Is needed asap - for every second this question isn't answered, you lose a day in your life - 15 points.

what sample size do you need to get a margin of error of 2.9%? (what do you have to remember when giving me a sample size? ) if the number is larger than 1000, don't type a comma.

don't write anything that isn't related to the question.

What does the comma have to do with anything? LOL

OK. You want a margin of error of 2.9%. But you do not have a confidence interval, which you need to find a sample size. I'm going to assume a confidence interval of 90%, because that's a popular one. If I chose the wrong one, hopefully you can repeat whatever I did.

For a 90% confidence interval, there is a Z-multiplier of 1.645. In your lessons, at least for now, you don't really have to know what a Z-multiplier represents. But keep in mind that (a) you need it, and (b) the day will eventually come when you need to understand why you need it.

The formula is

N = ((z-multiplier * standard deviation) / margin of error)^2

In this case that's

(1.645 * standard deviation / 0.029)^2

If I chose a standard deviation of 1, N would be

(1.645 / 0.029)^2 = about 3218. (

the correct answer to your question is a. 4^4

good luck my friend!

hope this !

maybe if i you could me by giving me brainliest?

! < 3

have a splendid day < 3 : )