What is the prime factorization for 112

Mathematics

1 answer:

lisov135 [29]3 years ago

5 0

112=2∗56, I believe would be the answer to your question. (The Prime Factors of 112: 2^4 • 7)

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) An instructor gives his class a set of 18 problems with the information that the next quiz will consist of a random selection

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Answer:

The probability the he or she will answer correctly is 1.5%

Step-by-step explanation:

In all, there are 18 problems. In this question, the order of which the problems are sorted for the quiz makes no difference. For example, if the question A of the quiz is P1 and question B P2, and question A P2 and question B P1, it is the same thing.

There are 18 problems and 9 are going to be selected. So, there is going to be a combination of 9 elements from a set of 18 elements.

A combination of n elements from a set of m objects has the following formula:

$C_{(m,n)} = \frac{m!}{n!(m-n)!}$

In this question, m = 18, n = 9. So the total number of possibilities is:

$T_{p} = C_{(18,9)} = \frac{18!}{9!(18-9)!} = 48620$

Now we have to calculate the number of desired outcomes. This number is a combination of 9 elements from a set of 13 elements(13 is the number of problems that the student has figured out how to do).

Now, m = 13, n = 9. The number of desired possibilities is:

$D_{p} = C_{(13,9)} = \frac{13!}{9!(13-9)!} = 715$