Answer:
A) Θ = 2A/r²
Step-by-step explanation:
A = Θr²/2
(Multiple both sides by 2)
2A = Θr²
(Divide both sides by r²)
2A/r² = Θ
Answer:
15.87% probability that a randomly selected individual will be between 185 and 190 pounds
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 is the probability that a randomly selected individual will be between 185 and 190 pounds?
This probability is the pvalue of Z when X = 190 subtracted by the pvalue of Z when X = 185. So
X = 190



has a pvalue of 0.8944
X = 185



has a pvalue of 0.7357
0.8944 - 0.7357 = 0.1587
15.87% probability that a randomly selected individual will be between 185 and 190 pounds
10 1/2 = 21/2
(21/2) / 3
= 21/2 x 1/3
= 21/6
= 3 1/2
answer
3 1/2 pages
The question is somehow incomplete but the answer is it in
the inferential stage of probability-based inference. It is in
complex networks of codependent variables is an lively theme in statistical
research, encouraged by such varied presentations as predicting, pedigree examination
and troubleshooting.
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