Answer: The solution is (4, -4)
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:
24
The LCM of 6 and 8 is 24. To find the least common multiple (LCM) of 6 and 8, we need to find the multiples of 6 and 8 (multiples of 6 = 6, 12, 18, 24; multiples of 8 = 8, 16, 24, 32) and choose the smallest multiple that is exactly divisible by 6 and 8, i.e., 24.