Algebra 1 Common Core Answers Chapter 12 Data Analysis and Probability Exercise 12.6

Algebra 1 Common Core Answers Student Edition Grade 8 – 9 Chapter 12 Data Analysis and Probability Exercise 12.6

Algebra 1 Common Core Solutions

Chapter 12 Data Analysis and Probability Exercise 12.6 1LC
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Chapter 12 Data Analysis and Probability Exercise 12.6 2LC
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Chapter 12 Data Analysis and Probability Exercise 12.6 3LC
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Chapter 12 Data Analysis and Probability Exercise 12.6 4LC
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Chapter 12 Data Analysis and Probability Exercise 12.6 5LC
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Chapter 12 Data Analysis and Probability Exercise 12.6 6LC
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Chapter 12 Data Analysis and Probability Exercise 12.6 7LC
The objective is to find the number of outfit that you can make with 6 shirt and 4 pairs of pants. If m and n are two positive integers, and there are m ways to make first selection and n ways to make second selection, then there are in•n ways to make the two selections Put the value 6 for m. and 4 for n.
64=24
Thus, there are 24 outfit you can make with 6 shirt and 4 pairs of pants

Chapter 12 Data Analysis and Probability Exercise 12.6 8LC
The objective is to determine whether permutations or combinations are used to find the possible arrangements 1o students in a line.
Explanation:
Definition of Permutation and Combination:
A permutation is the arrangement of objects in a specific order and a combination is the selection of objects in a specific order By the above definition. permutation are used to find the number of possible arrangement of 10 students in a line.

Chapter 12 Data Analysis and Probability Exercise 12.6 9LC
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From the definition of permutation and combination it is clear that. permutation and combination are different In permutation. the order of object is matter while arranging them. but in the combination or selection of object order is not required Therefore. permutation of used to count in situation where order is important. but in combination is used to count in situation where order is not important.

Chapter 12 Data Analysis and Probability Exercise 12.6 10LC
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Chapter 12 Data Analysis and Probability Exercise 12.6 11E
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Chapter 12 Data Analysis and Probability Exercise 12.6 12E
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(a)
To find the number of routes from A to C, use Multiplication counting Principle.
Now, the diagram indicated that there are two routes from A to B. and there are three routes
from B to C
Multiplication Counting Principle: If there are m ways to make first selection and n ways to make second selection, then there are ;n,, ways to make the selections
So. by using the Multiplication counting principle, put the value 2 for m and 3 for, then get the number of routes from A to C. The number of routes from A to C is given below:
2.3= 6
Therefore, the number of routes from A to C is 6 routes
To find the number of from A to D. use Multiplication counting Principle Now, the diagram indicated that there are two routes from A to B. there are three routes from B to C. and there are two routes from C to D.
Multiplication Counting Principle: If there are m ways to make first selection, n ways to make second selection, and k way to make third selection then there are m.n.k ways to make the selections. So, by using the Multiplication counting principle, put the value 2 for m. 3 for n, and 2 for k, then get the number of routs from A to D. The number of routes from A to D is given below:
2.3.2=12
Therefore, the number of routes from A to D is 12 routes.

Chapter 12 Data Analysis and Probability Exercise 12.6 13E
Let the number of skaters are participated in the final of ice-skating competition be .
To find the different orders are possible for the final program, use the concept of permutations.
Since there are 10 skaters in the final, so, there are 10 choices for the first skaters, 9 choices for the second, 8 choice for the third skaters, 7 choice for the fourth skaters, 6 choice for the fifth skaters, 5 choices for the sixth skaters, 4 choices for the seventh skaters, 3 choices for the eighth skaters, 2 choice for the ninth skaters, and one choice for the tenth skaters.
Multiply the number of ways to make the total number of different possible orders.
So, the different orders are possible for the final program can be calculated as follows:
10.9.8.7.6.5.4.3.2.1 = 3,628,800
Therefore, the total number of different possible orders is 3,628,800.

Chapter 12 Data Analysis and Probability Exercise 12.6 14E
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Chapter 12 Data Analysis and Probability Exercise 12.6 15E
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Chapter 12 Data Analysis and Probability Exercise 12.6 16E
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Chapter 12 Data Analysis and Probability Exercise 12.6 17E
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Chapter 12 Data Analysis and Probability Exercise 12.6 18E
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Chapter 12 Data Analysis and Probability Exercise 12.6 19E
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Chapter 12 Data Analysis and Probability Exercise 12.6 20E
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Chapter 12 Data Analysis and Probability Exercise 12.6 21E
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Chapter 12 Data Analysis and Probability Exercise 12.6 22E
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Chapter 12 Data Analysis and Probability Exercise 12.6 23E
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Chapter 12 Data Analysis and Probability Exercise 12.6 24E
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Chapter 12 Data Analysis and Probability Exercise 12.6 25E
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Chapter 12 Data Analysis and Probability Exercise 12.6 26E
Let the number of members in a student council is 24 members
Let the number of members selected for a committee be 3 students
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Chapter 12 Data Analysis and Probability Exercise 12.6 27E
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Chapter 12 Data Analysis and Probability Exercise 12.6 28E
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Chapter 12 Data Analysis and Probability Exercise 12.6 29E
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Chapter 12 Data Analysis and Probability Exercise 12.6 30E
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Chapter 12 Data Analysis and Probability Exercise 12.6 31E
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Chapter 12 Data Analysis and Probability Exercise 12.6 32E
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Chapter 12 Data Analysis and Probability Exercise 12.6 33E
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Chapter 12 Data Analysis and Probability Exercise 12.6 34E
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Chapter 12 Data Analysis and Probability Exercise 12.6 35E
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Chapter 12 Data Analysis and Probability Exercise 12.6 36E
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Chapter 12 Data Analysis and Probability Exercise 12.6 37E
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Chapter 12 Data Analysis and Probability Exercise 12.6 38E
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Chapter 12 Data Analysis and Probability Exercise 12.6 39E
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Chapter 12 Data Analysis and Probability Exercise 12.6 40E
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Chapter 12 Data Analysis and Probability Exercise 12.6 41E
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Chapter 12 Data Analysis and Probability Exercise 12.6 42E
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Chapter 12 Data Analysis and Probability Exercise 12.6 43E
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Chapter 12 Data Analysis and Probability Exercise 12.6 44E
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Chapter 12 Data Analysis and Probability Exercise 12.6 45E
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Chapter 12 Data Analysis and Probability Exercise 12.6 46E
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Chapter 12 Data Analysis and Probability Exercise 12.6 47E
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Creating a password. to use the four letters from name is not a good idea. Because, only 24 permutations are there, so it is not difficult to make 24 arrangements to Lena name, if any bode try it can find the password.

Chapter 12 Data Analysis and Probability Exercise 12.6 48E
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Chapter 12 Data Analysis and Probability Exercise 12.6 49E
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Chapter 12 Data Analysis and Probability Exercise 12.6 50E
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Chapter 12 Data Analysis and Probability Exercise 12.6 51E
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Chapter 12 Data Analysis and Probability Exercise 12.6 52E
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Chapter 12 Data Analysis and Probability Exercise 12.6 53E
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Chapter 12 Data Analysis and Probability Exercise 12.6 54E
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Chapter 12 Data Analysis and Probability Exercise 12.6 55E
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Let us find the different possible number of call sign for each station in US which begin with
letters W or K followed by 4 letters.
At the first place of the call sign is fixed either W or K.
So. to choose one letters, form 26 letters (A to Z) in 26 ways. which comes after W or K in the call sign. Similarly. to choose one letter from 26 letters (A to Z) in 26 ways which comes at the second place of the call sign after W or K. Choose one letter from 26 letters (A to Z) in 26 ways which comes at the third place of the call sign after W or K and to choose one letter from 26 letters (A to Z) in 26 ways which comes is the last letter of the call sign.
Therefore, the total possible number of ways to select 4 letters out of 26 letters, when repetitions are allowed, to make call sign which begin with either W or K is:
26x26x26x26+26x26x26x26 = 456976÷ 456976
= 913952
Thus, there are 19139521 different calls signs are possible. for each station in US which begin with letters W or K followed by 4 letters.

Chapter 12 Data Analysis and Probability Exercise 12.6 56E
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Chapter 12 Data Analysis and Probability Exercise 12.6 57E
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Chapter 12 Data Analysis and Probability Exercise 12.6 58E
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Chapter 12 Data Analysis and Probability Exercise 12.6 59E
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Chapter 12 Data Analysis and Probability Exercise 12.6 60E
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Chapter 12 Data Analysis and Probability Exercise 12.6 61E
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Chapter 12 Data Analysis and Probability Exercise 12.6 62E
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Chapter 12 Data Analysis and Probability Exercise 12.6 63E
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Chapter 12 Data Analysis and Probability Exercise 12.6 64E
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Chapter 12 Data Analysis and Probability Exercise 12.6 65E
The number of questions in a test is 1o.
The test is to answer total 7 questions, including exactly 4 of the first 6 question
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Chapter 12 Data Analysis and Probability Exercise 12.6 66E
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Chapter 12 Data Analysis and Probability Exercise 12.6 67E
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Chapter 12 Data Analysis and Probability Exercise 12.6 68E
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Chapter 12 Data Analysis and Probability Exercise 12.6 69E
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Chapter 12 Data Analysis and Probability Exercise 12.6 70E
Qualitative data: A data is said to qualitative, if it describes the quality that is the category of the data, and cannot measure, or numerical compared and it has no units
The data set of ZIP codes is represented the identification of the areas, hence it is qualitative The data set of ZIP codes describes the category of data and cannot be measured or numerically compared. it also have not units Therefore, the data set of ZIP codes is qualitative.

Chapter 12 Data Analysis and Probability Exercise 12.6 71E
Consider the data set of race times
The objective is to determine whether the data set of race times is quantitative or qualitative
Quantitative data: It measures the quantities of the data and can be describe numerically It has also the units and it can be numerically compared The data set of race times is represented by the numerical values and the race time has the unit in hours. The data set of race time are numerically described (countable number). and also it can be numerically compared So. the race times are quantitative data. Therefore, the data set of race times is quantitative.

Chapter 12 Data Analysis and Probability Exercise 12.6 72E
Consider the data set of heights of people.
The objective is to determine whether the data set of heights of people is quantitative or qualitative
Quantitative data: It measures the quantities of the data and can be describe numerically. It has also the units and it can be numerically compared The data set of heights of people is represented by the numerical value and the height of the people has the unit in cm or meter or feet. The data set of heights of people are numerically described (countable number). and also it can be numerically compared So. the heights of the people are quantitative data Therefore, the data set of heights of people quantitative.

Chapter 12 Data Analysis and Probability Exercise 12.6 73E
Consider the data set of emotions
The objective is to determine whether the data set of emotions is quantitative or qualitative.
Qualitative data: A data is said to qualitative, if it describes a category of the data. describes the quality, and cannot measure, or numerical compared and it has no units The data set of emotions is represented the quality data, since emotion is the attributes of the people. so it is a quality data The data set of emotions describes the category of data and cannot be measured or numerically compared. the emotions have not units Therefore, the data set of emotions is quaIitative.

Chapter 12 Data Analysis and Probability Exercise 12.6 74E
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Chapter 12 Data Analysis and Probability Exercise 12.6 75E
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Chapter 12 Data Analysis and Probability Exercise 12.6 76E
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Chapter 12 Data Analysis and Probability Exercise 12.6 77E
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Chapter 12 Data Analysis and Probability Exercise 12.6 78E
consider the decimal:
0.32.
The objective is to write the decimal in percent
To convert any decimal as a percent. move the decimal point two places to the right and write the percent sign.
So, 0.32=32%
Therefore, the decimal value 0.32 in Percent is 32% .

Chapter 12 Data Analysis and Probability Exercise 12.6 79E
Consider the decimal:
0.09
The objective is to write the decimal in percent
To convert any decimal as a percent. move the decimal point two places to the right and write the percent sign.
So, 0.09=9%.
Therefore, the decimal value 0.09 in percent is 9% .

Chapter 12 Data Analysis and Probability Exercise 12.6 80E
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Chapter 12 Data Analysis and Probability Exercise 12.6 81E
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