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2 years ago

What is halfway bewteen -23 and 23

1 answer:

7 0

Half way in between -23 & 23 would be 0 because if you start on both sides and carefully go towards the middle evenly you will get zero on a number line

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A company has a budget of $280,000 for computing equipment. Tablets cost $500 each, laptops cost $2000 each, and servers cost $5

Kazeer [188]

Answer:

The number of tablets 200

the number of laptops 40

and, the number of servers 20

Step-by-step explanation:

Let the number of tablets be 'x'

the number of laptops be 'y'

and, the number of servers be 'z'

According to the given question

Total budget = $280,000

Tablets cost = $500 each

laptops cost = $2000 each

servers cost = $5000 each

Thus,

500x + 2000y + 5000z = 280,000 ..............(1)

also,

x = 5y ................(2)

and,

y = 2z

z = $\frac{\textup{y}}{\textup{2}}$ ............(3)

substituting y from 2 and 3 in equation 1, we get

(500 × 5y) + (2000 × y) + (5000 × $\frac{\textup{y}}{\textup{2}}$ ) = 280,000

2500y + 2000y + 2500y = 280,000

7000y = 280,000

y = 40

substituting y in equation 2, we get

x = 5 × 40 = 200

substituting y in equation 3 we get

z = $\frac{\textup{40}}{\textup{2}}$

z = 20

hence,

the number of tablets 200

the number of laptops 40

and, the number of servers 20

7 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

2 years ago

Read 2 more answers

Shiloh has to earn at least $200 to meet her fundraising goal. She has only 100 cakes that she plans to sell at 5 dollars each.

kakasveta [241]

Here is the inequality you can enumerate:

5x >= 200, this means the money you sold for cakes must be larger or equal to your goal

x <= 100, this means you cannot sell more than 100 cakes, because you only have 100 cakes

Solve for them, you will get: 40 <= x <= 100

5 0

3 years ago

I’m putting the right answer which is “6” but the computer keeps telling me I’m wrong, I don’t understand, what else is there to

iragen [17]

You may have have forgotten to find the measure of A. Which would by using your 6 that you got would be 19.

16X-6=90

96=16X

X=6

8X6-29

48-29

3 0

2 years ago

#1. WILL REWARD BRAINLIEST

Gemiola [76]

c is the answer for 1'

6 0

2 years ago