You know that the average college student eats 0.75 pounds of food at lunch. If the standard deviation is 0.2 pounds of food, then what is the total amount of food that a cafeteria should have on hand to be 90 percent confident that it will not run out of food when feeding 50 college students?

$53.95

Step-by-step explanation:

Given: pounds

pounds

n= 50

Two tailed critical z-value for 90% confidence level = 1.645

Let x be the total amount of food that a cafeteria should have on hand to be 90 percent confident that it will not run out of food .

z-score =

Then,

Foe 50 students, total amount of food = 1.079 x 50 = $53.95

hence, the cafeteria should have food of amount $53.95 so that it is be 90 percent confident that it will not run out of food.

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b