We plan to choose a random sample of college students in Norman (OK) and Lubbock (TX) to examine how they spent their time on Friday evenings during the pandemic (from early April 2020 to late December 2020). Consider the population variable below: Cost of food, drink, and entertainment on a Friday evening. Let us assume that the population standard deviation is $12.5. Also, let us assume that we are willing to accept only a $1.5 margin of errors in our estimations. Level of confidence being set at 95%, how many observations do we need in order to estimate the population mean for the above cost?

267 observations are needed.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

That is z with a pvalue of , so Z = 1.96.

Now, find the margin of error M as such

Let us assume that the population standard deviation is $12.5.

This means that

Willing to accept only a $1.5 margin of errors in our estimations. How many observations do we need in order to estimate the population mean for the above cost?

This is n for which M = 1.5. So

Rounding up:

267 observations are needed.

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step-by-step explanation:

a.) list the factors of 40.

1/2/4/5/8/10/20/40

b.) list the factors of 40 that represent the number of cans that could be on each shelf.

5

c.) how many different arrangements of cans are possible?

just one because if you put more on one shelf you will go way over 40 and the cabinet will probably break.

d.) what are the possible arrangements of the cans?

put 5 cans on each self to reach 40 cans in total.

e.) which arrangement uses all the shelves?

5 because you can put 5 cans on each shelf, and since there are 8 shelfs you are equally going to get 40 cans.