I think 50% or 1/2 ( it’s the same just one is in fraction and other in percentage)
Step-by-step explanation:
please mark me as brainlest
Answer:
The sample size to obtain the desired margin of error is 160.
Step-by-step explanation:
The Margin of Error is given as

Rearranging this equation in terms of n gives
![n=\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2](https://tex.z-dn.net/?f=n%3D%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM%7D%5Cright%5D%5E2)
Now the Margin of Error is reduced by 2 so the new M_2 is given as M/2 so the value of n_2 is calculated as
![n_2=\left[z_{crit}\times \dfrac{\sigma}{M_2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{\sigma}{M/2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{2\sigma}{M}\right]^2\\n_2=2^2\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4n](https://tex.z-dn.net/?f=n_2%3D%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM_2%7D%5Cright%5D%5E2%5C%5Cn_2%3D%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM%2F2%7D%5Cright%5D%5E2%5C%5Cn_2%3D%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B2%5Csigma%7D%7BM%7D%5Cright%5D%5E2%5C%5Cn_2%3D2%5E2%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM%7D%5Cright%5D%5E2%5C%5Cn_2%3D4%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM%7D%5Cright%5D%5E2%5C%5Cn_2%3D4n)
As n is given as 40 so the new sample size is given as

So the sample size to obtain the desired margin of error is 160.

so our slope is undefined, that means a vertical line, check the picture below
Answer:
The minimum sample size needed is 125.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the zscore that has a pvalue of
.
The margin of error is:

For this problem, we have that:

99% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
What minimum sample size would be necessary in order ensure a margin of error of 10 percentage points (or less) if they use the prior estimate that 25 percent of the pick-axes are in need of repair?
This minimum sample size is n.
n is found when 
So






Rounding up
The minimum sample size needed is 125.