D (7) should be the correct answer.
Hi!
To compare this two sets of data, you need to use a t-student test:
You have the following data:
-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph
-Wednesday n2=20; </span>x̄2=56,3 mph; s2=4,4 mph
You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.

To calculate the degrees of freedom you need to use the following equation:

≈34
The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10
So, as the calculated value is higher than the critical tabulated one,
we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.
The answer would be A. He should use the mean because it is in the center of the data. I hope this helped ^^
Answer:
12,400
Step-by-step explanation:
Let the current employees be x
65% increase in number = 0.65x
If the total number of employees after the increase is 20,460, then;
x + 0.65x = 20,460
1.65x = 20,460
x = 20,460/1.65
x = 12,400
Hence the current number of employees is 12,400
C. 110. This is because all triangles add up to equal 180*. So you add the two angles you have and subtract from 180.