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Ierofanga [76]
3 years ago
15

Which solid figure has 5 faces,8 edges,and 5 vertices

Mathematics
1 answer:
Assoli18 [71]3 years ago
5 0
A square pyramid has 5 faces, 8 edges, and 5 vertices

hope it helps
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Yo what’s good brah <br> ✌
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A thief who is particularly knowledgeable about statistics and the legal system determines that if he commits a given heist the
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9514 1404 393

Answer:

  about 28.6%

Step-by-step explanation:

The conditional probability equation can be used:

  P(arrested | not in jail) = P(arrested & not in jail)/P(not in jail)

  = 0.24/0.84 = 2/7 ≈ 28.6%

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2 years ago
What is .725 as a decimal? (or) What is 29/40 as a decimal?
lisabon 2012 [21]

29/40 as a decimal would be .725 because you would have to divide 29 by 40 to get the answer.
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Help plsss!! thank you so mhcb
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The answer is 25 degrees
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HELP PLEASE!!! I need help with 94 if you could show the steps that would be very helpful!
aksik [14]
A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:

n!/((n-r)!r!)

In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.

5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5

5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10

5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10

5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5

Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.

Now we add together the combinations

1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.

The answer is 32.

-------------------------------

Note: There is also an equation for permutations which is:

n!/(n-r)!

Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.

We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.

I hope this helps! If you have any questions, let me know :)








7 0
3 years ago
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