Answer:
2.28% of tests has scores over 90.
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 proportion of tests has scores over 90?
This proportion is 1 subtracted by the pvalue of Z when X = 90. So



has a pvalue of 0.9772.
So 1-0.9772 = 0.0228 = 2.28% of tests has scores over 90.
Answer:
<h2>
43.4° and 46.6°</h2>
Step-by-step explanation:
x - measure of angle
x + 3.2° - measure of its complementary angle
Complementary angles adds to 90°
x + x + 3.2° = 90°
2x = 90° - 3.2°
2x = 86.8°
x = 86.8°÷2
x = 43.4°
x+3.2° = 43.4° + 3.2° = 46.6°
I'm not sure about this but will give it a try:
Let f(n) = 2xⁿ - 2
Then f(3) = 2x³ - 2
So, f(2) = 2x² - 2
22/28 converted to a percentage is 78.6%. (rounded to the nearest tenth)
78.6% of the class passed the test.
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Answer:
Step-by-step explanation:
22