Chicken biscuits and syrup
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.
3.5h+10=5.25h-11
21=1.75h
12=h
3.5(12)+10=5.25(12)-11
52=52
First, let’s all acknowledge that whoever comes up with problems like this WANTS kids to hate math...smh
I’m sure there is a prettier way to solve this, but here’s what I did:
8(2.25) + 3(22.50) =
18 + 67.50 = 85.50 per “set” of balls/jerseys
400/85.50 = 4.678 = number of “sets” he can buy. Round down to 4 so we have room for tax.
85.5 x 4 “sets”= $342
Tax on 342 is 0.06 x 342 = 20.52
$342 + 20.52 = $362.52 spent
Basketballs = 4 sets x 8 balls per set= 32
Jerseys = 4 sets x 3 jerseys per set= 12
32 basketballs, 12 jerseys, $362.52 spent