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miv72 [106K]
3 years ago
13

How do you simplify(n!)^2/(n+1)!(n-1)!

Mathematics
1 answer:
ANTONII [103]3 years ago
5 0
We haven n! = (n-1)! x n and (n+1)! = n! x (n + 1);
Then, (n!)^2 = n! x n! = n! x (n-1)! x n;
And (n+1)!(n-1)! = n! x (n + 1) x (n-1)!;
Finally, [n! x (n-1)! x n] / [n! x (n + 1) x (n-1)!] = (n+1)/n;
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Answer:

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Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that the first truck is available.

B is the probability that the second truck is available.

We have that:

A = a + (A \cap B)

In which a is the probability that the first truck is available and the second one is not A \cap B is the probability that both trucks are available.

By the same logic, we have that:

B = b + (A \cap B)

The probability that both trucks are available is .30.

This means that A \cap B = 0.3

The probability the second truck is available is .50

This means that B = 0.5. So

B = b + (A \cap B)

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The probability that at least one truck is available is:

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The probability that neither truck is available is:

1 - (A \cup B) = 1 - 0.95 = 0.05

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