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

The sample that would be required in order to estimate the fraction of people who black out at 6 or more Gs is of size 502.

Step-by-step explanation:

The (1 - α)% confidence interval for population proportion p is:

The formula to calculate the margin of error is,  

The given information is, = 0.32 and margin of error = 0.03.

The z-value for 85% confidence level is,

*Use a z-table for the z-value.

Then the sample size n is given by,

Thus, the sample that would be required in order to estimate the fraction of people who black out at 6 or more Gs is of size 502.

your answer is 23/2 - 13.5

step-by-step explanation:

7/2 evaluates to

multiply x and 4.5

multiply x and 1

the x just gets copied along.

the answer is x

x

4.5*x evaluates to 4.5x

multiply 3 and 4.5x

multiply 1 and x

the x just gets copied along.

x

4.5*x*3 evaluates to 13.5x

7/2-4.5*x*3 evaluates to

  + 8 =

the answer is

7/2-4.5*x*3+8 evaluates to

y=mx+b is the final answer!   : )