What is 62 times 160

A CD player is on sale for $90 at an electronics store. If the original price was $125, what was the percent of discount

krek1111 [17]

This is the working of how it is 28% discount

Find the amount of discount : $125 - $90 = $35
Divide by the original price : $35/125 = 0.28
Change in to percentage : 0.28 x 100 = 28%

Therefore the percentage discount is 28% off

:)

8 0

3 years ago

Find the equation of the circle with center (4,5) which passes through the y-intercept of the line 5x-2y+6=0

Elena-2011 [213]

Answer:

(x-4)^2 + (y-5)^2 = 20

Step-by-step explanation:

Rearrange the line equation of 5x - 2y + 6 = 0

So y = 5/2x +3

The y-intercept of the line equation is 3 as when x=0, y=3.

We know the Circle has the formula $(x-4)^2 + (y-5)^2 = ?^2$ from the question but with the intercept, we can find the entire equation as the y-intercept is (0,3) so we can substitute it into the equation to find the full equations so:

$(0-4)^2 + (3-5)^2 = ?^2$

This simplifies to:

$(-4)^2 +(-2)^2$ = $?^2$

16 + 4 = 20 = $?^2$

The answer is 20 so the equation of the circle is $(x-4)^2 + (y-5)^2 = 20$

3 0

2 years ago

What is 6/12 plus 3/12

Lilit [14]

Answer:

3/4

Step-by-step explanation:

Decimal form: 0.75

3 0

3 years ago

Read 2 more answers

A city has a population of 280,000 people. suppose that each year the population grows by 4.5% . what will the population be aft

Anastasy [175]

Given that population of the city is 280000.

Rate of growth of the population = r = 4.5% = 0.045

We have to find population after 15 years so time

t = 15

We can use growth formula to find the required population which is given by :

$A=P(1+r)^t$

Where P= starting population = 280000

r= rate of growth = 0.45

t= time = 15

now plug these values into above formula.

$A=280000(1+0.45)^{15}$

$A=280000(1.45)^{15}$

$A=280000(263.34194176)$

$A=73735743.6928$

Hence population after 15 years will be approx 73735743.

8 0

2 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

2 years ago

Read 2 more answers