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:
28
Step-by-step explanation:
In a regular 52 card deck, 26 are black (clubs or spades). Additionally, there are two red jacks (diamonds and hearts). This equals 28/52. As a probability, it would be 28/52=14/26=7/13. This could also be expressed as a percentage or decimal. 0.538 or 53.8%. But the question asks <em>how many ways</em> can she choose a jack or black card, so I’m assuming they don’t want the probability, just the number, which is 28.
The second bicycle would weigh 1.32142857143 pounds or rounded to 1.32.
You would find this by multiply 18.5 by 1/14
1 yard=36 inches
9 times 1=9
multiply both sides by 9
9 times 1 yard=9 times 36 inches
9 yard=324 inches