NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6 6 . In an earlier study, the population proportion was estimated to be 0.32 0.32 . How large a sample would be required in order to estimate the fraction of people who black out at 6 6 or more Gs at the 85% 85 % confidence level with an error of at most 0.03 0.03 ? Round your answer up to the next integer.

The sample size must be atleast 502 to have a margin of error of 0.03

Step-by-step explanation:

We are given the following in the question:

Proportion = 0.32

Level of significance = 0.15

Margin of error = 0.03

Formula for margin of error =

We have to find the sample size such that the margin of error is atmost 0.03.

Putting values, we get,

Thus, the sample size must be atleast 502 to have a margin of error of 0.03 in approximation of proportion of people who would black out at 6 or more Gs.

4.

step-by-step explanation:

it have the same area.brainliest

the 8 in 8,420 is in the thousands place so it's value is 8 thousand.

the 8 in 6,870 is in the hundreds place so it's value is 8 hundred.

divide 8000 by 800 to find the answer:

8000 / 800 = 10

the 8 in 8,420 is 10 times greater than the 8 in 6,870