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Charra [1.4K]
3 years ago
6

Help pleaseee?? Yuhh

Mathematics
2 answers:
Lemur [1.5K]3 years ago
7 0
The last one because x/y (4x/1) flips to (1/4x)
Lynna [10]3 years ago
4 0

Answer : Copy the question into brainly search bar and get youre answer.

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A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium d
nikitadnepr [17]

Answer:

a) k=2.08 1/hour

b) The exponential growth model can be written as:

P(t)=Ce^{kt}

c) 977,435,644 cells

d) 2.033 billions cells per hour.

e) 2.81 hours.

Step-by-step explanation:

We have a model of exponential growth.

We know that the population duplicates every 20 minutes (t=0.33).

The initial population is P(t=0)=58.

The exponential growth model can be written as:

P(t)=Ce^{kt}

For t=0, we have:

P(0)=Ce^0=C=58

If we use the duplication time, we have:

P(t+0.33)=2P(t)\\\\58e^{k(t+0.33)}=2\cdot58e^{kt}\\\\e^{0.33k}=2\\\\0.33k=ln(2)\\\\k=ln(2)/0.33=2.08

Then, we have the model as:

P(t)=58e^{2.08t}

The relative growth rate (RGR) is defined, if P is the population and t the time, as:

RGR=\dfrac{1}{P}\dfrac{dP}{dt}=k

In this case, the RGR is k=2.08 1/h.

After 8 hours, we will have:

P(8)=58e^{2.08\cdot8}=58e^{16.64}=58\cdot 16,852,338= 977,435,644

The rate of growth can be calculated as dP/dt and is:

dP/dt=58[2.08\cdot e^{2.08t}]=120.64e^2.08t=2.08P(t)

For t=8, the rate of growth is:

dP/dt(8)=2.08P(8)=2.08\cdot 977,435,644 = 2,033,066,140

(2.033 billions cells per hour).

We can calculate when the population will reach 20,000 cells as:

P(t)=20,000\\\\58e^{2.08t}=20,000\\\\e^{2.08t}=20,000/58\approx344.827\\\\2.08t=ln(344.827)\approx5.843\\\\t=5.843/2.08\approx2.81

3 0
3 years ago
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