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.
Answer:
(10, - 9 )
Step-by-step explanation:
Given the 2 equations
y = - 2x + 11 → (1)
y = - 3x + 21 → (2)
Substitute y = - 2x + 11 into (2)
- 2x + 11 = - 3x + 21 ( add 3x to both sides )
x + 11 = 21 ( subtract 11 from both sides )
x = 10
Substitute x = 10 into either of the 2 equations and evaluate fir y
Substituting into (1)
y = - 2(10) + 11 = - 20 + 11 = - 9
solution is (10, - 9 )
Answer:
56.7 mi/h
Step-by-step explanation:
The formula for average speed is:
Average Speed = 
Thus we have:
Average Speed = 
mph
<em>Rounding to the nearest tenth, we have </em>
Average Speed = 56.7 mi/h
Y = <span>b^x
when x = 1
y = b^1
y = b
Therefore, the value of b is the same as the value of y when x =1
From the graph,
When x = 1, y = 0.5
Therefore, b = 0.5
To confirm this
From the graph,
When x = -1, y = 2
Since </span>y = b^x<span>
2 = </span>b^-1
2 = 1/b
2b = 1
b = 0.5
When x = -2, y = 4
Since y = b^x
4 = b^-2
4 = 1/(b^2)
b^2 = 1/4
b = √(1/4)
b = 1/2
b = 0.5
Therefore, it is conformed that b = 0.5
