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3 years ago

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Cylinder A has a radius of 10 inches and a height of 5 inches. Cylinder B has a volume of 750π. What is the percentage change in

volume from cylinder A to cylinder B?

2 answers:

Firlakuza [10]3 years ago

6 0

Answer:

50% change in volume

Step-by-step explanation:

<h2>This problem bothers on the mensuration of solid shapes.</h2>

In this problem we are to find the volume of the first cylinder and compare with the second cylinder.

Given data

Volume v = ?

Diameter d= ?

Radius r = $10 in$

Height h= $5 in$

we know that the volume of a cylinder is expressed as

$volume = \pi r^{2}h$

Substituting our given data we have

$volume = \pi*10^{2}*5\\ volume= \pi *100*5\\volume= 500\pi in^{3} \\$

The first cylinder as a volume of $500\pi$

The change in volume is $750\pi - 500\pi = 250\pi$

percentage = $\frac{250\pi }{500\pi } *100$

$percentage=$ $0.5*100= 50$ %

Kobotan [32]3 years ago

4 0

Answer:

50% increase

Step-by-step explanation:

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Step-by-step explanation:

Since your cell phone plan is by the minute, and each minute of use cost $ 0.10, to create a relation that represents the amount spent, A, per minute, m, of call time, and then use the relation to find the amount spent if you talk 65 minutes, the following calculations must be performed:

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Step-by-step explanation:

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Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose

Furkat [3]

Answer:

a) 0.164 = 16.4% probability that a disk has exactly one missing pulse

b) 0.017 = 1.7% probability that a disk has at least two missing pulses

c) 0.671 = 67.1% probability that neither contains a missing pulse

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the binomial distribution(for item c).

Poisson distribution:

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.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Poisson mean:

$\mu = 0.2$

a. What is the probability that a disk has exactly one missing pulse?

One disk, so Poisson.

This is P(X = 1).

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

0.164 = 16.4% probability that a disk has exactly one missing pulse

b. What is the probability that a disk has at least two missing pulses?

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

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

$P(X = 0) = \frac{e^{-0.2}*0.2^{0}}{(0)!} = 0.819$

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.819 + 0.164 = 0.983$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.983 = 0.017$

0.017 = 1.7% probability that a disk has at least two missing pulses

c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

Two disks, so binomial with n = 2.

A disk has a 0.819 probability of containing no missing pulse, and a 1 - 0.819 = 0.181 probability of containing a missing pulse, so $p = 0.181$

We want to find P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.181)^{0}.(0.819)^{2} = 0.671$

0.671 = 67.1% probability that neither contains a missing pulse

8 0

3 years ago