More information is needed to answer you.
The sample proportion has an approximately normal distribution, with the mean being the population mean and the standard deviation (i.e. the standard error of the sampling proportion) equal to
SE = sqrt(p*q/n) where p is the population proportion q is 1-p, and n is the sample size.
<span>(Typically, for a confkdence interval, the sample proportion and its complement are substituted for p and q.) </span>
The margin of error will be Zc*SE, where Zc is the critical value of Z for the level of confidence required. The confidence interval, then, will be the sample proportion, plus or minus the margin of error.
For a 90% confidence interval, Zc is 1.645.
<span>So the margin of error is 1.645 * sqrt(.9*.1/400), roughly .025. So your confidence interval would be around [.875, .925]
The best and most correct answer among the choices provided by your question is the second choice or letter C which is 3%.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.</span>
1) add 7 to both sides and get -5....so -5 >x
2) add 7 to both sides.....so get b<-5
3) add 17 to both side...so get z>1
