Glencoe Algebra 2 Student Edition C2014
Glencoe Algebra 2 Student Edition C2014
1st Edition
ISBN: 9780076639908
Author: McGraw-Hill Glencoe
Publisher: MCG
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Chapter 0, Problem 29PR

a.

To determine

To calculate:Mean

a.

Expert Solution
Check Mark

Answer to Problem 29PR

The mean value of the data set is 84.0625

Explanation of Solution

Given information:

  85,100,92,36,74,88,92,86,88,82,98,70,78,92,84,100

Number of terms= 16

Used formula:

  Mean=sumoftermsNumberofterms

Calculation:

The mean of a data set is the sum of the terms divided by the total number of terms. Using math notation, we have:

  Mean=sumoftermsNumberofterms

Sum of term 85,100,92,36,74,88,92,86,88,82,98,70,78,92,84,100=1345

Number of terms= 16

  Mean=sumoftermsNumberofterms

  =134516=84.0625

b.

To determine

To calculate: Median

b.

Expert Solution
Check Mark

Answer to Problem 29PR

The median value of the data set is 87

Explanation of Solution

Given information:

  85,100,92,36,74,88,92,86,88,82,98,70,78,92,84,100

Number of terms= 16

Calculation:

The median is the middle number in a sorted list of numbers. So, to find the median, we need to place the numbers in value order and find the middle number.

Ordering the data from least to greatest, we get:

  36 ,70 ,74 ,78 ,82 ,84 ,85 ,86 ,88 ,88 ,92 ,92 ,92 ,98 ,100 ,100

As you can see, we do not have just one middle number but we have a pair of middle numbers, so the median is the average of these two numbers:

The median =86+882=87

c.

To determine

To calculate: Mode

c.

Expert Solution
Check Mark

Answer to Problem 29PR

The mode value of the data set is 92

Explanation of Solution

Given information:

  85,100,92,36,74,88,92,86,88,82,98,70,78,92,84,100

Number of terms= 16

Calculation:

The mode of a set of data is the value in the set that occurs most often.

Ordering the data from least to greatest, we get:

  36 ,70 ,74 ,78 ,82 ,84 ,85 ,86 ,88 ,88 ,92 ,92 ,92 ,98 ,100 ,100

We see that the mode is 92.

d.

To determine

To calculate: Range

d.

Expert Solution
Check Mark

Answer to Problem 29PR

The range value of the data set is 64

Explanation of Solution

Given information:

  85,100,92,36,74,88,92,86,88,82,98,70,78,92,84,100

Number of terms= 16

Calculation:

(iv) The range is the difference between the highest and lowest values in the data set.

Ordering the data from least to greatest, we get:

  36 ,70 ,74 ,78 ,82 ,84 ,85 ,86 ,88 ,88 ,92 ,92 ,92 ,98 ,100 ,100

The lowest value is 36.

The highest value is 100.

The range =10036=64.

e.

To determine

To calculate: Standard deviation

e.

Expert Solution
Check Mark

Answer to Problem 29PR

The standard deviation value of the data set is s: 14.951875927455

Explanation of Solution

Given information:

  85,100,92,36,74,88,92,86,88,82,98,70,78,92,84,100

Number of terms= 16

Used formula:

  σ=1Ni=1n(Xiμ)2

Calculation:

  Count, N:16Sum, Σ x:1345Mean, μ:84.0625Variance, σ2: 223.55859375

  σ=1Ni=1n(Xiμ)2σ2=(xiμ)2Nσ2=(85  84.0625)2+ ... + (100  84.0625)216σ2=3576.937516σ2=223.55859375σ = 223.55859375σ= 14.951875927455

f.

To determine

To calculate: Outliers

f.

Expert Solution
Check Mark

Answer to Problem 29PR

There outlier is 36.

Explanation of Solution

Given information:

  85,100,92,36,74,88,92,86,88,82,98,70,78,92,84,100

Number of terms= 16

Calculation:

There have outliers 36 because 36 have common number in the given question.

Chapter 0 Solutions

Glencoe Algebra 2 Student Edition C2014

Ch. 0.1 - Prob. 11ECh. 0.1 - Prob. 12ECh. 0.2 - Prob. 1ECh. 0.2 - Prob. 2ECh. 0.2 - Prob. 3ECh. 0.2 - Prob. 4ECh. 0.2 - Prob. 5ECh. 0.2 - Prob. 6ECh. 0.2 - Prob. 7ECh. 0.2 - Prob. 8ECh. 0.2 - Prob. 9ECh. 0.2 - Prob. 10ECh. 0.2 - Prob. 11ECh. 0.2 - Prob. 12ECh. 0.2 - Prob. 13ECh. 0.2 - Prob. 14ECh. 0.2 - Prob. 15ECh. 0.2 - Prob. 16ECh. 0.2 - Prob. 17ECh. 0.2 - Prob. 18ECh. 0.3 - Prob. 1ECh. 0.3 - Prob. 2ECh. 0.3 - Prob. 3ECh. 0.3 - Prob. 4ECh. 0.3 - Prob. 5ECh. 0.3 - Prob. 6ECh. 0.3 - Prob. 7ECh. 0.3 - Prob. 8ECh. 0.3 - Prob. 9ECh. 0.3 - Prob. 10ECh. 0.3 - Prob. 11ECh. 0.3 - Prob. 12ECh. 0.3 - Prob. 13ECh. 0.3 - Prob. 14ECh. 0.3 - Prob. 15ECh. 0.3 - Prob. 16ECh. 0.3 - Prob. 17ECh. 0.3 - Prob. 18ECh. 0.3 - Prob. 19ECh. 0.3 - Prob. 20ECh. 0.3 - Prob. 21ECh. 0.3 - Prob. 22ECh. 0.3 - Prob. 23ECh. 0.3 - Prob. 24ECh. 0.4 - Prob. 1ECh. 0.4 - Prob. 2ECh. 0.4 - Prob. 3ECh. 0.4 - Prob. 4ECh. 0.4 - Prob. 5ECh. 0.4 - Prob. 6ECh. 0.4 - Prob. 7ECh. 0.4 - Prob. 8ECh. 0.4 - Prob. 9ECh. 0.4 - Prob. 10ECh. 0.4 - Prob. 11ECh. 0.4 - Prob. 12ECh. 0.4 - Prob. 13ECh. 0.4 - Prob. 14ECh. 0.4 - Prob. 15ECh. 0.4 - Prob. 16ECh. 0.4 - Prob. 17ECh. 0.4 - Prob. 18ECh. 0.4 - Prob. 19ECh. 0.5 - Prob. 1ECh. 0.5 - Prob. 2ECh. 0.5 - Prob. 3ECh. 0.5 - Prob. 4ECh. 0.5 - Prob. 5ECh. 0.5 - Prob. 6ECh. 0.5 - Prob. 7ECh. 0.5 - Prob. 8ECh. 0.5 - Prob. 9ECh. 0.5 - Prob. 10ECh. 0.6 - Prob. 1ECh. 0.6 - Prob. 2ECh. 0.6 - Prob. 3ECh. 0.6 - Prob. 4ECh. 0.6 - Prob. 5ECh. 0.6 - Prob. 6ECh. 0.6 - Prob. 7ECh. 0.6 - Prob. 8ECh. 0.6 - Prob. 9ECh. 0.6 - Prob. 10ECh. 0.6 - Prob. 11ECh. 0.6 - Prob. 12ECh. 0.6 - Prob. 13ECh. 0.6 - Prob. 14ECh. 0.6 - Prob. 15ECh. 0.6 - Prob. 16ECh. 0.6 - Prob. 17ECh. 0.7 - Prob. 1ECh. 0.7 - Prob. 2ECh. 0.7 - Prob. 3ECh. 0.7 - Prob. 4ECh. 0.7 - Prob. 5ECh. 0.7 - Prob. 6ECh. 0.7 - Prob. 7ECh. 0.7 - Prob. 8ECh. 0.7 - Prob. 9ECh. 0.7 - Prob. 10ECh. 0.7 - Prob. 11ECh. 0.7 - Prob. 12ECh. 0.8 - Prob. 1ECh. 0.8 - Prob. 2ECh. 0.8 - Prob. 3ECh. 0.8 - Prob. 4ECh. 0.8 - Prob. 5ECh. 0.8 - Prob. 6ECh. 0.8 - Prob. 7ECh. 0.8 - Prob. 8ECh. 0.8 - Prob. 9ECh. 0.8 - Prob. 10ECh. 0.8 - Prob. 11ECh. 0.8 - Prob. 12ECh. 0.8 - Prob. 13ECh. 0.8 - Prob. 14ECh. 0.8 - Prob. 15ECh. 0.8 - Prob. 16ECh. 0.8 - Prob. 17ECh. 0.8 - Prob. 18ECh. 0.9 - Prob. 1ECh. 0.9 - Prob. 2ECh. 0.9 - Prob. 3ECh. 0.9 - Prob. 4ECh. 0.9 - Prob. 5ECh. 0.9 - Prob. 6ECh. 0.9 - Prob. 7ECh. 0.9 - Prob. 8ECh. 0.9 - Prob. 9ECh. 0.9 - Prob. 10ECh. 0.9 - Prob. 11ECh. 0 - Prob. 1PRCh. 0 - Prob. 2PRCh. 0 - Prob. 3PRCh. 0 - Prob. 4PRCh. 0 - Prob. 5PRCh. 0 - Prob. 6PRCh. 0 - Prob. 7PRCh. 0 - Prob. 8PRCh. 0 - Prob. 9PRCh. 0 - Prob. 10PRCh. 0 - Prob. 11PRCh. 0 - Prob. 12PRCh. 0 - Prob. 13PRCh. 0 - Prob. 14PRCh. 0 - Prob. 15PRCh. 0 - Prob. 16PRCh. 0 - Prob. 17PRCh. 0 - Prob. 18PRCh. 0 - Prob. 19PRCh. 0 - Prob. 20PRCh. 0 - Prob. 21PRCh. 0 - Prob. 22PRCh. 0 - Prob. 23PRCh. 0 - Prob. 24PRCh. 0 - Prob. 25PRCh. 0 - Prob. 26PRCh. 0 - Prob. 27PRCh. 0 - Prob. 28PRCh. 0 - Prob. 29PRCh. 0 - Prob. 30PRCh. 0 - Prob. 1POCh. 0 - Prob. 2POCh. 0 - Prob. 3POCh. 0 - Prob. 4POCh. 0 - Prob. 5POCh. 0 - Prob. 6POCh. 0 - Prob. 7POCh. 0 - Prob. 8POCh. 0 - Prob. 9POCh. 0 - Prob. 10POCh. 0 - Prob. 11POCh. 0 - Prob. 12POCh. 0 - Prob. 13POCh. 0 - Prob. 14POCh. 0 - Prob. 15POCh. 0 - Prob. 16POCh. 0 - Prob. 17POCh. 0 - Prob. 18POCh. 0 - Prob. 19POCh. 0 - Prob. 20POCh. 0 - Prob. 21POCh. 0 - Prob. 22POCh. 0 - Prob. 23POCh. 0 - Prob. 24POCh. 0 - Prob. 25POCh. 0 - Prob. 26POCh. 0 - Prob. 27POCh. 0 - Prob. 28POCh. 0 - Prob. 29POCh. 0 - Prob. 30PO
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