Question:
1. The females worked less than the males, and the female median is close to Q1.
2. There is a high data value that causes the data set to be asymmetrical for the males.
3. There are significant outliers at the high ends of both the males and the females.
4. Both graphs have the required quartiles.
Answer:
The correct option is;
1. The females worked less than the males, and the female median is close to Q1
Step-by-step explanation:
Based on the given data, we have;
For males
Minimum = 0
Q1 = 1
Median or Q2 = 20
Q3 = 25
Maximum = 50
For females;
Minimum = 0
Q1 = 5
Median or Q2 = 6
Q3 = 10
Maximum = 18
Therefore, the values of data that affect the statistical measures of spread and center are that
The females worked less than the males as such the statistical data for the females have less variability than the males in terms of interquartile range
Also the female median is very close to Q1, therefore it affects the definition of a measure of center.
Brodsky I wondering the same thing but good luck finding the answer
<h3>
Answer: 1</h3>
========================================================
Explanation:
Pick any two points you want from the blue line. I'll pick (0,1) and (1,2)
Apply the slope formula to those points
m = (y2-y1)/(x2-x1)
m = (2-1)/(1-0)
m = 1/1
m = 1
The slope is 1.
Notice how if we're at (0,1), then we move up 1 and over to the right 1 to arrive at (1,2).
slope = rise/run = 1/1
rise = 1, run = 1
The answer is most likely to be B
<h2><u>Solution</u> :-</h2>
Height of Eric = 6 + 1/6 feet
= 37/6 feet
Height of his brother = 5 + 3/4 feet
= 23/4 feet
Difference between their height :
= 37/6 - 23/4
= (74 - 69)/12
= 5/12
<u>Hence</u>,
Eric is 5/12 feet taller than his brother.
