Answer:
Whats the question?
Step-by-step explanation:
Srry i couldnt help
Cos(x)/(1+sin(x) + (1+sin(x))/cos(x)=
cos²(x)/cos(x)(1+sin(x)) + (1+sin(x))²/cos(x)(1+sin(x))=
cos²(x)+1+2sin(x)+sin²(x) /cos(x)(1+sin(x))=
2+2sin(x) / cos(x)(1+sin(x))=
2(1+sin(x)) / cos(x)(1+sin(x))=
2/cos(x)=
2*sec(x)
Answer:
The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).
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.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the z-score that has a p-value of
.
A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they'd received in the mail.
This means that ![n = 603, \pi = \frac{142}{603} = 0.2355](https://tex.z-dn.net/?f=n%20%3D%20603%2C%20%5Cpi%20%3D%20%5Cfrac%7B142%7D%7B603%7D%20%3D%200.2355)
95% confidence level
So
, z is the value of Z that has a p-value of
, so
.
The lower limit of this interval is:
![\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 - 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2016](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.2355%20-%201.96%5Csqrt%7B%5Cfrac%7B0.2355%2A0.7645%7D%7B603%7D%7D%20%3D%200.2016)
The upper limit of this interval is:
![\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 + 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2694](https://tex.z-dn.net/?f=%5Cpi%20%2B%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.2355%20%2B%201.96%5Csqrt%7B%5Cfrac%7B0.2355%2A0.7645%7D%7B603%7D%7D%20%3D%200.2694)
The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).
4/7 make into slope form which will get you that