Answer:
0.13% of customers spend more than 46 minutes
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

What percentage of customers spend more than 46 minutes?
This is 1 subtracted by the pvalue of Z when X = 46. So



has a pvalue of 0.9987
1 - 0.9987 = 0.0013
0.13% of customers spend more than 46 minutes
The answer is C because you do PEMDAS and then add the 5 from the other side
Just divide 40 by 100. Then simplify 40/100. The answer is 2/5.
Perform the indicated multiplication first: -15 + 5q + 4 = 5q - 11
Note that 5q appears on both sides of this equation. Cancelling, we get :
-11 = -11. This is always true. Thus, -5(3-q) +4=5q-11 is true for all q.