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3 years ago

There are 3,785 milliliters in 1 gallon, and there are 4 quarts in 1 gallon. For each question, explain or show reasoning.

Mathematics

2 answers:

irakobra [83]3 years ago

5 0

11356.235 946.353 is your answer

Novay_Z [31]3 years ago

4 0

Answer:

There are 11,355 milliliters in 3 gallons, and 946.25 milliliters in 1 quart.

Step-by-step explanation:

If there are 3,785 milliliters in 1 gallon, multiply each side by 3 because you want the product to equal 3 gallons.3,785 times 3 is equal to 11,355, and 1 times 3 is 3. So, 3 gallons are equal to 11,355 milliliters. If there are 3,785 milliliters in 1 gallon, and there are 4 quarts in 1 gallon, that means that there are 3,785 milliliters in 4 quarts. Now, you want to find the amount of milliliters in 1 quart so you should divide both the 3,785 and the 4 by 4. So, there are 946.25 milliliters in 1 quart.

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

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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:

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

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Step-by-step explanation:

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