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.
Using cash can help Ralph improve his credit score is just untrue in so many ways.
The answer is choice A because the statement does not justify Ralph's use of cash.
Answer:
Functions a(x) and b(x) intersect each other at x = 2.
Answer:
No solution
Step-by-step explanation:
To eliminate is to get rid of one of the variables.
You can choose to either add each term in the equations or subtract each term in the equation.
For a variable to be eliminated, there must one pair that have the same constant with it. Each equation already has the same constant with a variable.
Try adding them.
. y - x = 15
<u>+ y - x = 5</u>
2y - 2x = 20 No variables were eliminated.
Try subtracting.
. y - x = 15
<u>- y - x = 5</u>
0y - 0x = 10 All variables were eliminated.
. 0 = 10 This is false.
This system of equations cannot be solved.
Graphically, these two lines would have the same slope and are parallel. The solution to a system is same as the point of intersection. Parallel lines never meet, never intersect, therefore there is no solution.
