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2 years ago

How would you explain how to write 3610 as a decimal

1 answer:

4 0

The way that I would explain it is that you have 3610 and it is a whole number there basically 3 numbers so the 3 would be in the ones place 6 in the tenth and 1 in the hundredths place and your answer is 3.61 or another is .361

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Find the measure of 4

schepotkina [342]

Answer:

147º

Step-by-step explanation:

180-33=147

5 0

3 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

2 years ago

Read 2 more answers

6. Which is a unit rate?

Taya2010 [7]

it's A

Step-by-step explanation:

describes 55 per unit/the hour

3 0

3 years ago

The inverse variation equation shows the relationship between wavelength in meters, x, and frequency, y.

wolverine [178]

Answer:

$\lambda=10^{-10}\ m$

Step-by-step explanation:

Given that,

The frequency of x-rays, $f=3\times 10^{18}\ Hz$

We need to find the wavelength for x-rays. The relation between wavelength, frequency and speed of light is given by :

$v=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^{8}}{3\times 10^{18}}\\\\=10^{-10}\ m$

So, the wavelength for x-rays is $10^{-10}\ m$ .

7 0

3 years ago

What is 41x plus 21x

Free_Kalibri [48]

The answer to that question is 62x ()//‘’::;;

4 0

3 years ago