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.
Answer:
Step-by-step explanation:
Given points are: (-4,2) and (5,6)

Answer
(a) 0.94
(b)0.06
Step-by-step explanation:
P(husband is employed) = P(H) = 0.85
P(wife is employed) = P(W) = 0.59
P(both are employed) = P(H∩W) = 0.50
(a) At least one of them is employed
= P(H∪W)
= P(H) + P(W) - P(H∩W)
=0.85 + 0.59 - 0.50
=0.94
(b) Neither is employed,
=1 - P(H∪W)
= 1-0.94
=0.06