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3241004551 [841]
4 years ago
8

Juan is picking out some movies to rent, and he is primarily interested in horror films and mysteries. He has narrowed down his

selections to 77 horror films and 1313 mysteries. How many different combinations of 33 movies can he rent if he wants at least one horror film?
Mathematics
1 answer:
expeople1 [14]4 years ago
3 0

Answer:  854

Step-by-step explanation:

Given , Number of horror films = 7

Number of mysteries = 13

Total movies = 13+7=20

Number of combinations of r things taken out of n things : ^nC_r=\dfrac{n!}{r!(n-r)!}

Now , the number of  combinations of 3 movies can he rent if he wants at least one horror film = ^7C_1\times^{13}C_2+^7C_2\times^{13}C_1+^7C_3\times^{13}C_0

=(7)\times\dfrac{13!}{2!(13-2)!}+\dfrac{7!}{2!(7-2)!}\times(13)+\dfrac{7!}{3!(7-3)!}(1)\\\\=546+273+35\\\\=854

Hence, there are 854 different combinations of 3 movies can he rent if he wants at least one horror film .

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nlexa [21]
<h3>Answer:  9V</h3>

=============================================================

Reason:

The volume expression of a cone with radius r and height h is

\frac{1}{3}\pi*r^2*h

Let's plug in the given height h = 12 and we'd get

\frac{1}{3}\pi*r^2*h\\\\\frac{1}{3}\pi*r^2*12\\\\\left(\frac{1}{3}*12\right)\pi*r^2\\\\4\pi*r^2\\\\

This is the volume of the first cone. We're told the first cone has a volume of V, so we can say V = 4\pi r^2

We can't find the actual numeric volume because we don't know what value replaces r. So we leave it as is.

The second cone has the same height (h = 12) but the radius is now 3 times in size. Instead of r, we use 3r

Replace every copy of r with 3r. Then simplify

4\pi*r^2\\\\4\pi*(3r)^2\\\\4\pi*9r^2\\\\9(4\pi r^2)\\\\9V\\\\

The radius tripled which results in a volume that's 9 times bigger.

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2 years ago
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Answer:

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Step-by-step explanation:

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