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ohaa [14]
3 years ago
9

Round to the nearest ten thousand 905154

Mathematics
1 answer:
kolbaska11 [484]3 years ago
5 0
You want to round 905,154 to the nearest ten-thousands place. The ten-thousands place in your number is shown by the bold underlined digit here:

9<em><u>0</u></em>5,154

    To round 905,154 to the nearest ten-thousands place...

    The digit in the ten-thousands place in your number is the 0. To begin the rounding, look at the digit one place to the right of the 0, or the 5, which is in the thousands place.

    Since the 5 is greater than or equal to 5, we'll round our number up by

        Adding 1 to the 0 in the ten-thousands place, making it a 1.

    and by changing all digits to the right of this new 1 into zeros.

The result is: 910,000.

So, 905,154 rounded to the ten-thousands place is 910,000.

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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
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Answer:

a) k=2.08 1/hour

b) The exponential growth model can be written as:

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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

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