Answer:
We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.
Step-by-step explanation:
We have a sample of executives, of size n=160, and the proportion that prefer trucks is 26%.
We have to calculate a 95% confidence interval for the proportion.
The sample proportion is p=0.26.
The standard error of the proportion is:
The critical z-value for a 95% confidence interval is z=1.96.
The margin of error (MOE) can be calculated as:

Then, the lower and upper bounds of the confidence interval are:

The 95% confidence interval for the population proportion is (0.192, 0.328).
We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.
I just got this question. I think the answer is 5.
Width = x
Length = x+18
Assuming the table is rectangular:
Area = x(x + 18)
Therefore:
x(x + 18) <span>≤ 175
x^2 + 18x </span><span>≤ 175
Using completing the square method:
x^2 + 18x + 81 </span><span>≤ 175 + 81
(x + 9)^2 </span><span>≤ 256
|x + 9| </span><span>≤ sqrt(256)
|x + 9| </span><span>≤ +-16
-16 </span>≤ x + 9 <span>≤ 16
</span>-16 - 9 ≤ x <span>≤ 16 - 9
</span>-25 ≤ x <span>≤ 7
</span><span>
But x > 0 (there are no negative measurements):
</span><span>
Therefore, the interval 0 < x </span><span>≤ 7 represents the possible widths.</span><span>
</span>
Leftover per friend times number of friends=total left over
2/5 times number of friends=2
times both sides by 5/2
number of friends=10/2
number of friends=5
5 friends