Question 1 A b c d PLEASE HELP ASAP!!

A board 30 metres long is to be divided into 2pieces so that the larger piece is 3 metres longer than twice the shorter piece. H

yawa3891 [41]

Answer: 9 and 21

Step-by-step explanation:

board=30m

divide by 2 pieces

one piece, we call it a

b, the other piece is b= 2a+3

a+b=30

a+2a+3=30

3a=30-3

3a=27

a=9

b=21

6 0

3 years ago

Read 2 more answers

Explain why rational number is either a terminating or repeating decimal?

dedylja [7]

A rational number is a real number such as 5 or .25. An irrational number is a non realistic number such as pi

6 0

3 years ago

Read 2 more answers

Para calcular 1/7 de 2/5

dalvyx [7]

Answer: I'm willing to help you but, I don't get what your trying to ask.

4 0

3 years ago

Strawberry's are 3.50$ a pound and you bought 18.25$ worth of strawberries

yarga [219]

The correct answer is 5.21

3 0

4 years ago

Read 2 more answers

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