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GalinKa [24]
3 years ago
13

What is the approximate volume of a cylinder with a diameter of 10 meters and a height of 5 meters? Use 3.14 for pi. A. 392.5 m3

B. 1570 m3 C. 314 m3 D. 130.8 m3
Mathematics
2 answers:
EastWind [94]3 years ago
3 0
For cylinder
d=10m
So r=5m
h=5m
volume=pi×r^2×h
=3.14×5×5×5
=3.14×125
=392.5m^3
So option a is correct.
storchak [24]3 years ago
3 0
= (3.14)(20/2)²(5)
= 15.7(10)²
= 15.7(100)
= 1,570

Answer: D. 1,570 cubic meters
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Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights
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Step-by-step explanation:

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Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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.

Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.

This means that \mu = 85, \sigma = 8

a. Find the probability a randomly chosen apple exceeds 100 g in weight.

This is 1 subtracted by the p-value of Z when X = 100. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{100 - 85}{8}

Z = 1.875

Z = 1.875 has a p-value of 0.9697

1 - 0.9696 = 0.0304

0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b. What weight do 80% of the apples exceed?

This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.

Z = \frac{X - \mu}{\sigma}

-0.84 = \frac{X- 85}{8}

X - 85 = -0.84*8

X = 78.28

The weight that 80% of the apples exceed is of 78.28g.

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