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kati45 [8]
3 years ago
5

The table represents an exponential function.

Mathematics
2 answers:
White raven [17]3 years ago
7 0

Answer:

Letter answer is B. 3/4

Step-by-step explanation:

Tpy6a [65]3 years ago
3 0
TL;DR(too long didn't read): answer is 3/4


This question may look confusing, however, it is more easily understood once you see that the fractions appear to be changing more randomly than they are in a way you can recognize, however, they're not. Since they look like that it's because they're being multiplied by a fraction. Split the fractions into two to make it easier. 3/2 and 9/8, just look at them as '3' and '2' and '9' and '8'. 3 becomes 9. Which means either 6 was added or 3 was multiplied by 3. Compare to the next row, 27. 9-->27 can't be 6, so it's being multiplied by 3. Now for the bottom. 2 becomes 8, and knowing that the numerator of the fraction is being multiplied, so is the denominator then, so 2-->8 is 2 times 4. Put the numerator and denominator back together and you have 3/4. The answer is 3/4.
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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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