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Drupady [299]
2 years ago
6

Solve for the given time.

Mathematics
2 answers:
melamori03 [73]2 years ago
7 0

Answer:

45158

Step-by-step explanation:

Since the colony grows at a rate of 12°/• , after 2 years there will be 36000.

1.12 power of 2 = 45158.4 bacteria.

And you cant have part of a bacteria

you round it to

45158.

Fittoniya [83]2 years ago
5 0

Answer:

45158

Step-by-step explanation:

Since the colony grows at a rate of 12%, after 2 years there will be 36000\cdot 1.12^2 = 45158.4 bacteria. Since you probably can't have part of a bacteria, you round to 45158.

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The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E
nexus9112 [7]

Answer:

The projected enrollment is \lim_{t \to \infty} E(t)=10,000

Step-by-step explanation:

Consider the provided projected rate.

E'(t) = 12000(t + 9)^{\frac{-3}{2}}

Integrate the above function.

E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt

E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c

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2000=-8000+c

c=10,000

Therefore, E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

Now we need to find \lim_{t \to \infty} E(t)

\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

\lim_{t \to \infty} E(t)=10,000

Hence, the projected enrollment is \lim_{t \to \infty} E(t)=10,000

8 0
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