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:
The solution is (-2/3,7/3)
Step-by-step explanation:
1) What is the bottom part saying? I will just solve what I see on top since that may be the only part you need to solve, comment what the bottom part is if you need it, I'll help.
2) y=7/3 by solving. for y.
3) x = -4/6 by solving for x after plugging in the y.
4) Your answer is (-2/3,7/3).
Answer:
0 - 1
Step-by-step explanation:
U can do by simple division method to get the answer
I uploaded the answer to a file hosting. Here's link:
tinyurl.com/wpazsebu