Find the value of x.

What is 1080 miles on 15 gallons

yaroslaw [1]

Answer:

72 miles per gallons

Step-by-step explanation:

also can i have brainliest

7 0

3 years ago

PLEASE HELP ME I NEED HELP!!!!! ITS URGENT PLS HELP

Fudgin [204]

Answer:

it is blurry take it again plz

Step-by-step explanation:

7 0

3 years ago

Chin-li is building a brick wall along the front of his property. The wall will have 15 rows of bricks, with 32 bricks in each r

Finger [1]

Answer:

<h3>The solution is 480 bricks.</h3>

Step-by-step explanation:

We are given that number of rows of bricks in a wall = 15 rows.

Number of bricks in a row = 32 bricks.

In order to find the total number of bricks, we need to multiply number of rows with number of bricks in a row.

Total number of bricks in 15 rows = 15 × 32 = 480 bricks.

Therefore, will Chin-li will need 480 bricks.

<h3>The solution is 480 bricks.</h3>

3 0

3 years ago

Please help me!!!!!!!!!

nalin [4]

(y+20)=1(x+10); First, we can find the slope by using the equation y2-y1/x2-x1.

-9+20/1+10 (Accounting for double negatives).

11/11 = 1

The slope of the equation is 1, now we just need to pick a set of points from which we derived the slope from (doesn't matter which) as y1 and x1 in the point-slope equation. (y-y1)=m(x-x1)

Your final answer can either be (y+20)=1(x+10) or (y+9)=1(x-1) but in the context of this question it is (y+20)=1(x+10)

8 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

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