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3 years ago

The product of a number and ten is twelve more than the product of the number and five-eighths.

1 answer:

3 0

Let the unknown number be x.

Given,

$10 * x = 12 + \frac{5}{8} * x$

$10x = 12 + \frac{5x}{8}$

$10x - \frac{5x}{8} = 12$

$\frac{(10x*8)-5x}{8} = 12$

$80x-5x =12*8$

$75x = 96$

x = $\frac{96}{75}$ = 1.28

Hence, the number is 1.28

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curtis threw 15 darts at a dart board and 40% of his darts hit the bulls eye how many darts missed the bulls eye

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Answer:

Step-by-step explanation:

40 -100 = 60 and 60 percent of 15 is 9

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3 years ago

What is the average rate of change from -3 to 0 of the function represented by the graph?

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Average rate of change = change in y ÷ change in x

We already know the change in x because we are measuring the x values from -3 to 0.

change in x = 0-(-3) = 3

To find the change in y find the difference between the corresponding y-values when x=0 and when x=-3.

change in y = 0.5-2 = -1.5

average rate of change = -1.5/3 = -0.5

answer: -0.5

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3 years ago

A new shopping mall is considering setting up an information desk manned by one employee. Based upon information obtained from s

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

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:

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

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3 years ago

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3 years ago

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