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Anna [14]
3 years ago
12

A cylinder with radius r and height 2r+4 contains a cube with edge length r + 2, as shown.

Mathematics
1 answer:
marishachu [46]3 years ago
6 0

Volume is the amount of space in an object. The fraction occupied by the cube is: \frac{r\sqrt2}{\pi(r + 2)}

First, we calculate the volume of the cube.

Volume = Length^3

From the figure, we have:

Length = r\sqrt 2

So:

V_1 = (r\sqrt2)^3

V_1 = (2\sqrt2)r^3

Next, the volume of the cylinder

Volume=\pi r^2h

So, we have:

V_2 = \pi r^2 (2r + 4)

The fraction (n) occupied by the cube is:

n = \frac{V_1}{V_2}

So, we have:

n = \frac{(2\sqrt2)r^3}{\pi r^2(2r + 4)}

Factor out 2 in the denominator

n = \frac{(2\sqrt2)r^3}{2\pi r^2(r + 2)}

Cancel out common terms

n = \frac{r\sqrt2}{\pi(r + 2)}

<em>Hence, the fraction occupied by the cube is: </em>\frac{r\sqrt2}{\pi(r + 2)}<em />

Read more about volumes at:

brainly.com/question/15861918

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Hey there!

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What is the value of x​
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Answer:

45 degrees

Step-by-step explanation:

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4x=180

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\frac{4x}{x} =\frac{180}{4}

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Last year a video game cost $50. Price increase by 25% this year. What is the new price of this video game?
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student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
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