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Firdavs [7]
3 years ago
9

Representar las rectas a, b y c y determinar si son paralelas (o perpendiculares) dos a dos. A: y=−3x+5 | b: y =x/3+2 | c: y=−3x

+1
Mathematics
1 answer:
Anettt [7]3 years ago
7 0

Answer:

A y C: paralelas

B y A: perpendiculares

B y C: perpendiculares.

Step-by-step explanation:

Sea una recta:

f(x) = a*x + b

Donde a es la pendiente y b es la ordenada al origen.

Otra recta va a ser paralela a esta si tiene la misma pendiente, pero una diferente ordenada al origen, por ejemplo:

g(x) = a*x + c

Y una recta perpendicular a f(x) tendría una pendiente p = -(1/a)

Entonces podemos escribir la línea perpendicular a f(x) como:

h(x) = -(1/a)*x + c

Ahora veamos nuestras líneas:

A(x) =  y = −3*x+5

B(x) = y = x/3+2 = (1/3)*x + 2

C(x) = y = -3*x + 1

Primero veamos cuáles son paralelas:

A(x) y C(x) tienen la misma pendiente (-3) y diferente ordenada al origen, entonces  A y C son paralelas.

Sabiendo que la pendiente de A y C es -3, una línea perpendicular a esta tendría una pendiente:

-(1/-3) = 1/3

Que es justamente la pendiente que tiene la línea B, entonces podemos concluir que B es perpendicular a las líneas A y C.

A y C: paralelas

B y A: perpendiculares

B y C: perpendiculares.

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Answer:

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Step-by-step explanation:

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Solution to the problem

Alternative 1

Let X the random variable of interest, on this case we now that:

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We can find the probability of loss like this P(X=0) and if we find this probability we got this:

P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729

And the probability of loss with the first wersion is 0.729

Alternative 2

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Y \sim Binom(n=5, p=0.05)

The probability mass function for the Binomial distribution is given as:

P(Y)=(nCy)(p)^y (1-p)^{n-y}

Where (nCx) means combinatory and it's given by this formula:

nCy=\frac{n!}{(n-y)! y!}

We can find the probability of loss like this P(Y=0) and if we find this probability we got this:

P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.

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<h2>Hello!</h2>

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