Answer:
The projected enrollment is [tex]\lim_{t \to \infty} E(t)=10,000[/tex]
Step-by-step explanation:
Consider the provided projected rate.
[tex]E'(t) = 12000(t + 9)^{\frac{-3}{2}}[/tex]
Integrate the above function.
[tex]E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt[/tex]
[tex]E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c[/tex]
The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.
[tex]2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c[/tex]
[tex]2000=-\frac{24000}{3}+c[/tex]
[tex]2000=-8000+c[/tex]
[tex]c=10,000[/tex]
Therefore, [tex]E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000[/tex]
Now we need to find [tex]\lim_{t \to \infty} E(t)[/tex]
[tex]\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000[/tex]
[tex]\lim_{t \to \infty} E(t)=10,000[/tex]
Hence, the projected enrollment is [tex]\lim_{t \to \infty} E(t)=10,000[/tex]