Y=0.2x+0.3=0.5
y=0.05x+0.4=0.09
all you have to do is basically just add the numbers when you add the numbers that would be your fraction you have to add the zeros and which will make the number
Answer:
The answer is D
You have to distribute the x to the 2x + 3 equation
This will result in 2x^2 + 3x
Considering that you mean 6.5% interest yearly, the interest gained would be:
$146.25
and the total amount would be $896.25
Hope this helped!
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.

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 
95% confidence level
So
, z is the value of Z that has a p-value of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

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).