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3 years ago

-2x+5=4x-7 solve for x. Show all work

Mathematics

2 answers:

Neko [114]3 years ago

7 0

Answer:

2=x

Step-by-step explanation:

-2x+5=4x-7

+2x +2x

5=6x-7

+7 +7

12=6x

2=x

dmitriy555 [2]3 years ago

6 0

-2x+5=4x-7

+7 +7

-2x+12=4x

+2x +2x

12=6x

÷6 ÷6

2=x

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Step-by-step explanation:

Notation

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$p_x$ represent the probability that the employee is busy

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

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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".

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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

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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:

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

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

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

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