Algebra 1 Common Core Answers Student Edition Grade 8 – 9 Chapter 12 Data Analysis and Probability Exercise 12.2
Algebra 1 Common Core Solutions
Chapter 12 Data Analysis and Probability Exercise 12.2 1LC
Given, the data shown below is the battery life, in hours for different brand of batteries
12 9 10 14 10 11 10 18 21 10 14 22
We have to make a frequency table for the above data.
To make frequency table first we find the number of interval and the size of interval which is reasonable for the given data The minimum data value is Q and the maximum data value is 22. so we chose the total interval length is 16 (we chose any value for the length of interval which is near to the difference between maximum and minimum data value) We divide 16 values into 4 intervals of size 4 These intervals are:
Chapter 12 Data Analysis and Probability Exercise 12.2 2LC
Given, the data shown below is the battery life, in hours for different brand of batteries.
12 9 10 14 10 11 10 18 21 10 14 22
We have to make Histogram of the above data
Chapter 12 Data Analysis and Probability Exercise 12.2 3LC
Given. the data shown below is the battery life. in hours for different brand of batteries
12910141011101821101422
We have to make a cumulative frequency table for the above data
Chapter 12 Data Analysis and Probability Exercise 12.2 4LC
The Table 1 shows that the busiest hour for the store owner is 9-11 hour Thus we can say that frequency table helps to the store owner to determine its busiest business hour
Chapter 12 Data Analysis and Probability Exercise 12.2 5LC
Chapter 12 Data Analysis and Probability Exercise 12.2 6LC
Chapter 12 Data Analysis and Probability Exercise 12.2 7E
Chapter 12 Data Analysis and Probability Exercise 12.2 8E
Chapter 12 Data Analysis and Probability Exercise 12.2 9E
Chapter 12 Data Analysis and Probability Exercise 12.2 10E
Chapter 12 Data Analysis and Probability Exercise 12.2 11E
Chapter 12 Data Analysis and Probability Exercise 12.2 12E
Chapter 12 Data Analysis and Probability Exercise 12.2 13E
Chapter 12 Data Analysis and Probability Exercise 12.2 14E
From the above histogram we see that it has one peak this is not in the centre By the definition of skewed histogram. “if the histogram has one peak that is not in center then the histogram is skewed”. Thus the above histogram is skewed
Chapter 12 Data Analysis and Probability Exercise 12.2 15E
From the above histogram we see that it has one peak this is not in the Centre. By the definition of skewed histogram. “if the histogram has one peak that is not in center then the histogram is skewed” Thus the above histogram is skewed.
Chapter 12 Data Analysis and Probability Exercise 12.2 16E
From the above histogram we see that it has one peak this is not in the centre By the definition of skewed histogram, “if the histogram has one peak that is not in center then the histogram is skewed”. Thus the above histogram is skewed
Chapter 12 Data Analysis and Probability Exercise 12.2 17E
From the above histogram we see that it has one peak this is not in the centre By the definition of skewed histogram, “if the histogram has one peak that is not in center then the histogram is skewed”. Thus the above histogram is skewed
Chapter 12 Data Analysis and Probability Exercise 12.2 18E
Chapter 12 Data Analysis and Probability Exercise 12.2 19E
Chapter 12 Data Analysis and Probability Exercise 12.2 20E
Chapter 12 Data Analysis and Probability Exercise 12.2 21E
Chapter 12 Data Analysis and Probability Exercise 12.2 22E
Chapter 12 Data Analysis and Probability Exercise 12.2 23E
Chapter 12 Data Analysis and Probability Exercise 12.2 24E
Given, that the set of data which shows test score;
81 70 73 89 68 79 91 59 77 73 80 75 88 65 82 94 77 67 82
We have to make Histogram of the above data uses of interval 10.
Chapter 12 Data Analysis and Probability Exercise 12.2 25E
Given, that the set of data which shows test score;
81 70 73 89 68 79 91 59 77 73 80 75 88 65 82 94 77 67 82
We have to make Histogram of the above data uses of interval 20.
Chapter 12 Data Analysis and Probability Exercise 12.2 26E
The frequency table of all the three interval size shows that the interval size 20 makes a little variation in test score Since in the interval test score 59-78 the frequency is lo and in the interval of test score 79-98 the frequency is 9. Thus the interval size of 20 makes little variation in test score.
Chapter 12 Data Analysis and Probability Exercise 12.2 27E
From the above histogram we see that the last interval of the histogram is 80-99, and in this interval there are 4 Customers. The maximum value of the 80-99 is 99 Thus the upper limit on the amount of money that any customer spent is $999.
Chapter 12 Data Analysis and Probability Exercise 12.2 28E
From the above histogram we see that the last interval of the histogram is 80-99, and in this interval there are 4 Customers. The maximum value of the 80-99 is 99 Thus the upper limit on the amount of money that any customer spent is $999.
Chapter 12 Data Analysis and Probability Exercise 12.2 29E
From the above histogram we see that there are 9 customers which spent less than $20 This is corresponding to the first interval that O-19. Thus 9 customers spent less than $20.
Chapter 12 Data Analysis and Probability Exercise 12.2 30E
From the above histogram we see that there are 9 customers which correspond to the first interval O-19 spent less than $20. There are 15 customers which corresponds to the second interval 20-39 spent less than $40. There are 13 customers which is correspond to the first interval 40-59 spent less than $60, 9 customers which is correspond to the first interval 60-79 spent less than $80 and there are 4 customers which is correspond to the first interval 80-99 spent less than $100
Chapter 12 Data Analysis and Probability Exercise 12.2 31E
Chapter 12 Data Analysis and Probability Exercise 12.2 32E
Given, that a set of data which has 200 data values We have to make a histogram that must have 40°/b of the values lies in the interval 20-29 The remaining values should be evenly divided among the intervals 0-9, 10-19, 30-39, and 40-49.
Chapter 12 Data Analysis and Probability Exercise 12.2 33E
Chapter 12 Data Analysis and Probability Exercise 12.2 34E
Solution:
From the above histogram we see that it has one peak which is not in the centre. By the definition of skewed histogram. “if the histogram has one peak that is not in center then the histogram is skewed”. Thus the above histogram is skewed Therefore the correct option is (C).
The option (A) is not correct Because by the definition of symmetric histogram. “if a vertical line can divide the histogram into two parts that are close to mirror images, then the histogram is symmetric”. In the above histogram there is no any vertical line that divided histogram into two parts and each part is mirror image to other.
The option (B) is not correct. Because only three type of histogram, symmetric, skewed and uniform.
The option (D) is also not correct By the definition of a histogram, “if the bars are roughly the same then histogram is uniform”. Hence option (D) is not correct.
Chapter 12 Data Analysis and Probability Exercise 12.2 35E
Question:
What isthe solution of (—4x—6) + (6x+1)=I3?………. (1)
(F) —4 (G) 6 (H) 5 (I) 9
Chapter 12 Data Analysis and Probability Exercise 12.2 36E
Chapter 12 Data Analysis and Probability Exercise 12.2 37E
Chapter 12 Data Analysis and Probability Exercise 12.2 38E
Chapter 12 Data Analysis and Probability Exercise 12.2 39E
Chapter 12 Data Analysis and Probability Exercise 12.2 40E
Chapter 12 Data Analysis and Probability Exercise 12.2 41E
Given the numbers,
0.9,-0.2,1.2,5,-1,0,0.1,2……………….. (1)
We have to order the numbers from least to greatest
The least number in a numbers given numbers is … since it is the most negative value number The second least number —02. third least number O. fourth least number 0.1 . fifth least number 0.9 sixth least number 1.2. seventh least number 2 and 5 is the greatest number
Thus we arrange the given numbers as above, we get
-1,-0.2,0,0.1,O.9,1.2,2,5