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kicyunya [14]
3 years ago
9

The length of time (in months after maintenance) until failure of a bank’s surveillance television equipment has a weibull distr

ibution with α = 2 and β = 80 months. If the bank wants the probability of a breakdown before the next scheduled maintenance to be 0.02, how frequently should the equipment receive periodic maintenance?

Mathematics
1 answer:
miss Akunina [59]3 years ago
5 0

Answer:

There should be a periodic maintanance for every 1.27months

Step-by-step explanation:

The step by step calculation explaining how frequent the equipment should receive periodic maintenance <em><u>(Which are the activities carried out on equipment based on a stipulated time interval for the purpose of maintaining a smooth operation of a machine or equipment or on other asset.)</u></em> can be seen in the attached image below.

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Your friend is supposed to meet you at the beach between 12pm and 1:30pm. If you arrive at 12:25pm and you do not see you friend
Alexus [3.1K]

The probability that you will have missed your friend is 27.78%

<h3>How to determine the probability?</h3>

The time of meeting is given as:

Time = 90 minutes (i.e 12pm and 1:30pm)

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Arrival = 12:25pm

If you missed your friend, it means that your friend arrives earlier.

So, the time spent by your friend is:

Friend= 12:25 - 12 = 25 minutes

The probability that you will have missed your friend is:

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Read 2 more answers
En una hoja de papel cuyo perímetro es de 96 centímetros, se quiere imprimir un volante de manera que el área impresa sea un rec
elixir [45]

Answer:

El perímetro de la región impresa es 72 cm y su área es 288 cm².  

Step-by-step explanation:

1. Tenemos el perímetro de la hoja de papel:

P₁ = 96 cm = 2l₁ + 2a₁  (1)  

Como sabemos el margen superior, inferior, izquierdo y derecho podemos encontrar la relación entre el largo y ancho del rectángulo interno (región impresa) con el largo (l) y ancho (a) del rectángulo externo (hoja de papel):      

l_{2} = l_{1} - (m_{s} + m_{i}) = l_{1} - (3 cm + 2 cm) = l_{1} - 5 cm  (2)            

a_{2} = a_{1} - (m_{d} + m_{iz}) = a_{1} - (2 cm + 5 cm) = a_{1} - 7 cm   (3)    

El perímetro del rectángulo interno es:

P_{2} = 2l_{2} + 2a_{2}    (4)

Introduciendo la ecuación (2) y (3) en (4):

P_{2} = 2l_{2} + 2a_{2} = 2(l_{1} - 5 cm) + 2(a_{1} - 7 cm) = 2l_{1} + 2a_{1} - 10 cm - 14 cm = 96 cm - 24 cm = 72 cm  

Por lo tanto el perímetro del rectángulo interno (región impresa) es 72 cm.

 

2. Ahora para encontrar el área rectángulo interno debemos encontrar el largo y ancho del mismo, sabiendo que:

l_{2} = 2a_{2}     (5)

Introduciendo (5) en (4):

P_{2} = 2l_{2} + 2a_{2} = 2*2a_{2} + 2a_{2} = 6a_{2}

a_{2} = \frac{P_{2}}{6} = \frac{72 cm}{6} = 12 cm

l_{2} = 2a_{2} = 2*12 cm = 24 cm

Entonces el área es:

A_{2} = l_{2}*a_{2} = 12 cm*24 cm = 288 cm^{2}

Por lo tanto el área del rectágulo interno (región impresa) es 288 cm².      

Espero que te sea de utilidad!  

7 0
2 years ago
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