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3 years ago

Find the geometric mean of 4 and 20

Mathematics

2 answers:

lozanna [386]3 years ago

6 0

Answer:

Step-by-step explanation:

zysi [14]3 years ago

4 0

Answer:

5 is the geometric mean of 4 and 20

Step-by-step explanation:

I think... Don't know how to show work!!!!

Hope I helped!

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The area of a playground is 12 square yards. The length of the playground is 3 times longer than its width. Find the length and

marishachu [46]

Answer:

Width is 2 and Length is 6

Step-by-step explanation:

As the data is given, the area of a playground is 12 square yards.
The length of the playground is 3 times longer than its width.
Let width is x, length is 3 times therefore, length = 3x

As we know the formula for area,

Area=length x width.

So by putting values in this formula,

12sq = (3x)(x)

=> 3x2 = 12

=> x2 = 12/3,

=> x2 = 4 ,

by taking square root we get x= 2,

the width is 2 and length is 3 times hence, length is 3(2) = 6.

6 0

4 years ago

PLZZZ HELP!!! VERY CONFUSED!!! Will give brainliest!!!!

777dan777 [17]

By the intersecting chords theorem, you have

$7\cdot4=5\cdot x\implies 5x=28\implies\boxed{x=\dfrac{28}5}$

7 0

3 years ago

If f(x) = 3x5 + 4x3 + 3, then what is the remainder when f(x) is divided by<br> x - 3?

VikaD [51]

Answer:

840

Step-by-step explanation:

8 0

3 years ago

Match each justification to its corresponding step in the following solution. (Table shown above)

AveGali [126]

Answer:

Step-by-step explanation:

5 0

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:

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

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

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.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

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

4 years ago