Mathematics, 22.04.2020 01:41 murphyscott794

The count in a culture of bacteria was 600 after 2 hours and 38,400 after 6 hours. (a) What is the relative rate of growth of the bacteria population? Express your answer as a percentage. (Round your answer to the nearest whole number.) % (b) What was the initial size of the culture? (Round your answer to the nearest whole number.) bacteria (c) Find a function that models the number of bacteria n(t) after t hours. (Enter your answer in the form n0ert. Round your n0 value to the nearest whole number. Round your r value to two decimal places.) n(t) = (d) Find the number of bacteria after 4.5 hours. (Round your answer to the nearest hundred.) bacteria (e) After how many hours will the number of bacteria reach 75,000? (Round your answer to two decimal places.) hr

Answers: 3

Show answers

Answers

Answer from: dkhakdoust

The bacteria has an exponential growth rate and the population of the

bacteria increases rapidly with time.

(a) The relative rate of growth is $\underline{\dfrac{ln(64)}{4}}$ .

(b) The initial size of the culture is 75 bacteria.

(d) The number of bacteria after 4.5 hours is approximately 8,100 bacteria.

(e) The number of hours after which the bacteria will reach 75,000 is approximately 6.64 hours.

Reasons:

The count in the culture of bacteria after 2 hours = 600

The count after 6 hours = 38,400

(a) The relative rate of growth, k is given by the formula;

$y = \mathbf{C \cdot e^{k \cdot t}}$

Therefore, we get;

$600 = C \cdot e^{k \times 2}$

$38,400= C \cdot e^{k \times 6}$

Which gives;

$\dfrac{38,400}{600} = \mathbf{\dfrac{C \cdot e^{k \times 6}}{C \cdot e^{k \times 2}}}$

$64= \dfrac{e^{k \times 6}}{e^{k \times 2}} = e^{k \times 6- k \times 2} = e^{4\cdot k}$

$\mathbf{e^{4\cdot k}} = 64$

$ln\left(e^{4\cdot k}) = ln(64)$

4·k = ㏑(64)

$The \ relative \ rate \ of \ growth, \, \underline{ k = \dfrac{ln(64)}{4}}$

(b) The initial size of the culture, C, is given by the relation;

$C = \mathbf{\dfrac{y}{e^{k \cdot t}}}$

Therefore, we get;

$C = \dfrac{600}{e^{\dfrac{ln(64)}{4} \times 2}} = \mathbf{ 75}$

The initial size of the culture, C = 75

Where:

y = n(t)

C = n₀

k = r

We get;

$n(t) = \mathbf{n_0 \cdot e^{r \cdot t}}$

n₀ = C = 75

$r = k = \dfrac{ln(64)}{4} \approx 1.04$

Which gives the function as follows; $\underline{n(t) = 75 \cdot e^{1.04 \cdot t}}$

(d) The number of bacteria after 4.5 hours is $n(4.5) = 75 \cdot e^{1.04 \times 4.5}$ ≈ 8,100 bacteria

(e) At n(t) = 75,000, we have;

$n(t) = 75,000 = \mathbf{75 \cdot e^{1.04 \times t}}$

$t = \dfrac{ln\left(\dfrac{75,000}{75} \right)}{1.04} \approx \mathbf{6.64}$

The time at which the bacteria population will reach 75,000, t ≈ 6.64 hours.

Learn more here:

link

Answer from: shhells

A) 2033%

B) 75 bacteria

C) n(t) = 75(1 + 1.83%)^t R = 1.83%

D) 81.4 bacteria

E) 348.83 seconds

Step-by-step explanation:

Given that the count in a culture of bacteria was 600 after 2 hours and 38,400 after 6

A) growth (600 - 0)/2 = 300 bacteria/s

38400/6 = 6400 bacteria/s

Relative growth=(6400-300)/300 ×100

= 6100/300 × 100

= 2033%

B) initial size of the culture

Using exponential equation

P = I( 1 + R)^t

Where I = initial size

R = rate

600 = I(1 + R)^2

Log both sides

Log600 = log I(1+R)^2

2.778 = logI + 2log(1 +R)

2log (1+R) = 2.778 - logI (1)

Also,

38400 = I(1 + R)^6

Log both sides

Log 38400 = logI + 6log(1+R)

6Log(1+R) = 4.584 - logI (2)

Divide equation 2 by 1

6/2 = (4.584 - logI)/(2.778 - logI)

Cross multiply

16.668 - 6logI = 9.168 - 2logI

6logI - 2logI = 16.668 - 9.168

4logI = 7.5

LogI = 7.5/4

LogI = 1.875

I = 74.98 = 75 bacteria

C) A function that models the number of bacteria n(t) after t hours.

If I = 75 bacteria

Then n(t) = 75(1 + R)^t

600 = 75(1+R)^2

8 = (1+R)^2

Log both sides

Log8 = 2log(1+R)

0.903/2 = log(1+R)

0.45 = log(1+R)

1 + R = 2.83

R = 1.83%

The model function is therefore

n(t) = 75(1 + 1.83%)^t

D) the number of bacteria after 4.5 hours

n(t) = 75(1.02)^4.5

n(t) = 81.4 bacteria

E) After how many hours will the number of bacteria reach 75,000

n(t) = 75(1.02)^t

75000 = 75(1.02)^t

1000 = 1.02^t

Log both sides

Log 1000 = tlog 1.02

3 = 0.0086t

t = 348.83 seconds

Answer from: Quest

step-by-step explanation:

12 minutes

Answer from: Quest

Ralph is earning 2.8% commission

Another question on Mathematics

Mathematics, 21.06.2019 17:00

How do you find the area of a triangle

Answers: 1

Answer

Mathematics, 21.06.2019 18:30

Two angles are complementary. the large angle is 36 more than the smaller angle. what is the measure of the large angle?

Answers: 1

Answer

Mathematics, 21.06.2019 19:30

If you could answer these your a life saver

Answers: 1

Answer

Mathematics, 21.06.2019 20:40

Which questions would most a reader identify the author's purpose for writing? check all that apply. what type of text is this? what type of reviews did the book receive? how many copies of the book have been sold? what is the central idea? how does this text make one feel?

Answers: 2

Answer

You know the right answer?

The count in a culture of bacteria was 600 after 2 hours and 38,400 after 6 hours. (a) What is the r...

Questions