Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Get quick and reliable answers to your questions from a dedicated community of professionals on our platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
Given that the population can be represented by the equation;
[tex]P(t)=\frac{2t^2+75}{2t^2+150}[/tex]The current population (Initial population) is the population at time t=0;
Substituting;
[tex]t=0[/tex][tex]\begin{gathered} P(0)=\frac{2t^2+75}{2t^2+150}=\frac{2(0)^2+75}{2(0)^2+150}=\frac{75}{150} \\ P(0)=0.5\text{ million} \end{gathered}[/tex]Therefore, the current population of the habitat is;
[tex]0.5\text{ million}[/tex]The long term population would be the population as t tends to infinity;
[tex]\begin{gathered} \lim _{t\to\infty}P(t)=\frac{2t^2+75}{2t^2+150}=\frac{2(\infty)^2+75}{2(\infty)^2+150}=\frac{\infty}{\infty} \\ \lim _{t\to\infty}P(t)=\frac{4t}{4t}=1 \end{gathered}[/tex]Therefore, the long term population of the habitat is;
[tex]P(\infty)=1\text{ million}[/tex]
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
