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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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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Example Question #71 : Mean

Determine the mean of the numbers: [-3,-18,-3,99]

Possible Answers:

$\frac{75}{4}$

$\frac{123}{4}$

$\textup{There is no mean.}$

Correct answer:

$\frac{75}{4}$

Explanation:

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

Add all the numbers provided and divide the sum by a total of four, since there are four numbers given.

$\frac{(-3)+(-18)+(-3)+99}{4} = \frac{75}{4}$

This fraction is irreducible.

The answer is: $\frac{75}{4}$

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Example Question #261 : Algebra Ii

Determine the mean: [ -9,2,9,20,37]

Possible Answers:

$\frac{59}{5}$

$\frac{77}{5}$

$\textup{The mean is not given.}$

Correct answer:

$\frac{59}{5}$

Explanation:

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

Add all the numbers.

-9+2+9+20+37 = 59

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

The answer is: $\frac{59}{5}$

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Example Question #73 : Mean

Determine the mean of the numbers: $[\frac{1}{3},3\%, \frac{1}{30}]$

Possible Answers:

$\frac{119}{900}$

$\frac{119}{300}$

$\frac{19}{90}$

$\frac{19}{30}$

$\frac{1}{30}$

Correct answer:

$\frac{119}{900}$

Explanation:

The mean is the average of all the numbers. Convert the percentage to a fraction by removing the percentage and divide the number by 100.

$[\frac{1}{3},3\%, \frac{1}{30}]\rightarrow[\frac{1}{3}, \frac{3}{100},\frac{1}{30}]$

Convert all fractions to a common denominator. Write out the factors for the denominators.

3-[3,6,9,...,294,297,300]

30-[30,60,90,120,150,180,210,240,270,300]

100-[100,200,300]

The common denominator is 300.

$\frac{1}{3}+ \frac{3}{100}+\frac{1}{30}= \frac{1(100)}{3(100)}+ \frac{3(3)}{100(3)}+\frac{1(10)}{30(10)}$

Simplify the fractions.

$\frac{100}{300}+\frac{9}{300}+\frac{10}{300} = \frac{119}{300}$

There are three numbers in the data set. Divide the fraction by three.

$\frac{ \frac{119}{300}}{3}= \frac{119}{300} \div 3 = \frac{119}{300}\times \frac{1}{3} = \frac{119}{900}$

The answer is: $\frac{119}{900}$

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Example Question #262 : Algebra Ii

Determine the mean of the following numbers: [2,8,16,32,56]

Possible Answers:

$\frac{104}{5}$

$\frac{114}{5}$

Correct answer:

$\frac{114}{5}$

Explanation:

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

Add the numbers and divide by the total amount of numbers.

$\frac{2+8+16+32+56}{5}= \frac{114}{5}$

The answer is: $\frac{114}{5}$

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Example Question #192 : Basic Statistics

Determine the mean of the numbers: [2,9,1,-3,14,26]

Possible Answers:

$\frac{49}{6}$

$\frac{49}{5}$

$\frac{46}{5}$

$\frac{17}{3}$

$\frac{11}{2}$

Correct answer:

$\frac{49}{6}$

Explanation:

The mean is the sum of all the numbers in the data set over the total amount of numbers given.

Add the numbers.

2+9+1+(-3)+14+26 = 49

Divide this number by six since there are 6 numbers.

The answer is: $\frac{49}{6}$

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Example Question #194 : Basic Statistics

The top student of the class makes 95 and 97 on his first two tests. However, he was under pressure and scored a 25 on his third test. There are 2 more tests remaining, and the highest possible score he can make on each test is 100. Can he maintain at least a 90 in his class? If so, what is the mandatory average of his fourth and fifth tests?

Possible Answers:

$\textup{Yes, 95}$

$\textup{Yes, 98.5}$

$\textup{Yes, 100}$

$\textup{Yes, 99.5}$

$\textup{No, it is not possible to maintain the score.}$

Correct answer:

$\textup{No, it is not possible to maintain the score.}$

Explanation:

There are five test scores. Assume and represents the student's fourth and fifth tests.

Set up an equation such that all five scores will average a 90.

$\frac{95+97+25+x+y}{5} = 90$

Multiply by five on both sides.

The equation becomes:

Isolate the x and y-variables by subtracting all the integers on both sides.

95+97+25+x+y-(95+97+25)= 450-(95+97+25)

x+y=233

Divide both sides by two to determine the average score he will need for both tests.

$\frac{x+y}{2}=\frac{233}{2} =116.5$

The student cannot maintain an average score of 90 since he will need to score a grade over 100 on both tests.

The answer is: $\textup{No, the student will not maintain the score.}$

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #65 : Mean

Example Question #66 : Mean

Example Question #191 : Basic Statistics

Example Question #191 : Basic Statistics

Example Question #71 : Mean

Example Question #261 : Algebra Ii

Example Question #73 : Mean

Example Question #262 : Algebra Ii

Example Question #192 : Basic Statistics

Example Question #194 : Basic Statistics

All Algebra II Resources

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