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asambeis [7]
3 years ago
14

A cylinder has a base diameter of 18 feet and height of 15 feet . What is it’s volume in a cubic feet to the nearest tenths plac

e ?
Mathematics
2 answers:
agasfer [191]3 years ago
4 0
V = pi * r^2 * h

radius is half of diameter so 18/2 = 9. now plug that in :)

v = pi * 9^2 * 15
v = pi * (81)(15)
v = 1215(pi)
v = 3817.0

hope this helped :)
Fynjy0 [20]3 years ago
3 0

it's 3817.04 but you round to the nearest tenth

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The waiting time for a fire department to get called to a house fire is exponentially distributed with an average wait time of 1
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Answer:

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Step-by-step explanation:

To solve this question, we need to understand the exponential distribution and conditional probability.

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

f(x) = \mu e^{-\mu x}

In which \mu = \frac{1}{m} is the decay parameter.

The probability that x is lower or equal to a is given by:

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Which has the following solution:

P(X \leq x) = 1 - e^{-\mu x}

The probability of finding a value higher than x is:

P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: It has already taken 11 minutes.

Event B: It will take 16 more minutes.

Exponentially distributed with an average wait time of 14 minutes.

This means that m = 14, \mu = \frac{1}{14}

Probability of the waiting time being of at least 11 minutes:

P(A) = P(X > 11) = e^{-\frac{11}{14}} = 0.4558

Probability of the waiting time being of at least 11 minutes, and more than an additional 16 minutes:

More than 11 + 16 = 27 minutes. So

P(A \cap B) = P(X > 27) = e^{-\frac{27}{14}} = 0.1454

What is the probability that the wait time will be more than an additional 16 minutes?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1454}{0.4558} = 0.319

0.319 = 31.9% probability that the wait time will be more than an additional 16 minutes

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Answer:  =  7, so D.

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