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Mademuasel [1]
3 years ago
9

If Brian is driving 50 miles per hour. How far did he drive?

Mathematics
2 answers:
Vlad [161]3 years ago
4 0

The answer is 50 miles

RSB [31]3 years ago
4 0
Brian is driving 50 miles.
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Need help nowwww!!! With 1 threw 8
JulijaS [17]

1. B

2. B

3.C

4.B

5.B

6.A

6 0
3 years ago
A new shopping mall is considering setting up an information desk manned by one employee. Based upon information obtained from s
quester [9]

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
3 years ago
Read 2 more answers
Write 7/8 as a percent to the nearest tenth of the percent
Fiesta28 [93]

7/8 = 87.5%


I'm guessing it is meant to say round to the nearest 10th of a percent

If the question actually says something different comment so i can fix it

7 0
3 years ago
A drug company is testing a new blood pressure medication. 100 patients with high blood pressure will be divided into two groups
ohaa [14]
I would say C.
A and B make no sense, you want to divide the group randomly because you want to test the new medicine. D is a bit more logical than A and B but still can't, because that isn't random.
4 0
3 years ago
Read 2 more answers
Are all parallelograms similar
andreev551 [17]
Answer. No , we can say that all parallelograms are similar , because if we consider a rectangle and a square side by side , they are not similar to each other , we can take rhombus and square as another example . There are many types of parallelograms , with have something uncommon among themselves
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