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Zolol [24]
3 years ago
8

A water tank is in the shape of a cylinder with a diameter of 20 feet and a height of 20 feet. The tank is 70% full. About how m

any gallons of water are in the tank? Round your answer to the nearest whole number.
Mathematics
2 answers:
fomenos3 years ago
7 0

Answer:

32901 gallons

Step-by-step explanation:

First, we find the volume of the cylindrical water tank when it is full.

The volume of a cylinder is given as:

V = \pi r^2h

where r = radius and h = height of the cylinder

The diameter of the tank is 20 feet, hence, its radius is 10 feet.

Volume, V, is therefore:

V = \pi * 10^2*20 = 6283.19 ft^3

This is the volume of the tank, when it is full.

When it is 70% full, the volume will be:

\frac{70}{100} *  6283.19 = 4398.23 ft^3

1 cubic foot (ft^3) is 7.48 gallons.

Therefore, the volume of the 70% full tank in gallons will be:

4398.23 * 7.48 = 32901.05 gallons ≅ 32901 gallons

The tank contains 32901 gallons of water.

Mariulka [41]3 years ago
4 0

Answer:

32886.476 gal

Step-by-step explanation:

Given:

Diameter of tank 'd'= 20 ft

∴radius 'r'= d/2=> 20/2=> 10ft

Height 'h' = 20 ft

Volume of cylinder can be determined by

V= π r² h

Substituting the values in above equation

V= π (10)² (20) =>π(100)(20)

V=2000π ft³

When the tank is 70% full.

70% of 2000π =>0.7 x 2000π

1400π ft³

As we know that, 1ft³ = 7.481 gal

1400π ft³= 1400π x 7.481 gal

1400π ft³= 32886.476 gal

Therefore, if the tank is 70% full then 32886.476 gallons of water are in the tank

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The resting heart rate for an adult horse should average about µ = 47 beats per minute with a (95% of data) range from 19 to 75
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Answer:

a. 0.0582 = 5.82% probability that the heart rate is less than 25 beats per minute.

b. 0.1762 = 17.62% probability that the heart rate is greater than 60 beats per minute.

c. 0.7656 = 76.56% probability that the heart rate is between 25 and 60 beats per minute

Step-by-step explanation:

Empirical Rule:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

Normal Probability Distribution:

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.

Mean:

\mu = 47

(95% of data) range from 19 to 75 beats per minute.

This means that between 19 and 75, by the Empirical Rule, there are 4 standard deviations. So

4\sigma = 75 - 19

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a. What is the probability that the heart rate is less than 25 beats per minute?

This is the p-value of Z when X = 25. So

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

Z = \frac{25 - 47}{14}

Z = -1.57

Z = -1.57 has a p-value of 0.0582.

0.0582 = 5.82% probability that the heart rate is less than 25 beats per minute.

b. What is the probability that the heart rate is greater than 60 beats per minute?

This is 1 subtracted by the p-value of Z when X = 60. So

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

Z = \frac{60 - 47}{14}

Z = 0.93

Z = 0.93 has a p-value of 0.8238.

1 - 0.8238 = 0.1762

0.1762 = 17.62% probability that the heart rate is greater than 60 beats per minute.

c. What is the probability that the heart rate is between 25 and 60 beats per minute?

This is the p-value of Z when X = 60 subtracted by the p-value of Z when X = 25. From the previous two items, we have these two p-values. So

0.8238 - 0.0582 = 0.7656

0.7656 = 76.56% probability that the heart rate is between 25 and 60 beats per minute

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