Answer:
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A visit must be of at least 56.33 minutes to put a shopper in the longest 20 percent.
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula. In a set with mean
and standard deviation
, the z-score of a measure X is given by:

How long must a visit be to put a shopper in the longest 20 percent?
Here 
The 100-20=80th percentile, which is X when Z has p-value of 0.8, so Z=0.84


14.532=X-41.8
X=14.532+41.8
X=56.33
A visit must be of at least 56.33 minutes to put a shopper in the longest 20 percent.
Learn more about Normal Probability Distribution here:
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A = L * W
A = 12 * 8
A = 96 square feet
at 2.50 per square foot......96 * 2.5 = $ 240
You can arrange "PHONES" in 6 ways.
I hope this helps:)