Using the z-distribution, the 99% confidence interval to estimate the population proportion is: (0.2364, 0.4836).
<h3>What is a confidence interval of proportions?</h3>
A confidence interval of proportions is given by:

In which:
is the sample proportion.
In this problem, we have a 99% confidence level, hence
, z is the value of Z that has a p-value of
, so the critical value is z = 2.575.
The estimate and the sample size are given by:
.
Then the bounds of the interval are:
The 99% confidence interval to estimate the population proportion is: (0.2364, 0.4836).
More can be learned about the z-distribution at brainly.com/question/25890103
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A center at q scale factor of 1/2
Answer:
The percent markup is 91%.
Step-by-step explanation:
1. find the amount of profit per large bucket of popcorn
9 - 0.8 = $8.2
2. divide the amount of profit per large bucket of popcorn by the cost of a large bucket of popcorn
8.2 / 9 = <u>91%</u>
Answer:
8008
Step-by-step explanation:
This is the selection of 6 elements out of 16 elements. Applying the combination formula, the total number of selections is nCr with n =16 and r =6.
16C6 = 8008
Answer:
Step-by-step explanation: