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spayn [35]
3 years ago
14

Para reformar la cocina de su caja Berta compro la semana pasada 2 cajas de plaquetas para el suelo y 4 de azulejos para la pare

d por 200 €. Hoy la comprado 2 cajas más de plaquetas y otras 2 de azulejos per 130 € Cuanto cuesta la caja de plaquetas y la de azulejos
Mathematics
1 answer:
Anvisha [2.4K]3 years ago
6 0

Answer:

Caja de plaquetas = x = 30 €

Caja de azulejos = y = 35 €

Step-by-step explanation:

Deje que el costo de

Caja de plaquetas = x

Cuadro de mosaico = y

Por Berta

Para reformar su cocina, Berta compró la semana pasada 2 cajas de baldosas de plaquetas para el suelo y 4 cajas de baldosas para la pared por 200 €.

2x + 4y = 200 ...... Ecuación 1

Hoy

Hoy compré 2 cajas más de plaquetas y otras 2 de tejas por 130 €.

2x + 2y = 130 ..... Ecuación 2

Resolvemos usando el método de Eliminación

2x + 4y = 200 ...... Ecuación 1

2x + 2y = 130 ..... Ecuación 2

Restar la ecuación 2 de 1

2 años = 70

y = 70/2

y = 35 €

Resolviendo para x

2x + 4y = 200 ...... Ecuación 1.

2x + 4 × 35 = 200

2x + 140 = 200

2x = 200 - 140

2x = 60

x = 60/2

x = 30 €

el costo de

Caja de plaquetas = x = 30 €

Caja de azulejos = y = 35 €

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Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose
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Answer:

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Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the binomial distribution(for item c).

Poisson distribution:

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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x is the number of sucesses


e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Binomial 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.

Poisson mean:

\mu = 0.2

a. What is the probability that a disk has exactly one missing pulse?

One disk, so Poisson.

This is P(X = 1).

P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164


0.164 = 16.4% probability that a disk has exactly one missing pulse

b. What is the probability that a disk has at least two missing pulses?

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1)

In which

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}&#10;

P(X = 0) = \frac{e^{-0.2}*0.2^{0}}{(0)!} = 0.819

P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164&#10;

P(X < 2) = P(X = 0) + P(X = 1) = 0.819 + 0.164 = 0.983

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.983 = 0.017

0.017 = 1.7% probability that a disk has at least two missing pulses

c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

Two disks, so binomial with n = 2.

A disk has a 0.819 probability of containing no missing pulse, and a 1 - 0.819 = 0.181 probability of containing a missing pulse, so p = 0.181

We want to find P(X = 0).

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0.671 = 67.1% probability that neither contains a missing pulse

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