5x+y=-2 direct variation yes or no

Estimate the quotient by rounding the expression to relate to a one-digit fact. Explain your thinking in the space below. 1,275

madreJ [45]

Answer:

2 remainder 100

Step-by-step explanation:

1275=1300

588=600

1300/600=2 remainder 100

6 0

2 years ago

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Divide 82 in the ratio 1:3 will give brainliest to whoever answers first

Hatshy [7]

Answer:

20.5 : 61.5

Step-by-step explanation:

1:3

1 + 3 = 4

82 ÷ 4 = 20.5

ratio of 1:3

1 × 20.5 = 20.5

3 × 20.5 = 61.5

20.5 : 61.5

6 0

2 years ago

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I need help with this question.

LenaWriter [7]

Answer:

8/20, 5/20

Step-by-step explanation:

2/5 - 4/10 6/15 8/20

1/4 - 2/8 3/12 4/16 5/20

3 0

3 years ago

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450 g in the ratio 2:13

jonny [76]

450 g in 2:13

Total proportion asked for =2+13=15

Now comparing g with ratio, or whatever

450/15=30g $\implies$ Therefore 1 in ratio depicts of 30g

So final answer = $\boxed{60g : 390g}$

$\huge\mathfrak\colorbox{white}{}$

More to know, daily bit of knowledge: The Binomial Theorem:

$\boxed{{(a+b)} ^n= \sum_{k=0}^{n}\binom{n}{k} a^{n-k}b^{k}}$

3 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:

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

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2 years ago

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