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adoni [48]
3 years ago
5

Suppose customers in a certain queue are served one at a time sequentially, that the time for each individual customer is Expone

ntial, and the average service time per customer equals 12 minutes. (Assume service times are independent of each other.) The time, in minutes, until the next 3 services are completed is Gamma (more specifically, Erlang) with what shape and rate parameters?
Mathematics
1 answer:
alexandr402 [8]3 years ago
5 0

Answer:

The shape and rate parameters are \frac{1}{12} and 3.

Step-by-step explanation:

Let <em>X</em> = service time for each individual.

The average service time is, <em>β</em> = 12 minutes.

The random variable follows an Exponential distribution with parameter, \lambda=\frac{1}{\beta}=\frac{1}{12}.

The service time for the next 3 customers is,

<em>Z</em> = <em>X</em>₁ + <em>X</em>₂ + <em>X</em>₃

All the <em>X</em>_{i}'s are independent Exponential random variable.

The sum of independent Exponential random variables is known as a Gamma or Erlang random variable.

The random variable <em>Z</em> follows a Gamma distribution with parameters (<em>α</em>, <em>n</em>).

The parameters are:

\alpha =\lambda=\frac{1}{12}\\n=3

Thus, the shape and rate parameters are \frac{1}{12} and 3.

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Katena32 [7]

The question is incomplete. The complete question is :

The population of a certain town was 10,000 in 1990. The rate of change of a population, measured in hundreds of people per year, is modeled by P prime of t equals two-hundred times e to the 0.02t power, where t is measured in years since 1990. Discuss the meaning of the integral from zero to twenty of P prime of t, d t. Calculate the change in population between 1995 and 2000. Do we have enough information to calculate the population in 2020? If so, what is the population in 2020?

Solution :

According to the question,

The rate of change of population is given as :

$\frac{dP(t)}{dt}=200e^{0.02t}$  in 1990.

Now integrating,

$\int_0^{20}\frac{dP(t)}{dt}dt=\int_0^{20}200e^{0.02t} \ dt$

                    $=\frac{200}{0.02}\left[e^{0.02(20)}-1\right]$

                   $=10,000[e^{0.4}-1]$

                    $=10,000[0.49]$

                    =4900

$\frac{dP(t)}{dt}=200e^{0.02t}$

$\int1.dP(t)=200e^{0.02t}dt$

$P=\frac{200}{0.02}e^{0.02t}$

$P=10,000e^{0.02t}$

$P=P_0e^{kt}$

This is initial population.

k is change in population.

So in 1995,

$P=P_0e^{kt}$

   $=10,000e^{0.02(5)}$

   $=11051$

In 2000,

$P=10,000e^{0.02(10)}$

   =12,214

Therefore, the change in the population between 1995 and 2000 = 1,163.

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3 years ago
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raketka [301]
5x-3-7=9+2x-16
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3 years ago
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UNO [17]

Answer:

53.33 mph

Step-by-step explanation:

Speed is the rate of change of distance with time. Mathematically,

speed = distance/time

and is measured in miles per hour or meter per second.

To get the average speed over the entire trip, we must get the total time  spent on trip and the total distance, then apply the speed formula

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Total distance covered

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Step-by-step explanation:

△BAC ≅ △EDF (given)

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algol [13]

Answer:

3

Step-by-step explanation:

6 0
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