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.
It is 61 degrees bec u j add them then subrteact that by 189
Answer:
m = 
Step-by-step explanation:

Answer:
12 units
Step-by-step explanation:
Notice that as we go from Q to R, x (the horizontal distance) increases by 12 and y (the vertical distance) does not change. Thus, the distance between the two points is merely the horizontal distance, 12 units.
Answer:
A
Step-by-step explanation:
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