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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Question
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Chapter 0.5, Problem 1E

(a)

To determine

To add a column showing the experimental probability of landing on each given color with the next spin.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given Information:

The following is the table showing the results of various spins

    ColorFrequency
    RedGlencoe Algebra 2 Student Edition C2014, Chapter 0.5, Problem 1E , additional homework tip  1
    BlueGlencoe Algebra 2 Student Edition C2014, Chapter 0.5, Problem 1E , additional homework tip  2
    Yellow9
    Orange12
    Purple5
    Green11
    Total 50

The total number of spins is 50 .

The probability for any event A is given by P(A)=favorablecasestotalcases

For the red color favorable cases are 6 and the total cases are 50 , therefore, the probability is given by P(A)=650 which is equal to 0.12 .

Make the table again by calculating the probabilities as shown.

    ColorFrequencyProbability
    RedGlencoe Algebra 2 Student Edition C2014, Chapter 0.5, Problem 1E , additional homework tip  3650=0.12
    BlueGlencoe Algebra 2 Student Edition C2014, Chapter 0.5, Problem 1E , additional homework tip  4750=0.14
    Yellow9950=0.18
    Orange121250=0.24
    Purple5550=0.10
    Green111150=0.22
    Total 50

The table shows the probabilities for different color spinners.

(b)

To determine

Draw the bar diagram showing the experimental probabilities.

(b)

Expert Solution
Check Mark

Explanation of Solution

GRAPH:

The following bar graph shows the experimental probabilities.

Take color on x - axis and probability on y - axis and draw the bar graph as shown below.

  Glencoe Algebra 2 Student Edition C2014, Chapter 0.5, Problem 1E , additional homework tip  5

Bar graph for the experimental probabilities.

(c)

To determine

Draw a table by adding a column that showing the theoretical probability of the spinner.

(c)

Expert Solution
Check Mark

Explanation of Solution

There are six colors and all colors have equal probabilities.

Since, the total probability is 1 and there are 6 colors, Therefore, probability of each color will be 0.16 .

The following table shows the theoretical probabilities in the table.

    ColorFrequencyProbability
    RedGlencoe Algebra 2 Student Edition C2014, Chapter 0.5, Problem 1E , additional homework tip  60.16
    BlueGlencoe Algebra 2 Student Edition C2014, Chapter 0.5, Problem 1E , additional homework tip  70.16
    Yellow90.16
    Orange120.16
    Purple50.16
    Green110.16
    Total 50

The table showing the theoretical probabilities.

(d)

To determine

Create a bar diagram showing the theoretical probabilities.

(d)

Expert Solution
Check Mark

Explanation of Solution

GRAPH:

To plot the bar graph, proceed as follows.

Take color on x - axis and probability on y - axis and draw the bar graph as shown below.

  Glencoe Algebra 2 Student Edition C2014, Chapter 0.5, Problem 1E , additional homework tip  8

Bar graph for the theoretical probabilities.

(e)

To determine

Interpret the graph by comparing the graphs created in above parts (b) and (d)

(e)

Expert Solution
Check Mark

Explanation of Solution

In the first bar graph the experimental probabilities are given which is maximum for orange. Therefore, the probability of spinner landing on orange is more than green. So according the bar graph higher the length of bar, higher the probability of landing on that color.

But in the second bar graph all the bars have equal length implying that probability of landing on all the colors is same. All the colors have equal probability.

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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