Is 6xy2 is the greatest common factor of 12xy2, 36x2y2, and 42xy4

What is 2/9 divide 9/10=

Ipatiy [6.2K]

To divide fractions:

Flip the second fraction

Multiply the numerators

Multiply the denominators

Simplify

2/9*10/9

2*10=20

9*9=81

20/81

Hope this helps!!

6 0

3 years ago

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HELP DUE IN 15 MINS!

Stells [14]

Answer:

n = 18

Step-by-step explanation:

<u>Sum of interior angles of a polygon:</u>

(n - 2)180 = 160n
180n - 360 = 160n
20n = 360
n = 18

8 0

3 years ago

What formula do i use when it tells me to find the area with a given circumference? If you know please give an example as well,

liberstina [14]

Answer:

Step-by-step explanation:

A = C^2 / 4 x pi

so if the circumfernce was 10 you would square it to make it 100 and then you would do 4 x pi to get 12.57 ( rounded).

you would then just divide 100 by 12.57 = 7.96 ( rounded)

hope it helped

6 0

3 years ago

you must attend a minimum of 85% of the practices in order to play in the playoffs you have made 37 of the 42 practices will you

Deffense [45]

Yes, you'd be able to play.
37/42 = .88 = 88%
So you've attended more than 85% of the practices.

3 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

Read 2 more answers