Step-by-step explanation:
$75 30% 0.70(75) = x $52.50
$18 65% 0.35(18) = x $6.30
$60 30% 0.70x = 42 $42
$35 20% 0.80x = 28 $28
$150 25% 0.75(150) = x $112.50
Divide the time he traveled by the distance to find how fast he travels 1 mile, then multiply that by 5 miles:
12 minutes / 2 miles = 6 minutes per mile.
6 minutes per mile x 5 miles = 30 total minutes.
121 is big enough to assume normality and not worry about the t distribution. By the 68-95-99.7 rule a 95% confidence interval includes plus or minus two standard deviations. So 95% of the cars will be in the mph range

The question is a bit vague, but it seems we're being asked for the 95% confidence interval on the average of 121 cars. The 121 is a hint of course.
The standard deviation of the average is in general the standard deviation of the individual samples divided by the square root of n:

So repeating our experiment of taking the average 121 cars over and over, we expect 95% of the averages to be in the mph range

That's probably the answer they're looking for.