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Scrat [10]
2 years ago
9

Instructions: Find the volume of each figure. Round your answers to the nearest tenth, if necessary.

Mathematics
1 answer:
Ilia_Sergeevich [38]2 years ago
7 0

The volume of the cone is  84.9 km³

<h3>How to find the volume of a cone?</h3>

Volume of a cone = 1 / 3 πr²h

where

  • h = height
  • r = radius

Therefore,

radius = 3 km

height = √9.5² - 3²

height = √81.25

height = 9.01387818866

height = 9.014 km

Volume of a cone = 1 / 3 × 3.14 × 3² × 9.014

volume of the cone = 254.732197612 / 3

volume of the cone = 84.9107325372

volume of the cone = 84.9 km³

learn more on volume here: brainly.com/question/17075150

#SPJ1

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(Literal equations) The formula for the volume of a cylinder is V = πr^2h. Solve V = πr^2h for h, the height of the cylinder. A.
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A study of long-distance phone calls made from General Electric's corporate headquarters in Fairfield, Connecticut, revealed the
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Answer:

a) 0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

b) 0.0668 = 6.68% of the calls last more than 4.2 minutes

c) 0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

d) 0.9330 = 93.30% of the calls last between 3 and 5 minutes

e) They last at least 4.3 minutes

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 3.6, \sigma = 0.4

(a) What fraction of the calls last between 3.6 and 4.2 minutes?

This is the pvalue of Z when X = 4.2 subtracted by the pvalue of Z when X = 3.6.

X = 4.2

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

Z = \frac{4.2 - 3.6}{0.4}

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

X = 3.6

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

Z = \frac{3.6 - 3.6}{0.4}

Z = 0

Z = 0 has a pvalue of 0.5

0.9332 - 0.5 = 0.4332

0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

(b) What fraction of the calls last more than 4.2 minutes?

This is 1 subtracted by the pvalue of Z when X = 4.2. So

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

Z = \frac{4.2 - 3.6}{0.4}

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% of the calls last more than 4.2 minutes

(c) What fraction of the calls last between 4.2 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 4.2. So

X = 5

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

Z = \frac{5 - 3.6}{0.4}

Z = 3.5

Z = 3.5 has a pvalue of 0.9998

X = 4.2

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

Z = \frac{4.2 - 3.6}{0.4}

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

0.9998 - 0.9332 = 0.0666

0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

(d) What fraction of the calls last between 3 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 3.

X = 5

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

Z = \frac{5 - 3.6}{0.4}

Z = 3.5

Z = 3.5 has a pvalue of 0.9998

X = 3

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

Z = \frac{3 - 3.6}{0.4}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

0.9998 - 0.0668 = 0.9330

0.9330 = 93.30% of the calls last between 3 and 5 minutes

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At least X minutes

X is the 100-4 = 96th percentile, which is found when Z has a pvalue of 0.96. So X when Z = 1.75.

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

1.75 = \frac{X - 3.6}{0.4}

X - 3.6 = 0.4*1.75

X = 4.3

They last at least 4.3 minutes

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Divide f(x) by x + 15

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