Mathematics: The time between surface finish problems in a galvanizing process is exponentially distributed with a mean of 42 hours. a single plant operates three galvanizing lines that are assumed to operate independently. round your answers to four decimal places (e.g. 98.7654). (a) what is the probability that none of the lines experiences a surface finish problem in 42 hours of operation

The time between surface finish problems in a

Ответ:

NetherisIsTheQueen

08.01.2022

Mathematics

0.0498

Step-by-step explanation:

given,

The time between surface finish problems in a galvanizing process is exponentially distributed with a mean of 42 hours.

$\lambda = \dfrac{1}{42}$

a) probability that none of the lines experiences a surface finish problem in 42 hours of operation

= $\dfrac{1}{42}\dfrac{1}{42}\dfrac{1}{42} \int_{42}^{\infty} \int_{42}^{\infty} \int_{42}^{\infty} (e^{\dfrac{-x}{42}})\ (e^{\dfrac{-y}{42}})\ (e^{\dfrac{-z}{42}})dx\ dy\ dz$

= $(\dfrac{1}{42})^3\int_{42}^{\infty} \int_{42}^{\infty} \int_{42}^{\infty} (e^{\dfrac{-x}{42}})\ (e^{\dfrac{-y}{42}})\ [\dfrac{e^{\dfrac{-z}{42}}}{\dfrac{-1}{42}}]_{42}^{\infty}dx\ dy\ dz$

= $(\dfrac{1}{42})^2\int_{42}^{\infty} \int_{42}^{\infty} (e^{\dfrac{-x}{42}})\ (e^{\dfrac{-y}{42}})\ e^{-1}dx\ dy$

= $(\dfrac{1}{42})^2\ e^{-1}[\dfrac{e^{\dfrac{-y}{42}}}{\dfrac{-1}{42}}]_{42}^{\infty}$

= $(\dfrac{1}{42}) \ e^{-1}\ e^{-1}[\dfrac{e^{\dfrac{-x}{42}}}{\dfrac{-1}{42}}]_{42}^{\infty}$

= $e^{-1}\ e^{-1}\ e^{-1}$

= $e^{-3}$

= 0.0498

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Ответ:

Joxnny8757

17.05.2023

Mathematics

find y=mx+b

y=-2/3x+4

so m= -2/3

b= 4

what you want to do is start at 4 since b and go down 2 times then right 3 times or go up 2 times and left 3 times or just plug in the equation in desmos graphing calculator

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