors among 13 students.
There are 4 seniors among 12 remaining
students.
4
P(senior after junior) = 12
P(junior then senior) = P(junior) • P(senior after junior)
4
4
= 13 • 12
16
4
= 156 = 39
Substitute.
Simplify.
4
The probability that the teacher will choose a junior then a senior is 39.
4-6
Probability of Compound Events
ALGEBRA 1 LESSON 4-6
pages 222–224 Exercises
1.
2.
3.
4.
5.
6.
1
36
1
18
1
18
1
9
1
4
25
36
10.
11.
12.
13.
14.
15.
7. 1
16.
8. 0
17.
4
9. 81
18.
2
27
1
9
16
81
4
27
4
27
2
11
3
11
1
55
3
11
19. 0
20. 1
30. Dep.; with one name gone
the data set changes.
31. Indep.; the data set hasn’t changed.
32. Answers may vary. Sample: For
dep. events, the outcome of the first
event affects the outcome of the
second (example: picking a marble
out of a bag, and then picking a
second marble without replacing
the first one). For independent
events, the outcomes do not affect
each other (example: picking the
second marble after replacing
the first).
21. 2
22.
23.
24.
25.
26.
7
3
22
1
6
2
9
1
9
1
15
27. 0
28. 1
45
29. Indep.; you still have 2
choices for each coin with
or without the other coin.
4-6
Probability of Compound Events
ALGEBRA 1 LESSON 4-6
33. a. 0.58
b. 0.003248
34. a-c. Check students’ work.
35. 0.0036
c. 20
77
d. 20
77
e. Answers may vary.
36. 1
37.
38.
39.
40.
41.
6
1
10
1
12
1
5
1
15
1
18
42. a. 2
7
b. 15
77
Sample: 1; 2 + 15 + 20 + 20 = 1
7
43. a. 1
3125
44. a. 1
36
45. a. 12
46. C
47. F
48. B
4-6
77
1
c. 5
15,625
b. 1
c. 1
36
6
b. 5
c. 1
6
3
b.
77
77
Probability of Compound Events
ALGEBRA 1 LESSON 4-6
49. [2] P(green, green) = 3 • 2 = 6 = 1 ,
8
72 12
P(red, red) = 4 • 3 = 12 = 1 ,
72
6
9 8
55. 2
9
P(r, r) is twice as likely as P(g, g).
[1] correct calculations for both
probabilities but incorrect statement
OR correct calculations for one
probability and correct statement
based on that answer
50. 11
21
51. 4
21
52. 2
7
53. 8
21
54. 4, –4
4-6
56. all real numbers
57. No solution; abs. value
can’t be negative.
58. No solution; abs. value
can’t be negative.
59. t < 17 or t > 35
Probability of Compound Events
ALGEBRA 1 LESSON 4-6
You roll two number cubes. Find each probably.
1. P(odd and even)
1
4
2. P(1 or 2 and less than 5)
2
9
You select letters from the following: A A B B B C D D E F G G G
and do not replace them. Find each probability.
3. P(A then B)
1
26
4. P(vowel then G)
3
52
4-6
Solving and Applying Proportions
ALGEBRA 1 CHAPTER 4
9
14
b. 1
2
1. 15
13. 10%; decrease 23. a.
2. 7.5
14. 33.3%; increase
3. 2.4
15. 3
4. 20
5. 40
6. 64%
7. 20
8. 12 cm
9. 4%
10. $7.80
c. 0
5
16. 1
5
24. 12 carnations for $6.99
25. Answers may vary.
Sample: Four cards have
one letter each: A, B, C, or D.
What is the probability that the
first card you select is A and
the second is B, if you don’t
replace the first card before
selecting the second card? 1
17. 2.24
18.
1
6
19. 3080
20. 1
21. 162.5 mi
12
22. 12.5 ft
11. 11.1%; increase
12. 25%; decrease
4-A
Solving and Applying Proportions
ALGEBRA 1 CHAPTER 4
26. a. about 1143%
b. Sample: Use the second row. Subtract the amount in the first column
from the amount in the second column. Divide the result by the amount
in the first column and multiply by 100.
27. a. 1
4
b. 4
15
c. 1
4
4-A