Asimple random sample of 1000 american adults found that the average number of hours spent watching television during a typical week was 13.8. a simple random sample of 500 canadians yielded an average of 12.5 hours per week of television viewing. assume that the american and canadian distributions for weekly television, viewing times have the same standard deviations. the sampling variability associated with the sample means (described above) is

think the answer would beb

a) at most means it can cost $12 or less

so the inequality sign would be ≤

you would then need to multiply the number of toppings by the 0.35 for each one and add it to the cost of the cheese pizza.

the equation would be 9.99 + 0.35x ≤ 12

b) using the above equation:

9.99 + 0.35x ≤ 12

subtract 9.99 from each side:

0.35x ≤2.01

divide both sides by 0.35:

x ≤ 2.01 / 0.35

x ≤5.74

at most you could have 5 toppings.

the 5 topping pizza would cost: 0.35 x 5 = 1.75 + 9.99 = $11.74