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Posttest
Use set notation to write the elements of each set. Then determine whether the statement about the set
is true or false .
1. M is the set of natural number multiples of 5 that are less than 50. 12 M
SOLUTION: M = {5, 10, 15, 20, 25, 30, 35, 40, 45}; 12 is not an element of this set since 12 is not a multiple of 5. Therefore, the
statement 12 M is false.
ANSWER: M = {5, 10, 15, 20, 25, 30, 35, 40, 45}; false
2. S is the set of integers that are less than –40 but greater than −50. −49
S
SOLUTION: S = {–49, –48, –47, –46, –45, –44, –43, –42, –41}; –49 is an element of this set. Therefore, the statement −49
true.
S is
ANSWER: S = {–49, –48, –47, –46, –45, –44, –43, –42, –41}; true
Let B = {0, 1, 2, 3}, C = {0, 1, 2, 3, 4, 5, 6}, D = {1, 3, 5, 7, 9}, E = {0, 2, 4, 6, 8, 10}, and F = {0, 10}.
Find each of the following.
3. SOLUTION: means the intersection of D and C. Identify the elements that belong to both D and C. D = {1, 3, 5, 7, 9}
and C = {0, 1, 2, 3, 4, 5, 6}. The elements found in both sets are 1, 3 and 5. So
= {1, 3, 5}.
ANSWER: {1, 3, 5}
4. SOLUTION: means the intersection of D and F. Identify the elements that belong to both D and F. D = {1, 3, 5, 7, 9}
and F = {0, 10}. There are no elements found in both sets. So the intersection of D and F is the empty set. That is,
= .
ANSWER: 5. SOLUTION: means the union of D and B. Identify the elements that belong to D, B, or to both sets. E = {0, 2, 4, 6, 8, 10}
and B = {0, 1, 2, 3}. The elements found in E, B, or both sets are 0, 1, 2, 3, 4, 6, 8, and 10. So
= {0, 1, 2, 3, 4, 6, 8, 10}.
ANSWER: {0, 1, 2, 3, 4, 6, 8, 10}
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6. SOLUTION: Page 1
and F = {0, 10}. There are no elements found in both sets. So the intersection of D and F is the empty set. That is,
= .
ANSWER: Posttest
5. SOLUTION: means the union of D and B. Identify the elements that belong to D, B, or to both sets. E = {0, 2, 4, 6, 8, 10}
and B = {0, 1, 2, 3}. The elements found in E, B, or both sets are 0, 1, 2, 3, 4, 6, 8, and 10. So
= {0, 1, 2, 3, 4, 6, 8, 10}.
ANSWER: {0, 1, 2, 3, 4, 6, 8, 10}
6. SOLUTION: means the union of D and F. Identify the elements that belong to D, F, or to both sets. D = {1, 3, 5, 7, 9}
and F = {0, 10}. The elements found in D, F, or both sets are 0, 1, 3, 5, 7, 9, and 10. So
= {0, 1, 3, 5, 7, 9, 10}.
ANSWER: {0, 1, 3, 5, 7, 9, 10}
Simplify each expression.
7. (1 + 4i) + (–2 − 3i)
SOLUTION: ANSWER: −1 + i
8. (2 + 4i) – (–1 + 5i)
SOLUTION: ANSWER: 3–i
9. (6 + 7i)(−5 + 3i)
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ANSWER: –51 – 17i
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ANSWER: Posttest
3–i
9. (6 + 7i)(−5 + 3i)
SOLUTION: ANSWER: –51 – 17i
10. (–1 + i)(–6 + 2i)
SOLUTION: ANSWER: 4 – 8i
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12. Page 3
ANSWER: Posttest
12. SOLUTION: ANSWER: Determine whether each function has a maximum or minimum value. Then find the value of the
maximum or minimum, and state the domain and range of the function.
13. SOLUTION: 2
For the function f (x) = –x + 5x – 1, a = –1. Because a is negative, the graph opens down and the function has a
maximum value.
The maximum value of the function is the y-coordinate of the vertex. To find the vertex, first find equation of the
axis of symmetry.
Because the equation of the axis of symmetry is x =
, the x-coordinate of the vertex is
. Find the y-coordinate of
the vertex.
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ANSWER: Posttest
Determine whether each function has a maximum or minimum value. Then find the value of the
maximum or minimum, and state the domain and range of the function.
13. SOLUTION: 2
For the function f (x) = –x + 5x – 1, a = –1. Because a is negative, the graph opens down and the function has a
maximum value.
The maximum value of the function is the y-coordinate of the vertex. To find the vertex, first find equation of the
axis of symmetry.
Because the equation of the axis of symmetry is x =
, the x-coordinate of the vertex is
. Find the y-coordinate of
the vertex.
or Therefore, f (x) has a maximum value at
.
The domain of the function is all real numbers, so the domain is R. The range of the function is all real numbers less
than or equal to the maximum value
, so the range is y ≤
, for y
R.
ANSWER: maximum;
; domain: R; range: y ≤
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, for y
R
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ANSWER: maximum;
Posttest
; domain: R; range: y ≤
, for y
R
14. SOLUTION: 2
For the function f (x) = 3x + 4x + 1, a = 3. Because a is positive, the graph opens up and the function has a
minimum value.
The minimum value of the function is the y-coordinate of the vertex. To find the vertex, first find equation of the axis
of symmetry.
Because the equation of the axis of symmetry is x =
, the x-coordinate of the vertex is
. Find the y-
coordinate of the vertex.
Therefore, f (x) has a minimum value at
or .
The domain of the function is all real numbers, so the domain is R. The range of the function is all real numbers
greater than or equal to the minimum value
, so the range is y ≥
, for y
R.
ANSWER: minimum;
; domain: R; range: y ≥
, for y
R
Solve each equation.
15. x2 – x – 72 = 0
SOLUTION: Factor the quadratic expression on the left side of the equation and then apply the Zero Product Property.
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ANSWER: Posttest
minimum;
; domain: R; range: y ≥
, for y
R
Solve each equation.
15. x2 – x – 72 = 0
SOLUTION: Factor the quadratic expression on the left side of the equation and then apply the Zero Product Property.
ANSWER: −8, 9
16. x2 – 6x + 4 = 0
SOLUTION: Solve by completing the square.
ANSWER: 3±
17. 2x2 – 5x + 4 = 0
SOLUTION: 2
In the equation 2x – 5x + 4 = 0, a = 2, b = –5, and c = 4. Apply the Quadratic Formula.
ANSWER: 18. 2x2 – x – 3 = 0
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SOLUTION: Factor the quadratic expression on the left side of the equation and then apply the Zero Product Property.
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ANSWER: Posttest
18. 2x2 – x – 3 = 0
SOLUTION: Factor the quadratic expression on the left side of the equation and then apply the Zero Product Property.
ANSWER: –1,
19. RECREATION The current value C and the original value v of a recreational vehicle are related by C = v(1 – r)n,
where r is the rate of depreciation per year and n is the number of years. If the current value of a recreational
vehicle is $47,500, what would be the value of the vehicle after 75 months at an annual depreciation rate of 15%?
SOLUTION: From the information given, we know that v = 47,500 and r = 0.15. The time is given in months but the depreciation
formula requires that the time be in years. Since 30 months is equivalent to
or 6.25 years, n = 6.25. Evaluate the
given equation to find the value of the car V.
The value of the car would be $17,202.32.
ANSWER: $17,202.32
Simplify each expression.
20. SOLUTION: Because you are taking an even root of an even power and the result is an odd power, you must use the absolute
3
3
value of x and of y .
ANSWER: 21. eSolutions Manual - Powered by Cognero
SOLUTION: Page 8
value of x and of y .
ANSWER: Posttest
21. SOLUTION: Because the index is odd, it is not necessary to use absolute value.
ANSWER: 22. SOLUTION: Even though you are taking even roots of even powers, the results are even powers, so it is not necessary to use
absolute value.
ANSWER: 4 8
4t u
23. SOLUTION: Because the index is odd, it is not necessary to use absolute value.
ANSWER: Simplify each expression.
24. SOLUTION: eSolutions Manual - Powered by Cognero
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Because the index is odd, it is not necessary to use absolute value.
ANSWER: Posttest
Simplify each expression.
24. SOLUTION: ANSWER: y
25. SOLUTION: Because the index is odd, it is not necessary to use absolute value.
ANSWER: 26. SOLUTION: eSolutions Manual - Powered by Cognero
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Because the index is odd, it is not necessary to use absolute value.
ANSWER: Posttest
26. SOLUTION: Because the even root of the even power does not result in an odd power, therefore it is not necessary to use
absolute value.
ANSWER: 27. SOLUTION: We can simplify further by rationalizing the denominator.
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absolute value.
ANSWER: Posttest
27. SOLUTION: We can simplify further by rationalizing the denominator.
ANSWER: 28. JOBS Leah babysits during the day for $3 per hour and at night for $5 per hour. If she worked 5 hours and earned
$19, how many hours did she babysit during the day? How many at night?
SOLUTION: Use the information given to write a system of equations. Let x be the number of hours babysitting during the day
and y be the number of hours babysitting during at night. If Leah worked for a total of 5 hours, then x + y = 5. If she
makes $3 per hour during the day and $5 per at night and made a total of $19, then 8x + 5y = 19. Therefore, a
system of equations representing this situation is
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To solve this system, solve the first equation for y.
.
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ANSWER: Posttest
28. JOBS Leah babysits during the day for $3 per hour and at night for $5 per hour. If she worked 5 hours and earned
$19, how many hours did she babysit during the day? How many at night?
SOLUTION: Use the information given to write a system of equations. Let x be the number of hours babysitting during the day
and y be the number of hours babysitting during at night. If Leah worked for a total of 5 hours, then x + y = 5. If she
makes $3 per hour during the day and $5 per at night and made a total of $19, then 8x + 5y = 19. Therefore, a
system of equations representing this situation is
.
To solve this system, solve the first equation for y.
Then substitute this expression for y into the other equation and solve for x.
Substitute this value for x into the equation you solved for y to find the value of y.
The solution is (3, 2), which means that Leah worked 3 hours babysitting during the day and 2 hours babysitting at
night.
ANSWER: 3 day, 2 night
Solve each system of equations. State whether the system is consistent and independent, consistent and
dependent, or inconsistent.
29. SOLUTION: Eliminate the variable y in the system
by adding multiplying the second equation by –1.5 and then
adding the two equations together.
Because 0 = 0 is always true, there are an infinite number of solutions. Therefore, the system is consistent and
dependent.
ANSWER: infinitely many; consistent and dependent
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SOLUTION: Page 13
dependent.
ANSWER: Posttest
infinitely many; consistent and dependent
30. SOLUTION: To solve the system
by substitution, first solve one equation for x or y. In this case, x is easiest to solve
for in the second equation.
Then substitute this expression for x into the other equation and solve for y.
Substitute this value for y into the equation you solved for y to find the value of x.
The solution is (4, –18).
ANSWER: (4, −18); consistent and independent
31. SOLUTION: Eliminate one variable in two pairs of the system
.
Multiply the first equation by –
, and then add the two equations.
, multiply the second equation by
Because 0 = 0.5 is not a true statement, this system has no solutions. Therefore, the system is inconsistent.
ANSWER: no solution; inconsistent
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SOLUTION: Page 14
no solution; inconsistent
32. Posttest
SOLUTION: Eliminate one variable in two pairs of the system
.
Multiply the second equation by –2 and then add it to the second equation.
Multiply the third equation by 2 and then add it to the first equation.
Solve this system of two equations by multiplying the first by –5, multiplying the second by 3, and then adding the
two equations together.
Substitute these two values into one of the original equations to find z.
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The solution is
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. The system is consistent and independent because it has exactly one solution.
Posttest
Substitute these two values into one of the original equations to find z.
The solution is
. The system is consistent and independent because it has exactly one solution.
ANSWER: ; consistent and independent
Solve each system of inequalities. If the system has no solution, state no solution.
33. SOLUTION: Use a solid line to graph each related equation,
and
, since each inequality contains either ≥ or ≤.
Points on or above the line
make the inequality
true, so shade the region above the line
. Points on or to the right of the line
true, so shade the region to the right of
make the inequality the line
.
The solution of
is Regions 1 and 2.
The solution of
is Regions 2 and 3.
Region 2 contains points that are solutions to both inequalities, so this region is the solution to the system.
ANSWER: eSolutions Manual - Powered by Cognero
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ANSWER: Posttest
; consistent and independent
Solve each system of inequalities. If the system has no solution, state no solution.
33. SOLUTION: Use a solid line to graph each related equation,
and
, since each inequality contains either ≥ or ≤.
Points on or above the line
make the inequality
true, so shade the region above the line
. Points on or to the right of the line
true, so shade the region to the right of
make the inequality the line
.
The solution of
is Regions 1 and 2.
The solution of
is Regions 2 and 3.
Region 2 contains points that are solutions to both inequalities, so this region is the solution to the system.
ANSWER: 34. SOLUTION: Use a dashed line to graph each related equation,
and
, since each inequality contains either <
, first rewrite the equation in slope intercept form,
.
or >. To graph eSolutions Manual - Powered by Cognero
Page 17
Posttest
34. SOLUTION: Use a dashed line to graph each related equation,
and
, since each inequality contains either <
, first rewrite the equation in slope intercept form,
.
or >. To graph Points below the line
Points to the right of the line
line
.
make the inequality
true, so shade the region below the line
. true, so shade the region to the right of the
make the inequality The solution of
is Regions 1 and 2.
The solution of
is Regions 2 and 3.
Region 2 contains points that are solutions to both inequalities, so this region is the solution to the system.
ANSWER: 35. SOLUTION: Use a dashed line to graph each related equation,
or >. To graph and
and
, since each inequality contains either <
, first rewrite the equation in slope intercept form,
and
.
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Posttest
35. SOLUTION: Use a dashed line to graph each related equation,
or >. To graph and
and
, since each inequality contains either <
, first rewrite the equation in slope intercept form,
and
.
Points on or above the line
make the inequality
line
. Points on or to the right of the line region to the right of the line
.
true, so shade the region above the
true, so shade the
make the inequality The solution of
is Regions 1 and 2.
The solution of
is Regions 2 and 3.
Region 2 contains points that are solutions to both inequalities, so this region is the solution to the system.
ANSWER: 36. SOLUTION: eSolutions
Manual
Cognero
Graph
each- Powered
related by
equation,
and inequality contains ≤. Use a dashed line since for
. Use a solid line for
Page 19
since its related since its related inequality contains >. To graph
36. Posttest
SOLUTION: and Graph each related equation,
. Use a solid line for
inequality contains ≤. Use a dashed line since for
since its related inequality contains >. To graph
, first rewrite the equation in slope intercept form,
Points on or below the line
below
above
make the inequality
. Points above the line
since its related .
true, so shade the region make the inequality
true, so shade the region
.
The solution of
The solution of
is Region 2.
is Region 1.
Since these solutions do not contain points that are common to both inequalities, this system of inequalities has no
solution.
ANSWER: no solution
Find each of the following for A =
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37. A + B + C
,B=
, and C =
.
Page 20
Posttest
no solution
Find each of the following for A =
,B=
, and C =
.
37. A + B + C
SOLUTION: ANSWER: 38. B – C
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Posttest
38. B – C
SOLUTION: ANSWER: 39. 2A – B
SOLUTION: ANSWER: Find each permutation or combination.
10C3
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SOLUTION: Page 22
ANSWER: Posttest
Find each permutation or combination.
40. 10C3
SOLUTION: ANSWER: 120
41. 10P3
SOLUTION: ANSWER: 720
42. 6P6
SOLUTION: ANSWER: 720
43. 6C6
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SOLUTION: Page 23
ANSWER: Posttest
720
43. 6C6
SOLUTION: ANSWER: 1
44. 8P4
SOLUTION: ANSWER: 1680
45. 8C4
SOLUTION: ANSWER: eSolutions
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70
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ANSWER: Posttest
1680
45. 8C4
SOLUTION: ANSWER: 70
46. CARDS Four cards are randomly drawn from a standard deck of 52 cards. Find each probability.
a. P(1 ace and 3 kings)
b. P(2 odd and 2 face cards)
SOLUTION: a. There are 4 aces and 4 kings in a standard deck of cards. Since the order of the cards is not important, use
combinations and the Fundamental Counting Principle to find the number of ways to choose 1 out of 3 aces and 3
out of 4 kings.
Then use combinations to find the number of ways to choose 4 cards out of 52 from the deck.
To find the probability that the 4 cards chosen include 1 ace and 3 kings, divided the number of ways to choose 1
ace and 3 kings by the number of ways to choose 4 cards.
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The probability that the four cards chosen include 1 ace and 3 kings is
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or about 0.006%.
Posttest
To find the probability that the 4 cards chosen include 1 ace and 3 kings, divided the number of ways to choose 1
ace and 3 kings by the number of ways to choose 4 cards.
The probability that the four cards chosen include 1 ace and 3 kings is
or about 0.006%.
b. There are 5(4) or 20 odd cards and 3(4) or 12 face cards. Since the order of the cards is not important, use
combinations and the Fundamental Counting Principle to find the number of ways to choose 2 out of 20 odd cards
and 2 out of 12 face cards.
.
Then use combinations to find the number of ways to choose 4 cards out of 52 from the deck.
To find the probability that the 4 cards chosen include 2 odd and 2 face cards, divided the number of ways to choose
2 odd and 2 face cards by the number of ways to choose 4 cards.
The probability that the four cards chosen include 2 odd and 2 face cards is
or about 4.6%.
ANSWER: a.
b.
or about 0.006%
or about 4.6%
Find the mean, median, and mode for each set of data. Then find the range, variance, and standard
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deviation
eachbypopulation.
47. {1, 1, 1, 2, 2, 3}
Page 26
a.
b.
Posttest
or about 0.006%
or about 4.6%
Find the mean, median, and mode for each set of data. Then find the range, variance, and standard
deviation for each population.
47. {1, 1, 1, 2, 2, 3}
SOLUTION: Mean To find the mean of the set of data, divide the sum of the data by the number of pieces of data.
So the mean of the data is about 1.7
Median To find the median, order the data and find the middle number in the set of data or the average of the two
middle numbers.
The middle two numbers of this set are 1 and 2. The mean of these two numbers is 1.5. Therefore, the median of
the set of data is 1.5.
Mode To find the mode, determine which piece or pieces of data appear most often. Since 1 appears the most
often, 1 is the mode.
Range The range is the difference between the greatest and least data values, so the range of the data is 3 – 1 or
2.
Variance The population variance is calculated by taking the mean of the sum of the squares of the deviations from
the population mean.
The variance of the population data is about 0.6.
Standard Deviation The standard deviation of the population data is the square root of the variance.
The standard deviation of the population data is about 0.7.
ANSWER: mean:
1.7, median: 1.5, mode: 1, range: 2, variance:
0.6, standard deviation:
0.7
48. {0.8, 0.9, 0.4, 0.8, 0.6, 0.8, 0.6}
SOLUTION: Mean To find the mean of the set of data, divide the sum of the data by the number of pieces of data.
So the mean of the data is 0.7.
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Median To find the median, order the data and find the middle number in the set of data or the average of the two
middle numbers. When arranged in ascending order, the data are as follows.
0.4, 0.6, 0.6, 0.8, 0.8, 0.8, 0.9
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The standard deviation of the population data is about 0.7.
ANSWER: Posttest
mean:
1.7, median: 1.5, mode: 1, range: 2, variance:
0.6, standard deviation:
0.7
48. {0.8, 0.9, 0.4, 0.8, 0.6, 0.8, 0.6}
SOLUTION: Mean To find the mean of the set of data, divide the sum of the data by the number of pieces of data.
So the mean of the data is 0.7.
Median To find the median, order the data and find the middle number in the set of data or the average of the two
middle numbers. When arranged in ascending order, the data are as follows.
0.4, 0.6, 0.6, 0.8, 0.8, 0.8, 0.9
The middle number of this set is 0.8. Therefore, the median of the set of data is 0.8.
Mode To find the mode, determine which piece or pieces of data appear most often. Since 0.8 appears the most
often, 0.8 is the mode.
Range The range is the difference between the greatest and least data values, so the range of the data is 0.9 – 0.4
or 0.5.
Variance The population variance is calculated by taking the mean of the sum of the squares of the deviations from
the population mean.
The variance of the population data is about 0.26.
Standard Deviation The standard deviation of the population data is the square root of the variance.
The standard deviation of the population data is about 0.16.
ANSWER: mean: 0.7, median: 0.8, mode: 0.8, range: 0.5, variance:
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0.26, standard deviation:
0.16
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