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insens350 [35]
3 years ago
15

In meters what is the value of x? Pls i really dont know how to do this

Mathematics
1 answer:
Hitman42 [59]3 years ago
6 0

Answer:

x = 75 m

Step-by-step explanation:

This is a problem of similar triangles solved using proportions.

In similar triangles, corresponding side lengths are proportional.

Here the small and large triangles are similar because the have all three angles congruent.

By proportions, identifying corresponding sides,

x/45 = 50/30

solve for x

x = 45*(50/30) = 75 m

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a) Poisson distribution

b) 99.33% probability that in any one minute at least one purchase is​ made

c) 0.05% probability that seven people make a purchase in the next four ​minutes

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

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

5 people per minute check out from their shopping cart and make an online purchase.

This means that \mu = 5

a) What model might you suggest to model the number of purchases per​ minute? ​

The only information that we have is the mean number of an event(purchases) in a time interval. Each event is also independent fro each other. So you should suggest the Poisson distribution to model the number of purchases per​ minute.

b) What is the probability that in any one minute at least one purchase is​ made? ​

Either no purchases are made, or at least one is. The sum of the probabilities of these events is 1. So

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

We want to find P(X \geq 1)

So

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

In which

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

P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067

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99.33% probability that in any one minute at least one purchase is​ made

c) What is the probability that seven people make a purchase in the next four ​minutes?

The mean is 5 purchases in a minute. So, for 4 minutes

\mu = 4*5 = 20

We have to find P(X = 7).

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

P(X = 0) = \frac{e^{-20}*(20)^{7}}{(7)!} = 0.0005

0.05% probability that seven people make a purchase in the next four ​minutes

8 0
3 years ago
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