Answer:
We are 95% confident that the percent of executives who prefer trucks is between 19.43% and 33.06%
Step-by-step explanation:
We are given that in a group of randomly selected adults, 160 identified themselves as executives.
n = 160
Also we are given that 42 of executives preferred trucks.
So the proportion of executives who prefer trucks is given by
p = 42/160
p = 0.2625
We are asked to find the 95% confidence interval for the percent of executives who prefer trucks.
We can use normal distribution for this problem if the following conditions are satisfied.
n×p ≥ 10
160×0.2625 ≥ 10
42 ≥ 10 (satisfied)
n×(1 - p) ≥ 10
160×(1 - 0.2625) ≥ 10
118 ≥ 10 (satisfied)
The required confidence interval is given by

Where p is the proportion of executives who prefer trucks, n is the number of executives and z is the z-score corresponding to the confidence level of 95%.
Form the z-table, the z-score corresponding to the confidence level of 95% is 1.96







Therefore, we are 95% confident that the percent of executives who prefer trucks is between 19.43% and 33.06%
Answer:
6(y+1) Hope this helps!!
Step-by-step explanation:
Answer: The correct answer is Choice C.
Step-by-step explanation: In order to solve this problem you need to look at the equation’s parts.
The first part is multiplying 5 by the contents of the parenthesis. So, 5(2x-2) equals 10x - 10. Now, but the equation together and simplify it:
10x - 10 + 10x =
20x - 10
Answer:
9
Step-by-step explanation:
The equation says y = x + 9
Therefore, y = 9 since x = 0 on the y axis.
Picture of graph also attached :)
Correct me if I’m wrong, I believe you have to add all of those numbers together. Then divide after and plug in those remaining digits