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3 years ago

Find the volume of a cube with a side length of 4r3s2.

Mathematics

1 answer:

andrey2020 [161]3 years ago

8 0

Answer:

V = 64 r^(9) * s^(6)

Step-by-step explanation:

The volume of a cube is given by

V = s^3 where s is the side length

V = ( 4r^3s^2) ^3

We know that ( ab) ^c = a^c * b^c

V = 4^3 * r^3^3 * s^2^3

V = 64 * r^3^3 * s^2^3

We know that a^b^c = a^(b*c)

V = 64 r^(3*3) * s^(2*3)

V = 64 r^(9) * s^(6)

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bagirrra123 [75]

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2 years ago

One senior from the high schools will be selected at random. What is the probability that the senior selected will not be from h

DochEvi [55]

Using the probability concept, it is found that there is a 0.7361 = 73.61% probability that the senior selected will not be from high school b given that the senior responded with a choice other than college.

<h3>What is a probability? </h3>

A <em>probability </em>is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

Researching the problem on the internet, it is found that:

538 seniors responded with a choice other than college.

Of those students, 99 + 83 + 49 + 31 + 63 + 71 = 396 are not from high school b.

Hence:

$p = \frac{396}{538} = 0.7361$

0.7361 = 73.61% probability that the senior selected will not be from high school b given that the senior responded with a choice other than college.

You can learn more about the probability concept at brainly.com/question/26148436

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2 years ago

Peyton buys cookies and apples at the store.

Dmitrij [34]

Answer:

Hope this helps sorry I'm in class right now

Step-by-step explanation:

6 0

2 years ago

Determine whether the improper integral converges or diverges, and find the value of each that converges.

Ksju [112]

Answer:converge at $I=\frac{1}{3}$

Step-by-step explanation:

Given

Improper Integral I is given as

$I=\int^{\infty}_{3}\frac{1}{x^2}dx$

integration of $\frac{1}{x^2}$ is - $\frac{1}{x}$

$I=\left [ -\frac{1}{x}\right ]^{\infty}_3$

substituting value

$I=-\left [ \frac{1}{\infty }-\frac{1}{3}\right ]$

$I=-\left [ 0-\frac{1}{3}\right ]$

$I=\frac{1}{3}$

so the value of integral converges at $\frac{1}{3}$

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3 years ago

What is the answer I'm so stuck this is so hard for dumb people like me helpppppppp

miskamm [114]

Well I don't remember the equation with all the numbrical stuff but if you times 5 and 5 than times 4 and 5 which is 25 and 20 take 20 away from 25 and you have your answer hope it helps

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3 years ago