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Amiraneli [1.4K]

3 years ago

7

can someone help me please the problems are above i think or below whatever way they show up for people who help me

1 answer:

Phoenix [80]3 years ago

3 0

Answer

a. The volume is multiplied by 4

b. The volume is doubled

Explanation

a. The volume of a cone with radius r and height h is $\pi r^2 h/3$ . If the radius is doubled, the new volume will be $\pi (2r)^2h/3$ , which is equal to $4\pi r^2h/3$ . This means that the volume is multiplied by 4.

b. The volume of a cone with radius r and height h is $\pi r^2 h/3$ . If the height is doubled, the new volume will be $\pi r^2(2h)/3$ , which is equal to $2\pi r^2 h/3$ . This means that the volume is multiplied by 2.

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The number of typing errors made by a typist has a Poisson distribution with an average of three errors per page. If more than t

damaskus [11]

Answer:

0.6472 = 64.72% probability that a randomly selected page does not need to be retyped.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Poisson distribution with an average of three errors per page

This means that $\mu = 3$

What is the probability that a randomly selected page does not need to be retyped?

Probability of at most 3 errors, so:

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240$

Then

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0498 + 0.1494 + 0.2240 + 0.2240 = 0.6472$

0.6472 = 64.72% probability that a randomly selected page does not need to be retyped.

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3 years ago