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ioda
4 years ago
13

two forces act at 60 degrees and their resultant is 14 neutons, if they act at 90 degrees their resultant is 90 degrees. find th

e mangitude of forces?
Engineering
1 answer:
g100num [7]4 years ago
5 0

A2A.

Let the magnitude of the two forces be x and y.

Let resultant at right angle be p(= √15N) and at 60 degrees be q(= √18N.

Now, p = √(x2 + y2) = √15,

q = √(x2 + y2 +2xycos60) = √18.

So x2 + y2 = 15,

x2 + y2 + xy = 18,

therefore xy = 3 or y = 3/x.

Now, x2 + (3/x)2 = 15,

x4 + 9 = 15x2,

x4 - 15x2 + 9 = 0.

Now find the roots of this quadratic equation, the two values of x will correspond to the magnitudes of the two vectors.

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4 years ago
El tiempo hasta que falle un sistema informático sigue una distribución Exponencial con media de 600hs. (Utilice 3 decimales par
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La probabilidad pedida es 0.717

Explanation:

Primero comencemos definiendo la variable aleatoria. Para nuestro problema, la variable aleatoria es la siguiente :

X: '' El tiempo (en horas) hasta que falle un sistema informático ''

La variable aleatoria X será entonces una variable aleatoria continua.

Sabemos que sigue una distribución exponencial con una media de 600 hs.

Esto se escribe :

X ~ ε ( λ ) (I)

En donde λ es igual a la inversa de la media. Esto se escribe :

λ =\frac{1}{E(X)}

En donde E(X) es la media de la variable. Por ende, si reemplazamos los datos del ejercicio obtenemos ⇒

λ =\frac{1}{E(X)} ⇒ λ =\frac{1}{600}

Si reemplazamos el valor de λ en (I) obtenemos :

X ~ ε (\frac{1}{600})

La función de distribución de X (por ser una variable aleatoria exponencial) es :

F_{X}(x)=P(X\leq x)=  1 - e ^ ( - λx) con x > 0 y F_{X}(x)=0 en caso contrario.

Si reemplazamos el valor de λ en la función de distribución de X obtenemos :

F_{X}(x)=P(X\leq x)=1-e^{-\frac{x}{600}}  

Dado que la variable aleatoria X se distribuye de manera exponencial, el hecho de saber que el sistema ha estado funcionando sin fallas durante 400 hs no nos aporta ninguna información sobre lo que ocurrirá después. Esta característica se conoce como propiedad de perdida de memoria de la variable aleatoria exponencial. Entonces, la probabilidad pedida se reduce a calcular :

P(X>200)    

Dado que saber que el sistema ha estado funcionando sin fallas durante 400 hs no nos dice nada sobre lo que ocurrirá instantes posteriores a esas 400 hs.

Calculamos entonces la probabilidad pedida :

P(X>200)=1-P(X\leq 200)=1-F_{X}(200)=1-(1-e^{-\frac{200}{600}})=e^{-\frac{1}{3}}=0.717

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