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.
Start with the given equation:

Substitute known values:

Evaluate:

Similarly:


<span>If you have an equilateral triangle, the median is also the altitude so - That means that if you draw altitude, it will bisect the base of the triangle and meet at two right angles. That gives you two of the measurements for the right triangle - the side (6) hypotenuse, and the base (3). You can then figure out the height using the Pythagorean as you have the a and the c for the theorem. Then you can use 1/2 base times height to find the area.
</span>