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Ganezh [65]
3 years ago
15

A suburban specialty restaurant has developed a single drive-thru window. Customers order, pay, and pick up their food at the sa

me window. Arrivals follow a Poisson distribution, while service times follow an exponential distribution. If the average number of arrivals is 6 per hour and the service rate is 3 every 15 minutes:What is the average number of customers in the system?
A. 0.50B. 1.00C. 2.25D. 3.00E. None of the above
Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
8 0

Answer:  Option 'c' is correct.

Step-by-step explanation:

Since we have given that

Average number of arrivals = 6 per hour

Service rate = 3 every 15 minutes

Service rate is given by

\mu=\dfrac{60\times 3}{15}=12

So, the average number of customers in the system is given by

\dfrac{\mu}{\mu-\lambda}\\\\=\dfrac{6}{12-6}\\\\=\dfrac{6}{6}\\\\=1

Hence, Option 'c' is correct.

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This is due like right now, please help me!!!!
Varvara68 [4.7K]

\bold{\huge{\orange{\underline{ Solution }}}}

\bold{\underline{ Given \: Rules}}

  • <u>The </u><u>sum</u><u> </u><u>of </u><u>the </u><u>number </u><u>in </u><u>each </u><u>of </u><u>the </u><u>four </u><u>rows </u><u>is </u><u>the </u><u>same </u>
  • <u>The </u><u>sum </u><u>of </u><u>the </u><u>numbers </u><u>in </u><u>each </u><u>of </u><u>the </u><u>three </u><u>columns </u><u>is </u><u>the </u><u>same</u>
  • <u>The </u><u>sum </u><u>of </u><u>any </u><u>row </u><u>does </u><u>not </u><u>equal </u><u>the </u><u>sum </u><u>of </u><u>any </u><u>column </u>

\bold{\underline{ Let's \: Begin}}

<u>According </u><u>to </u><u>the </u><u>Second</u><u> </u><u>rule </u><u>:</u><u>-</u>

\sf{ 75+b+83=76+80+d=a+81+85+78+c+e }

\sf{ 158 + b = 156 + d = 166 + a = 78 + c + e ...(1)}

<u>According </u><u>to </u><u>the </u><u>first </u><u>rule </u><u>:</u><u>-</u><u> </u>

\sf{ 75+76+a+78 = b+80+81+c = 83+86+d+e}

\sf{ 229 + a = 161 + b + c = 168 + d + e ...(2)}

<u>From </u><u>(</u><u> </u><u>1</u><u> </u><u>)</u><u> </u><u>we </u><u>got </u><u>:</u><u>-</u>

\sf{ 158 + b = 166 + a, 156 + d = 166 + a }

\sf{ b = 166 - 158 + a,  d = 166 - 156 + a }

\sf{ b = 8 + a,  d = 10 + a ...(3)}

<u>Subsitute </u><u>(</u><u>3</u><u>)</u><u> </u><u>in </u><u>(</u><u> </u><u>2</u><u> </u><u>)</u><u> </u><u>:</u><u>-</u>

\sf{229+a = 161+8+a+c = 168+10+a+e}

\sf{ 229+a = 169+a+c = 178+a+e}

<u>We</u><u> </u><u>can </u><u>write </u><u>it </u><u>as </u><u>:</u><u>-</u>

\sf{ 229+a = 169+a+c \:or\:229+a = 178+a+e}

\sf{ c = 299-169+a-a\:or\:e = 229-178+a-a}

\sf{ c = 60 \: and \: e = 51 }

<u>Subsitute </u><u>the </u><u>value </u><u>of </u><u>c </u><u>and </u><u>e </u><u>in </u><u>(</u><u> </u><u>1</u><u> </u><u>)</u><u>:</u><u>-</u>

\sf{ 158 + b = 156 + d = 166 + a = 78 + 60 + 51 }

\sf{ 158 + b = 156 + d = 166 + a = 189}

<u>Now</u><u>, </u>

\sf{ For \: b,  158 + b = 189 }

\sf{ b = 189 - 158 }

\sf{ b = 31}

\sf{ For \: d ,  156 + b = 189 }

\sf{ d = 189 - 156 }

\sf{ d = 33}

\sf{ For \: a,  166 + a = 189 }

\sf{ a = 189 - 166 }

\sf{ a = 23 }

Hence, The value of a, b, c, d and e is 23, 31 ,60 ,33 and 51 .

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Bezzdna [24]

Answer:

10.5 is the answer

Step-by-step explanation:

4(5y−8−2)=185−15

(4)(5y)+(4)(−8)+(4)(−2)=185+−15(Distribute)

20y+−32+−8=185+−15

(20y)+(−32+−8)=(185+−15)(Combine Like Terms)

20y+−40=170

20y−40=170

20y−40+40=170+40

20y=210

20y/20=210/20

y=21/2

y=10.5

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I have two fair spinners. The first has the numbers 1, 2, 3 and 4. The second has the numbers 2, 3, 4 and 5. I spin both spinner
Snezhnost [94]

Step-by-step explanation:

There are 4 * 4 = 16 outcomes from spinning these spinners. You get an odd number by multiplying an odd number by an odd number. Since there are 2 odd numbers on both spinners the successful outcomes are 2 * 2 = 4 so probability is 4 / 16 = 1 / 4.

3 0
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There were nine red roses and four white roses in the vase. Melanie cut some more roses from her flower garden. There are now si
Eva8 [605]

Answer:

7 roses

Step-by-step explanation:

1. 9 red roses and 4 white roses

2. cuts more red roses and now has 16 so subtract

3. 16-9=7

4. She cut 7 red roses

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3 years ago
Jennifer prepares 2 kilograms of dough every hour she works at the bakery.How many hours did Jennifer work if she prepared 4 kil
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Answer:

2 kilograms dough per hour

4 kilograms produced in x hours

cross multiply 2 and x and 4 by 1

4

2 = 2 hours

therefore Jennifer worked for 2 hrs.

8 0
4 years ago
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