Y= 850-18x
since you start with 850 ounces that will be your y intercept
each time you go down 18 ounces so the slope is -18
Sum is
![S_{n}=\frac{a_{1}(1-r^{n})}{1-r}](https://tex.z-dn.net/?f=%20S_%7Bn%7D%3D%5Cfrac%7Ba_%7B1%7D%281-r%5E%7Bn%7D%29%7D%7B1-r%7D%20)
r=common ratio
a1=first term
it looks like 2^0=1 is the first term aka a1
it goes to the 9th term (2^9)
sub
![S_{9}=\frac{1(1-(2)^{9})}{1-2}](https://tex.z-dn.net/?f=%20S_%7B9%7D%3D%5Cfrac%7B1%281-%282%29%5E%7B9%7D%29%7D%7B1-2%7D%20)
![S_{9}=\frac{1-512}{-1}](https://tex.z-dn.net/?f=%20S_%7B9%7D%3D%5Cfrac%7B1-512%7D%7B-1%7D%20)
![S_{9}=\frac{-511}{-1}](https://tex.z-dn.net/?f=%20S_%7B9%7D%3D%5Cfrac%7B-511%7D%7B-1%7D%20)
Equation:
2x + 6 = 4x/2 + 12/2
To solve this, we need to transpose like terms to the same side.
2x - 4x/2 = 12/2 - 6
2x - 2x = 6 - 6
0 = 0
Since both sides are zero, it means that the equation has infinite number of solutions.
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).