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soldi70 [24.7K]
3 years ago
15

Which of the following is the greatest common factor of 38 and 57?

Mathematics
2 answers:
Ivenika [448]3 years ago
7 0

the answer is 19

hope this helps.

Masja [62]3 years ago
5 0

<u>Answer: </u>

The greatest common factor of 38 and 57 is 19.

<u>Solution: </u>

Given that the two numbers are 38 and 57. We need to find the greatest common factor of these two numbers.

G.C.D means the greatest factor of the two given numbers.

To find the greatest common factor, we need to find the prime factors of the individual numbers.

The prime factors of 38 = 2 \times 19

The prime factors of 57 = 3 \times 19

Here 19 is a common factor in both 38 and 57

Therefore the greatest common factor of 38 and 57 is 19

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Hospitals typically require backup generators to provide electricity in the event of a power outage. Assume that emergency backu
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Answer:

a) There is a 10.24% probability that both generators fail during a power outage.

b) There is an 89.76% probability of having a working generator in the event of a power outage, which is not high enough for the hospital.

Step-by-step explanation:

For each emergency backup generator, there are only two possible outcomes. Either they work correctly, or they fail. So we use the binomial probability distribution to solve this problem.

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}.p^{x}.(1-p)^{n-x}

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

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

And p is the probability of X happening.

In this problem we have that:

Assume that emergency backup generators fail 32% of the times when they are needed. So they work correctly 100-32 = 68% of the time. So p = 0.68

There are two generators, so n = 2

a. Find the probability that both generators fail during a power outage

This is P(X = 0)

So

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

P(X = 0) = C_{2,0}.(0.68)^{0}.(0.32)^{2} = 0.1024

There is a 10.24% probability that both generators fail during a power outage.

b. Find the probability of having a working generator in the event of a power outage. Is that probability high enough for the hospital? Assume the hospital needs both generators to fail less than 1% of the time when needed.

Either there are no working generators, or there is at least one working generator. The sum of the probabilities of these events is decimal 1. So

P(X = 0) + P(X \geq 1) = 1

We want P(X \geq 1). So

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1024 = 0.8976

There is an 89.76% probability of having a working generator in the event of a power outage, which is not high enough for the hospital.

To be high enough for the hospital, this probability should be at least of 99%.

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Since each slice of pizza is $2 and Wyatt bought 14 slices of pizza, you would multiply 14 by 2 then add it onto the cost of the membership.

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The monthly cost for 14 slices of pizza is $48.

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Answer:

Check the explanation

Step-by-step explanation:

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Of the 60 and ups: 28 of 122 = 23.0% order three or more.

The older you are, the less likely you are to order lots of condiments.

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