Can someone help me translate this into a mathematical notation

What is the simplified expression for-<br> HURRY PLEASE

borishaifa [10]

Answer:

3/2

Step-by-step explanation:

you basically just simplify the exponents first, and then simplify from there.

5 0

3 years ago

a punch recipe says to mix 4 cups pineapple juice,8 cups orange juice,and 12 cups seltzer in order to make 18 servings of punch.

patriot [66]

2/3 cups of pineapple juice, 4/3 cups of orange juice and 2 cups of seltzer for 3 servings.

8 0

3 years ago

Someone help me pls

stepladder [879]

Answer:

1 rubber cube dude give since we'll

6 0

3 years ago

6.1.3 quiz pythagorean theorem question 9

Olegator [25]

Answer: 14.73

Step-by-step explanation:

The given triangle is a right angle triangle.

EF^2 + DF^2 = ED^2

The hypotenuse is |ED| while the two shorter legs are |EF| and |DF|.

We can then apply the Pythagoras Theorem to find the length of EF.

(EF)^2 + (DF)^2 = (ED)^2

(EF)^2 + (12)^2 = (19)^2

(EF)^2 + 144 = 361

(EF)^2 = 361 - 144

(EF)^2 = 217

EF = 14.73

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

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