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valentina_108 [34]
3 years ago
12

The number of E.coli bacteria cells in a pond of stagnant water can be represented by the function below, where A represents the

number of E.coli bacteria cells per 100 mL of water and t represents the time, in years, that has elapsed.
A(t)=136(1.123)^4t

Based on the model, by approximately what percent does the number of E.coli bacteria cells increase each year?


A.

60%

B.

59%

C.

41%

D.

40%
Mathematics
1 answer:
DiKsa [7]3 years ago
3 0

Answer:c 41 %

Step-by-step explanation:

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Rad 68

Step-by-step explanation:

84- 4^2=68

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Put these decimals in order from least to greatest.<br> 8.009<br> 8.90<br> 8.09<br> 8.913
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8.009, 8.09, 8.90, 8.913

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Point Q' is the image of Q(5, -3) after a translation down 2 units and right 4 units. What are the coordinates of Q'?
ra1l [238]

Answer:

A. (9,-5)

Step-by-step explanation:

Translation down is -2 to the y coordinate, which is (5,-5). Then you translate it right 4 units, which is +4 to the x coordinate, so the final coordinate is (9,-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
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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Please help with the bottom questions not filled in &lt;3
bezimeni [28]

Answer:

Given below

Step-by-step explanation:

y = 4 , m = -1 , x = 2

y = mx + b

4 = -1 (2) + b

4 = -2 + b

6 = b

4 0
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