Which of the following equations can represent the function graphed below? Select all the possible

seraphim [82]

I hope you speak english, because although i understood your question I don't know enough spanish to respond in it. You multiply polynomials by multiplying each part of one polynomial by all the other parts of the others, and then adding it all. For example: (ax+b) * (x+c) = (ax*x)+(ax*c)+(b*x)+(b*c) = $a x^{2} + acx + bx+bc$

Comprehende?

7 0

3 years ago

Uyuuuuuyyyuuuyyyyyy answer

Lapatulllka [165]

Answer:

e.b=20

f. t=-20

g. x=2

h. h=3

3 0

3 years ago

How would I find all real solutions in x^4/3 -10^2/3 +21= 0

Oxana [17]

Answer:

Step-by-step explanation:

This equation looks complicated.We have to make it easier

let's say x^2/3 = t and x^4/3 = t^2

t^2-10t+21=0 [ we can factorize this equation as a (t-3)(t-7) ]

(t-3)(t-7)=0 [ that means , t can be 3 or 7 ]

But don't forget we have to find x not t so,

t=x^2/3=3 ∛ x^2 = 3 x^2 = 9 x=3 or x= -3

t=x^2/3=7 ∛x^2 = 7 x^2 = 343 x ~18.5 or x ~ -18.5

7 0

2 years ago

A volunteer group picked up 3 times as many plastic bottles as cans in their community. The number of papers they picked up was

sweet-ann [11.9K]

try 4,320 for your answer

4 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

3 years ago

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