Justin recently drove to visit his parents who live 300 miles away. On his way there his average speed was 11 miles per hour fas
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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%
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▹ Answer
<em>1/18 of a gallon of juice in each cup.</em>
▹ Step-by-Step Explanation
Since we have 4/9 of a gallon of juice for 8 cups, we will divide to find the amount of juice in each cup.
Since this operation is division, we have to change 8 to it's reciprocal which is 1/8:
Our final answer will be 1/18 of a gallon of juice in each cup.
Hope this helps!
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