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4 years ago

Find the geometric mean of 4 and 12

Mathematics

2 answers:

jonny [76]4 years ago

7 0

Answer:

The Geometric Mean of 4 and 12 is 6.9

Step-by-step explanation:

Given Numbers are 4 and 12

To find : Geometric mean of the given No.

The Geometric Mean is a type of average where we multiply the nos. together and then take a square root (for two nos), cube root (for three nos) etc.

Formula for Geometric Mean is given by,

$Geometric\,Mean\,of\,x_1\,,\,x_2\,,\,x_3..x_n=\sqrt[n]{x_1\times x_2\times x_3\times\,...\,\times x_n}$

⇒ Geometric Mean of 4 and 12 = $\sqrt{4\times12}$

= $\sqrt{48}$

= $4\sqrt{3}$

= 6.92820323028

= 6.9

Therefore, The Geometric Mean of 4 and 12 is 6.9

kondaur [170]4 years ago

6 0

The geometric mean of 4 and 12 is 8.0000000 hope this helps :P

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A functionf(x) is graphed on the coordinate plane.

ollegr [7]

Answer:

y = -1/2x +1

Step-by-step explanation:

The y intercept is 1 (this is where it crosses the y intercept)

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The slope is given by

m = (y2-y1)/(x2-x1)

= (0-1)/(2-0)

= -1/2

The slope is -1/2

The slope intercept form is y = mx+b where m is the slope and b is the y intercept

y = -1/2x +1

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3 years ago

Read 2 more answers

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

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Mariana [72]

Depending on what the number is in for the bag

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3 years ago

(PLS help I’ll give brainliest to whoever can answer right nobody has been able to. )

natali 33 [55]

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= 3.14 x 0.5625 x 2/3
= 1.76625 x 2/3
= 1.1775 = 1.18 feet^3

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2 years ago

Three six-sided fair dice are rolled. The six sides of each die are numbered 1; 2; : : : ; 6. Let A be the event that the first

valkas [14]

Answer:

1.) [A, B, C]

2.) [A', B', C']

3.) [A', B, C] or [A, B', C] or [A, B, C'] or [ A', B', C] or [A', B, C'] or [A, B', C'] or [A', B', C']

4.) A', B', C] or [A', B, C'] or [A, B', C'] or [A', B, C] or [A, B', C] or [A, B, C'] or [A, B, C]

Step-by-step explanation:

If A is the event that the first die shows an even number, then A' is the event that first die DOES NOT show an even number.

If B is the event that the Second die shows and even number, then B' is the event that second die DOES NOT show an even number.

If C is the event that the third die shows an even number, then C' is the event that third die DOES NOT show an even number.

Hence, our possible events are A, A', B, B', C, C'.

For question 1, in the event that all three dice show and even number, the expression of the event = [A, B, C]

For question two, in the event that no die shows an even number, the expression of the event = [A', B', C']

For question 3,In the event that at least one die shows an odd number, the expression of the event = [A', B, C] or [A, B', C] or [A, B, C'] or [ A', B', C] or [A', B, C'] or [A, B', C'] or [A', B', C']

For question 4,in the event that at most two dice show odd numbers, the expression of the event = [A', B', C] or [A', B, C'] or [A, B', C'] or [A', B, C] or [A, B', C] or [A, B, C'] or [A, B, C]

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3 years ago