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zalisa [80]
3 years ago
12

The prices of tickets for the local rock concert were $16 for Friday night and $28 for Saturday night. The total attendance for

the two nights was 1090. If the total receipts were $24,232, how many people attended Saturday night?
Mathematics
2 answers:
Leni [432]3 years ago
8 0

Answer:C

Step-by-step explanation:

The prices of tickets for the local rock concert were $16 for Friday night and $28 for Saturday night. The total attendance for the two nights was 1090. If the total receipts were $24,232, how many people attended Saturday night?

A) 251 people

B) 524 people

C) 566 people

D) 1132 people

ASHA 777 [7]3 years ago
4 0
Let x: number of tickets that sold on Friday, y: number of tickets that sold on Saturday night

16x+28y=24232
x+y=1090
x=1090-y,
so 16(1090-y)+28y=24232
17440-16y+28y=24232
12y=6792
y=566. As a result, there will be 566 tickets sold on Saturday night. Hope it help!
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Answer:

a) P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33 and that represent the 33%

b) p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66

c) L_s =\frac{20}{30-20}=\frac{20}{10}=2 people

d) L_q =\frac{20^2}{30(30-20)}=1.333 people

e) W_s =\frac{1}{\lambda -\mu}=\frac{1}{30-20}=0.1hours

f) W_q =\frac{\lambda}{\mu(\mu -\lambda)}=\frac{20}{30(30-20)}=0.0667 hours

Step-by-step explanation:

Notation

P represent the probability that the employee is idle

p_x represent the probability that the employee is busy

L_s represent the average number of people receiving and waiting to receive some information

L_q represent the average number of people waiting in line to get some information

W_s represent the average time a person seeking information spends in the system

W_q represent the expected time a person spends just waiting in line to have a question answered

This an special case of Single channel model

Single Channel Queuing Model. "That division of service channels happen in regards to number of servers that are present at each of the queues that are formed. Poisson distribution determines the number of arrivals on a per unit time basis, where mean arrival rate is denoted by λ".

Part a

Find the probability that the employee is idle

The probability on this case is given by:

In order to find the mean we can do this:

\mu = \frac{1question}{2minutes}\frac{60minutes}{1hr}=\frac{30 question}{hr}

And in order to find the probability we can do this:

P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33 and that represent the 33%

Part b

Find the proportion of the time that the employee is busy

This proportion is given by:

p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66

Part c

Find the average number of people receiving and waiting to receive some information

In order to find this average we can use this formula:

L_s= \frac{\lambda}{\lambda -\mu}

And replacing we got:

L_s =\frac{20}{30-20}=\frac{20}{10}=2 people

Part d

Find the average number of people waiting in line to get some information.

For the number of people wiating we can us ethe following formula"

L_q =\frac{\lambda^2}{\mu(\mu-\lambda)}

And replacing we got this:

L_q =\frac{20^2}{30(30-20)}=1.333 people

Part e

Find the average time a person seeking information spends in the system

For this average we can use the following formula:

W_s =\frac{1}{\lambda -\mu}=\frac{1}{30-20}=0.1hours

Part f

Find the expected time a person spends just waiting in line to have a question answered (time in the queue).

For this case the waiting time to answer a question we can use this formula:

W_q =\frac{\lambda}{\mu(\mu -\lambda)}=\frac{20}{30(30-20)}=0.0667 hours

6 0
2 years ago
Read 2 more answers
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