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3 years ago

Can you help me please?

Mathematics

1 answer:

IrinaK [193]3 years ago

7 0

Answer:

down bellow-

Step-by-step explanation:k ask ur parent/teacher

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7. The length of time (in months after maintenance) until failure of a bank’s surveillance television equipment follows a Weibul

cupoosta [38]

Answer:

The frequency by which the equipment should receive periodic maintenance is $x= 13.59$

Step-by-step explanation:

A Wellbull distribution is mathematically represented as

$f(x,\alpha, \beta ) = \left \{ { {\frac{\alpha }{\beta^{\alpha } }x^{\alpha -1} e^{-[\frac{x}{\beta } ]^\alpha} \ \ \ \ \ \ \\ \\ , x \ge0} \atop {0 \ \ \ \ \ \ \ \ Otherwise }} \right.$

Where x is the frequency that the equipment should receive periodic maintenance

probability of a breakdown before the next scheduled maintenance is mathematically represented as

$P(x) = \int\limits^x_0 {\frac{\alpha }{\beta^{\alpha } } x^{\alpha -1 } e^{-[\frac{x}{\beta } ]^{\alpha }} } \, dx$

Substituting 2 for $\alpha$ and 60 for $\beta$ 0.05 for P(x)

$0.05 = \int\limits^x_0 {\frac{2}{60^2} x^{2-1} e^{-[\frac{x}{60}]^2 } \, dx$

$0.05 = \frac{2}{3600}\int\limits^x_0 {xe^{[-\frac{x^2}{3600} ]}} \, dx$

Let $v = - \frac{x^2}{3600}$

=> $dv = \frac{-xdx}{1800}$

=> $-xdx = 1800\ dv$

Multiply both sides by minus

=> $xdx = -1800dv$

Substituting this into the equation

$0.05 = \frac{2}{3600} \int\limits^x_0 {-1800 e^u} \, du$

Integrating and substituting back for x

$0.05 = -\frac{1800}{1800} [e^{[-\frac{x^2}{3600} ]}]\left {{x} \atop {0}} \right.$

substituting for the range of x and 0

$0.05 = - [e^{-[\frac{x^2}{3600}] } -1]$

$0.05 = -e^{[-\frac{x^2}{3600} ]} +1$

$0.05 -1 = -e^{-\frac{x^2}{3600} }$

$-0.95 = -e^{[-\frac{x^2}{3600} ]}$

$0.95 = e^{[-\frac{x^2}{3600} ]}$

Taking ln of both sides

$-0.05129 = - \frac{x^2}{3600}$

making x the subject of the formula

$x = \sqrt{0.05129 * 3600 }$

$x= 13.59$

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