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Algebra II : Mean

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #61 : Mean

Solve for the mean: [0,0,10,-20,30,1,4,9]

Possible Answers:

$\frac{17}{8}$

$\frac{11}{2}$

$\frac{17}{2}$

$\frac{17}{4}$

$\frac{17}{3}$

Correct answer:

$\frac{17}{4}$

Explanation:

The mean is the average of all the numbers in the data set.

Add all the numbers.

0+0+10+(-20)+30+1+4+9 =34

Divide this number by the total amount of numbers in the data set.

$\frac{34}{8}$

Reduce this fraction.

The answer is: $\frac{17}{4}$

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Example Question #61 : Mean

Determine the mean of the set of data: $[1,\frac{1}{3}, -\frac{1}{4}, 3]$

Possible Answers:

$\frac{49}{12}$

$\frac{55}{4}$

$\frac{55}{12}$

$\frac{55}{48}$

$\frac{49}{48}$

Correct answer:

$\frac{49}{48}$

Explanation:

Add all the numbers in the data set.

$1+\frac{1}{3}+ (-\frac{1}{4})+ 3$

Convert all numbers and fractions to a least common denominator.

The least common denominator is 12 because this is the least number divisible by every denominator.

$1+\frac{1}{3}-\frac{1}{4}+ 3 =\frac{12}{12} +\frac{4}{12}-\frac{3}{12}+\frac{36}{12}$

Since all the denominators are common, we can add and subtract the numerators.

$\frac{49}{12}$

Divide this number by four, or multiply by one-fourth, since there are four numbers in the data set.

$\frac{49}{12}\times \frac{1}{4} =\frac{49}{48}$

The mean is: $\frac{49}{48}$

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Example Question #62 : Mean

Sarah is to compete in 4 areas at a gymnastics competition . On high bar, floor, and on vault, she scored a 9.4, 8.9, and a 9.8. The maximum score is 10. If placing top three in the competition requires her to score at least a 9.5 overall, what must she score on balance beam?

Possible Answers:

9.85

9.8

$\textup{Sarah will need to score higher than 10 to earn awards.}$

9.9

9.95

Correct answer:

9.9

Explanation:

The average is already known. What is not known is Sarah's balance beam score.

Set up an equation such that the quantity of the four scores including the unknown score divided by four is equal to 9.5. Let the score of the balance beam routine be .

$\frac{9.4+8.9+9.8+x}{4}=9.5$

Multiply both sides by four to eliminate the denominator and simplify the numerator.

$(\frac{9.4+8.9+9.8+x}{4})\times 4=9.5\times 4$

28.1+x=38

Subtract 28.1 from both sides to determine the unknown variable.

x=9.9

Sarah's score must be a 9.9 or higher to earn awards.

The answer is: 9.9

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Example Question #63 : Mean

Bob scores a 5, 6, and 7 on three of four tests. The highest possible score is a ten. What must he score on his next test to average an overall score of 8?

Possible Answers:

9.9

9.5

$\textup{Bob cannot average an overall score of 8.}$

Correct answer:

$\textup{Bob cannot average an overall score of 8.}$

Explanation:

Set up an equation such that the average of Bob's three grades and his unknown fourth test grade is equal to eight.

$\frac{5+6+7+x}{4}= 8$

Multiply both sides of the equation by four.

$\frac{5+6+7+x}{4}\times 4= 8\times 4$

Simplify both sides.

5+6+7+x = 32

18+x=32

Subtract 18 from both sides.

18+x-18=32-18

x=14

Bob is unable to average an 8 overall score since he would have to make a 14 on his final test grade, which is impossible since 10 is the highest score he can earn.

The answer is: $\textup{Bob cannot average an 8 overall score.}$

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Example Question #252 : Algebra Ii

Determine the mean of the numbers: [2,9,32,88,10,-5,7]

Possible Answers:

$\frac{143}{7}$

$\frac{148}{7}$

Correct answer:

$\frac{143}{7}$

Explanation:

In order to determine the mean of the set of numbers, sum the numbers provided and divide the sum by the total amount of numbers.

2+9+32+88+10+(-5)+7=143

Divide this number by seven.

The mean is: $\frac{143}{7}$

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Example Question #161 : Data Properties

If Bob scores a 60, 80, and 90 on three of five tests in his class, what is the minimum average of Bob's fourth and fifth test scores in order for Bob to earn an overall score of 80 in his class?

Possible Answers:

$\textup{Bob cannot average an 80 in his class.}$

Correct answer:

Explanation:

The test scores of Bob's fourth and fifth tests are unknown. Let Bob's fourth test score be and fifth test score be . Average all test scores.

$\frac{60+80+90+x+y}{5} = 80$

Simplify the equation. Multiply by five on both sides.

$\frac{60+80+90+x+y}{5} \cdot 5= 80\cdot 5$

230+x+y = 400

Subtract 230 from both sides.

x+y=170

The sum of Bob's fourth and fifth test must add up to this value.

To average the test scores, simply divide both sides by two. This will be the minimum average of both test four and five.

$\frac{x+y}{2}=\frac{170}{2} = 85$

The answer is:

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Example Question #65 : Mean

Determine the mean of the following numbers: [-5,9,18,30,-6]

Possible Answers:

$\textup{There is no mean.}$

$\frac{46}{5}$

$\frac{68}{5}$

Correct answer:

$\frac{46}{5}$

Explanation:

In order to determine the mean of the numbers, sum the numbers and divide the sum by the total amount of numbers provided in the data set.

(-5)+9+18+30+(-6) = 46

Divide this number by five since there are five numbers.

The mean is: $\frac{46}{5}$

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Example Question #66 : Mean

Determine the mean of the numbers: [9,11,15,28,34]

Possible Answers:

$\frac{59}{2}$

$\frac{97}{5}$

$\textup{There is no mean.}$

Correct answer:

$\frac{97}{5}$

Explanation:

To determine the average of the numbers in the data set, we will need to add all the numbers in the data set and divide the sum by the total numbers in the data set.

$\frac{9+11+15+28+34}{5}$

Simplify this fraction by adding the numerator.

The answer is: $\frac{97}{5}$

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Example Question #191 : Basic Statistics

Determine the mean of the following data set: [-1,-9,-19,39]

Possible Answers:

$\frac{5}{2}$

$\textup{There is no mean.}$

Correct answer:

$\frac{5}{2}$

Explanation:

The mean is the average of all the numbers provided in the data set.

Add all the numbers.

(-1)+(-9)+(-19)+39 = 10

Divide this value by four since there are four numbers given.

$\frac{10}{4} = \frac{5}{2}$

The mean is: $\frac{5}{2}$

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Example Question #191 : Basic Statistics

Billy scores 3, 5, and 7 on three of his four tests in his class. The maximum score possible for each test is ten. What must his fourth test be in order to average at least a 7 in his class?

Possible Answers:

9.5

$\textup{Billy cannot average a 7 in his class.}$

Correct answer:

$\textup{Billy cannot average a 7 in his class.}$

Explanation:

Let Billy's unknown score be . Add the known scores including the unknown score and divide by four. Set this expression equal to seven.

$\frac{3+5+7+x}{4}=7$

Multiply by four on both sides. The equation becomes:

3+5+7+x = 28

Add the terms on the left.

15+x=28

Subtract 15 on both sides.

15+x-15=28-15

x=13

Since this score exceeds the maximum score possible, Billy cannot achieve the average score of seven.

The correct answer is: $\textup{Billy cannot average a 7 in his class.}$

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Algebra II : Mean

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #61 : Mean

Example Question #61 : Mean

Example Question #62 : Mean

Example Question #63 : Mean

Example Question #252 : Algebra Ii

Example Question #161 : Data Properties

Example Question #65 : Mean

Example Question #66 : Mean

Example Question #191 : Basic Statistics

Example Question #191 : Basic Statistics

All Algebra II Resources

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