Classify the polygon using as many names as possible. *
1 answer:
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Answer:
a. This probability is the p-value of Z when X = 60.
b. This probability is the p-value of Z when X = B subtracted by the p-value of Z when X = A.
c. The proportion is the p-value of Z when X is the length of the examination period. How many students is this proportion multiplied by the number of students.
Step-by-step explanation:
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:
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 p-value, we get the probability that the value of the measure is greater than X.
In this question:
Mean , standard deviation
.
a. What is the probability of completing the exam in one hour or less (to 4 decimals)?
This probability is the p-value of Z when X = 60.
b. What is the probability that a student will complete the exam in more than minutes A but less than B minutes (to 4 decimals)?
This probability is the p-value of Z when X = B subtracted by the p-value of Z when X = A.
c. Assume that the class has students and that the examination period is minutes in length. How many students do you expect will be unable to complete the exam in the allotted time (to nearest whole number)?
The proportion is the p-value of Z when X is the length of the examination period. How many students is this proportion multiplied by the number of students.
Rewrite the limand as
(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = (1 - sin(<em>x</em>)) / (cos²(<em>x</em>) / sin²(<em>x</em>))
… = ((1 - sin(<em>x</em>)) sin²(<em>x</em>)) / cos²(<em>x</em>)
Recall the Pythagorean identity,
sin²(<em>x</em>) + cos²(<em>x</em>) = 1
Then
(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = ((1 - sin(<em>x</em>)) sin²(<em>x</em>)) / (1 - sin²(<em>x</em>))
Factorize the denominator; it's a difference of squares, so
1 - sin²(<em>x</em>) = (1 - sin(<em>x</em>)) (1 + sin(<em>x</em>))
Cancel the common factor of 1 - sin(<em>x</em>) in the numerator and denominator:
(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = sin²(<em>x</em>) / (1 + sin(<em>x</em>))
Now the limand is continuous at <em>x</em> = <em>π</em>/2, so