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goldfiish [28.3K]
2 years ago
14

Does anyone have a clue lol. I know it has something to do with proportion but to me it makes 0 sense.

Mathematics
1 answer:
katrin [286]2 years ago
8 0
57508.5 and 120, it gives you ounces for first one and then asks for pounds and it gives you the number of ounces per pound on the side so you do 920136/16 and for the second you do the same thing but with the number of feet per yard and you would do 360/3
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(f - g)(x) = ƒ(x) -g(x) = -5^x -(-3x -2) = -5^x + 3x + 2

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Assume a warehouse operates 24 hours a day. Truck arrivals follow Poisson distribution with a mean rate of 36 per day and servic
kirill [66]

The expected waiting time in system for typical truck is 2 hours.

Step-by-step explanation:

Data Given are as follows.

Truck arrival rate is given by,   α  = 36 / day

Truck operation departure rate is given,   β= 48 / day

A constructed queuing model is such that so that queue lengths and waiting time can be predicted.

In queuing theory, we have to achieve economic balance between number of customers arriving into system and that of leaving the system whether referring to people or things, in correlating such variables as how customers arrive, how service meets their requirements, average service time and extent of variations, and idle time.

This problem is solved by using concept of Single Channel Arrival with exponential service infinite populate model.

Waiting time in system is given by,

w_{s} = \frac{1}{\alpha - \beta  }

        where w_s is waiting time in system

                   \alpha is arrival rate described Poission distribution

                   \beta is service rate described by Exponential distribution

w_{s} = \frac{1}{\alpha - \beta  }

w_{s} = \frac{1}{48 - 36 }

w_{s} = \frac{1}{12 } day

w_{s} = \frac{1}{12 }  \times 24  hour        ...it is due to 1 day = 24 hours

w_{s} = 2 hours

Therefore, time required for waiting in system is 2 hours.

           

                   

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4 years ago
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