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Gennadij [26K]
4 years ago
9

Tim gets $5 for delivering the paper. How much money will he have after delivering the paper 3 times?

Mathematics
2 answers:
kirza4 [7]4 years ago
5 0

Answer:

Tim will get 15$

Step-by-step explanation:

5$ x 3 = 15$

Taya2010 [7]4 years ago
4 0

Answer:

Step-by-step explanation:

so add the 5  3 times the answer is for this is 15

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Answer:

2800 in^3

Step-by-step explanation:

Calculate the volume as normal, but take 1/10 of the result. Since volume is length*width*height, we have 50*28*20 which is 28000. 1/10 of that is 2800. So the volume when scaled down by a factor of 1/10 is 2800 in^3

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Simplify <br> -5(3(8-6)+7)
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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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$16

Step-by-step explanation:

4 books would cost $6 and 5 tapes would be $10. Add them to get 16.

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