Download Chapter 0 -0 (Post-Test) 29-48 Alll - MOC-FV

Posttest
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.
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).
31. SOLUTION: Eliminate one variable in two pairs of the system
.
Multiply the first equation by –
, and then add the two equations.
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, multiply the second equation by
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Substitute this value for y into the equation you solved for y to find the value of x.
Posttest
The solution is (4, –18).
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.
32. 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.
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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.
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
.
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Posttest
The solution is
. The system is consistent and independent because it has exactly one solution.
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.
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
.
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make the inequality
true, so shade the region below the line
. true, so shade the region to the right of the
make the inequality Page 4
The solution of
is Regions 1 and 2.
The solution of
is Regions 2 and 3.
Posttest
Region 2 contains points that are solutions to both inequalities, so this region is the solution to the system.
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.
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
.
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true, so shade the region above the
true, so shade the
make the inequality Page 5
The solution of
is Regions 1 and 2.
The solution of
is Regions 2 and 3.
Posttest
Region 2 contains points that are solutions to both inequalities, so this region is the solution to the system.
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.
36. SOLUTION: Graph each related equation,
and inequality contains ≤. Use a dashed line since for
. Use a solid line for
since its related inequality contains >. To graph
, first rewrite the equation in slope intercept form,
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since its related .
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The solution of
is Regions 1 and 2.
The solution of
is Regions 2 and 3.
Posttest
Region 2 contains points that are solutions to both inequalities, so this region is the solution to the system.
36. 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.
Find each of the following for A =
,B=
, and C =
.
37. A + B + C
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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
Posttest
solution.
Find each of the following for A =
,B=
, and C =
.
37. A + B + C
SOLUTION: 38. B – C
SOLUTION: 39. 2A – B
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Posttest
39. 2A – B
SOLUTION: Find each permutation or combination.
40. 10C3
SOLUTION: 41. 10P3
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Posttest
41. 10P3
SOLUTION: 42. 6P6
SOLUTION: 43. 6C6
SOLUTION: 44. 8P4
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45. 8C4
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Posttest
44. 8P4
SOLUTION: 45. 8C4
SOLUTION: 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.
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Posttest
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.
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.
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The probability that the four cards chosen include 2 odd and 2 face cards is
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or about 4.6%.
Posttest
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%.
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.
48. {0.8,Manual
0.9, 0.4,
0.8, 0.6,
0.8, 0.6}
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SOLUTION: Mean To find the mean of the set of data, divide the sum of the data by the number of pieces of data.
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Posttest
The standard deviation of the population data is about 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.
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