Answer:
6.18% of the class has an exam score of A- or higher.
Step-by-step explanation:
When the distribution is normal, we use 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 question, we have that:

What percentage of the class has an exam score of A- or higher (defined as at least 90)?
This is 1 subtracted by the pvalue of Z when X = 90. So



has a pvalue of 0.9382
1 - 0.9382 = 0.0618
6.18% of the class has an exam score of A- or higher.
25=
[{50 divided by 5} plus 3]times 2 -1=25
So your equation is a composite function. It means to plug 1.5 as x into g(x), then the answer you get for that, plug into h(x).
So.
g(1.5)=8-3(1.5)
=8-4.5
=3.5
Then.
H(3.5)=2x+5
=2(3.5)+5
=7+5=12
Your final answer is 12.