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Monica [59]
3 years ago
8

Can some please explain how to do the math

Mathematics
1 answer:
UNO [17]3 years ago
8 0

Answer:

The enrollment drop is 12 students per year. The equation is 60=5d. The complete sentence would be the enrollment drop at the after school program dropped 12 students per year, and the equation is 60=5d

Step-by-step explanation:

So I will start with explaining the equation. 60 is the amount of students that dropped, 5 is the year, and d is the enrollment drop rate per year that you are trying to find. The way to use this equation is to isolate d. To do that you divide both sides of the equation by 5, this leaves you with 12 = d. Therefore your yearly enrollment drop is 12 students.


60 = 5d

60/5 = d

60/5 = 12

12 = d

Hopefully this explained your answer.

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The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per
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Answer:

1.76% probability that in one hour more than 5 clients arrive

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

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The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that \mu = 2

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

P(X \leq 5) + P(X > 5) = 1

We want P(X > 5). So

P(X > 5) = 1 - P(X \leq 5)

In which

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353

P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707

P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707

P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804

P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902

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