student enrollment the registrar of kellogg university estimates that the student enrollment in the…

student enrollment the registrar of kellogg university estimates that the student enrollment in the continuing education division will grow at the rate of n(t)=2500(1 + 0.2t)^(-3/2) students/year, t years from now. if the current student enrollment is 1,000, find an expression giving the enrollment t years from now. n(t)= what will be the enrollment 5 years from now? (round your answer to the nearest student.) students

student enrollment the registrar of kellogg university estimates that the student enrollment in the continuing education division will grow at the rate of n(t)=2500(1 + 0.2t)^(-3/2) students/year, t years from now. if the current student enrollment is 1,000, find an expression giving the enrollment t years from now. n(t)= what will be the enrollment 5 years from now? (round your answer to the nearest student.) students

Answer

Explanation:

Step1: Integrate $N'(t)$

We know that $N(t)=\int N'(t)dt=\int 2500(1 + 0.2t)^{-3/2}dt$. Let $u = 1+0.2t$, then $du=0.2dt$ and $dt=\frac{du}{0.2}=5du$. So $N(t)=2500\int u^{-3/2}\times5du = 12500\int u^{-3/2}du$. Using the power - rule for integration $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $N(t)=12500\times\frac{u^{-3/2 + 1}}{-3/2+1}+C=12500\times\frac{u^{-1/2}}{-1/2}+C=-25000u^{-1/2}+C=- \frac{25000}{\sqrt{1 + 0.2t}}+C$.

Step2: Find the value of $C$

Given that when $t = 0$, $N(0)=1000$. Substitute $t = 0$ and $N(0)=1000$ into $N(t)=-\frac{25000}{\sqrt{1 + 0.2t}}+C$. We get $1000=-\frac{25000}{\sqrt{1+0}}+C$, so $1000=-25000 + C$, and $C = 26000$. Thus, $N(t)=26000-\frac{25000}{\sqrt{1 + 0.2t}}$.

Step3: Calculate $N(5)$

Substitute $t = 5$ into $N(t)=26000-\frac{25000}{\sqrt{1+0.2\times5}}$. First, calculate the denominator: $\sqrt{1 + 0.2\times5}=\sqrt{1 + 1}=\sqrt{2}$. Then $N(5)=26000-\frac{25000}{\sqrt{2}}\approx26000 - 17678=8322$.

Answer:

$N(t)=26000-\frac{25000}{\sqrt{1 + 0.2t}}$ 8322