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Lady bird [3.3K]
3 years ago
10

If Y has exactly 2 positive integer factors then Y is a prime number. if so how?

Mathematics
1 answer:
Setler79 [48]3 years ago
7 0

Step-by-step explanation:

yes, if your teacher meant to include the default factors of every positive integer number : 1 and the number itself.

these are the 2 positive integer factors.

if a positive number can only be divided without remainder by 1 or by itself (with result 1), then it is per definition a prime number.

if the number can be divided without remainder by another number, then we are not dealing with a prime number.

but if your teacher did not have these default factors in mind, then no :

example : 15

15 has only 2 factors : 3 and 5.

both are positive integer numbers.

and yet 15 is not a prime number.

but back to my first point, 15 also has formally the factors 1 and 15. so, if we count them too, then we have suddenly 4 factors. and the original statement fits again.

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3 years ago
Blood type AB is the rarest blood type, occurring in only 4% of the population in the United States. In Australia, only 1.5% of
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Answer:

There is a 27.62% probability that exactly 2 of the U.S. residents have blood type AB.

Step-by-step explanation:

For each U.S. resident, there are only two outcomes possible. Either they have blood type AB, or they do not. This means that we can solve this problem using binomial probability distribution concepts.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And \pi is the probability of X happening.

In this problem, we have that:

50 U.S residents are sampled, so n = 50

4% of the U.S population has blood type AB, so p = 0.04.

What is the probability that exactly 2 of the U.S. residents have blood type AB?

This is P(X = 2). So:

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{50,2}.(0.04)^{2}.(0.96)^{48} = 0.2762

There is a 27.62% probability that exactly 2 of the U.S. residents have blood type AB.

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