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Burka [1]
3 years ago
14

I need this to be solved plzzzz

Mathematics
1 answer:
tatiyna3 years ago
4 0

Answer:

ln2

Step-by-step explanation:

n --> infinity

Integral of 1/N as N varies from n+1 to 2n

Integral of 1/N = ln(N)

Upper limit - lower limit

ln(2n) - ln(n + 1)

ln[2n/(n + 1)]

ln[2/(1 + 1/n)]

As n --> infinity,

1/n --> 0

ln[2/(1 + 1/n)] --> ln2

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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

2000=-\frac{24000}{3}+c

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c=10,000

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

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

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