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3 years ago

Find the volume of this cylinder. Give your answer to one decimal place.

Mathematics

2 answers:

Alika [10]3 years ago

6 0

$V = \pi r^{2} \times h$

$r = 11 \div 2 = 5.5 \\ h = 14 \\ \pi = 3.14$

$V = 3.14 \times 5.5 ^{2} \times 14 = 1,329.79$

Answer: V=1,329.8

swat323 years ago

5 0

Answer:

Step-by-step explanation:

Given

diameter (d) = 11 cm

Height(h) = 14 cm

first calculating the radius

radius = d/2 = 11/2 = 5.5 cm

Now

Volume of the cylinder

= $\pi r^{2}h$

= 3.14 * (5.5)^2 * 14

= 1329.8 cm^3

Hope it will help

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3 years ago

The probability that two people have the same birthday in a room of 20 people is about 41.1%. It turns out that

salantis [7]

Answer:

a) Let X the random variable of interest, on this case we know that:

$X \sim Binom(n=20, p=0.411)$

This random variable represent that two people have the same birthday in just one classroom

b) We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

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And we can find the individual probability:

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And then:

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And since we want the probability in the 3 classes we can assume independence and we got:

$P= 0.99997^3 = 0.9992$

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

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The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

Solution to the problem

Part a

Let X the random variable of interest, on this case we know that:

$X \sim Binom(n=20, p=0.411)$

This random variable represent that two people have the same birthday in just one classroom

Part b

We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

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

And we can find the individual probability:

$P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253$

And then:

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

And since we want the probability in the 3 classes we can assume independence and we got:

$P= 0.99997^3 = 0.9992$

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

4 0

3 years ago

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Answer:

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Step-by-step explanation:

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This gives the required equation

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