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MA_775_DIABLO [31]
3 years ago
13

Considerando las siguientes variables a = ancho y l = longitud, determina su valor, si se sabe que se cuenta con un rectángulo q

ue tiene como perímetro 35 metros, y se sabe que el doble de su ancho es igual a 3 veces su longitud.
Mathematics
1 answer:
kogti [31]3 years ago
8 0

Answer:

La longitud = 7 metros

La anchor = 10.5 metros

Step-by-step explanation:

La fórmula para el perímetro de un rectángulo es

P = 2L + 2W

L = longitud

A = ancho

P = 35 metros

se sabe que el doble de su ancho es igual a 3 veces su largo.

2A = 3L

Ancho = 3L / 2 = 1.5L

Sustituimos 1.5L por W

35 = 2 L + 2 (1.5 L)

35 = 2L + 3L

35 = 5 litros

L = 35/5

L = 7 metros

Por lo tanto, la longitud = 7 metros

Recuerda

A = 1.5 L

A = 1.5 × 7

A = 10.5 metros

La anchor = 10.5 metros

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To solve this question, we need to understand the Poisson and the binomial probability distributions.

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

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e = 2.71828 is the Euler number

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Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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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.

Poisson distribution with a mean of 0.08 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel.

So \mu = 10*0.08 = 0.8

(a) What is the probability that there are no surface flaws in an auto's interior?

Single car, so Poisson distribution. This is P(X = 0).

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

P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493

44.93% probability that there are no surface flaws in an auto's interior

(b) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws?

For each car, there is a p = 0.4493 probability of having no surface flaws. 10 cars, so n = 10. This is P(X = 10), binomial, since there are multiple cars and each of them has the same probability of not having a surface defect.

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

P(X = 10) = C_{10,10}.(0.4493)^{10}.(0.5507)^{0} = 0.0003

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(c) If 10 cars are sold to a rental company, what is the probability that at most 1 car has any surface flaws?

At least 9 cars without surface flaws. So

P(X \geq 9) = P(X = 9) + P(X = 10)

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

P(X = 9) = C_{10,9}.(0.4493)^{9}.(0.5507)^{1} = 0.0041

P(X = 10) = C_{10,10}.(0.4493)^{10}.(0.5507)^{0} = 0.0003

P(X \geq 9) = P(X = 9) + P(X = 10) = 0.0041 + 0.0003 = 0.0044

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