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MA_775_DIABLO [31]
3 years ago
13

How to simplify (n+2)!/n! I will mark brainliest and give a thanks!

Mathematics
1 answer:
Anastaziya [24]3 years ago
3 0
\dfrac{(n+2)!}{n!}=\dfrac{(n+2)(n+1)n!}{n!}=(n+2)(n+1)=n^2+3n+2
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Triangles A and B are similar. A has side lengths 2, 4, and 6. B has corresponding side lengths ???, 16, and 24
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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
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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

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

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3 years ago
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