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katrin [286]
4 years ago
12

Resolver el siguiente sistema de ecuación:

Mathematics
1 answer:
snow_tiger [21]4 years ago
6 0

Answer:

x=-1\\y=4

Step-by-step explanation:

\frac{5x+2y}{3}=1

\frac{2x+y}{2}=1

Vamos a utilizar el método de sustitución. Vamos a despejar la segunda ecuación por alguna de las variables. Elegiré la y porque esta sola.

\frac{2x+y}{2}=1

Vamos a empezar por eliminar ese 2 del denominador. Para hacer esto, multiplicamos por 2 ambos lados de la ecuación.

2x+y=1*2

2x+y=2

Como vamos a despejar la y, restamos 2x en ambos lados.

y=2-2x

Listo.

------------------------------------------------------------------------

Procedemos a reemplazar el valor de y en la primera ecuación.

\frac{5x+2y}{3}=1

\frac{5x+2(2-2x)}{3}=1

\frac{5x+4-4x}{3} =1

Procedemos a operar términos semejantes.

\frac{x+4}{3} =1

Nos deshacemos del 3 multiplicando.

x+4=1*3\\x+4=3

Restamos 4

x=3-4\\x=-1

-----------------------------------------------------------------------------

Una vez encontrado el valor de x, procedemos a reemplazar en cualquier ecuación para encontrar el valor de y. Usaré la segunda ecuación.

\frac{2x+y}{2}=1

\frac{2(-1)+y}{2}=1

Procedemos a resolver;

\frac{-2+y}{2} =1

Multiplicamos por 2.

-2+y=1*2

-2+y=2

Sumamos 2

y=2+2\\y=4

-------------------------------------------------------------------------------

Para comprobar que los valores son correctos, reemplazamos ambos valores en cualquier ecuación y el resultado debe ser igual.

Usaré la primera.

\frac{5x+2y}{3} =1

\frac{5(-1)+2(4)}{3} =1

Resolvemos;

\frac{-5+8}{3} =1

\frac{3}{3}=1

1=1

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(b) The probability that all of the selected consumers recognize the brand name is 0.0041.

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(d) The events of 5 customers recognizing the brand name is unusual.

Step-by-step explanation:

Let <em>X</em> = number of consumer's who recognize the brand.

The probability of the random variable <em>X</em> is, P (X) = <em>p</em> = 0.40.

A random sample of size, <em>n</em> = 6 consumers are selected.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

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Compute the value of P (X = 5) as follows:

P(X=5)={6\choose 5}0.40^{5}(1-0.40)^{6-5}\\=6\times0.01024\times0.60\\=0.036864\\\approx0.0369

Thus, the probability that exactly 5 of the 6 consumers recognize the brand name is 0.0369.

(b)

Compute the value of P (X = 6) as follows:

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Thus, the probability that all of the selected consumers recognize the brand name is 0.0041.

(c)

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P (X ≥ 5) = P (X = 5) + P (X = 6)

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Thus, the probability that at least 5 of the selected consumers recognize the brand name is 0.041.

(d)

An event is considered unusual if the probability of its occurrence is less than 0.05.

The probability of 5 customers recognizing the brand name is 0.0369.

This probability value is less than 0.05.

Thus, the events of 5 customers recognizing the brand name is unusual.

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