The mean amount of money spent on lunch per week for a sample of 458458 college students is $43$ 43. If the margin of error for the population mean with a 99%99% confidence interval is 1.601.60, construct a 99%99% confidence interval for the mean amount of money spent on lunch per week for all college students.

The 99% confidence interval for the mean amount of money spent on lunch per week for all college students is between $41.40 and $44.60.

Step-by-step explanation:

A confidence interval has two bounds, a lower bounds and an upper bound.

These bounds depend on the sample mean and the margin of error.

The lower bound is the sample mean subtracted by the margin of error.

The upper bound is the margin of error added to the sample mean.

In this problem, we have that:

The sample mean is $43.

The margin of error is $1.60 for a 99% confidence interval.

Lower bound: 43 - 1.60 = $41.40

Upper bound: 43 + 1.60 = $44.60

The 99% confidence interval for the mean amount of money spent on lunch per week for all college students is between $41.40 and $44.60.

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet