Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Connect with a community of experts ready to provide precise solutions to your questions quickly and accurately. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
9514 1404 393
Answer:
n(t) = 20,000/(1 +99e^(-0.7033t))
Step-by-step explanation:
One way to write the logistic function is ...
n(t) = L/(1 +a·e^(-kt))
where L is the maximum value and parameters 'a' and k depend on boundary conditions.
Here, we have n(∞) = L = 20,000 and n(0) = L/(1+a) = 200. We also have n(1) = L/(1+a·e^-k) = 400.
Solving for 'a', we get ...
n(0) = 20000/(1+a) = 200
20000/200 -1 = a = 99
Solving for k, we get ...
n(1) = 20000/(1 +99e^-k) = 400
20000/400 -1 = 99e^-k = 49
e^-k = 49/99
k = -ln(49/99) ≈ 0.7033
So, the desired function is ...
n(t) = 20000/(1 +99e^(-0.7033t))

We appreciate your time. Please come back anytime for the latest information and answers to your questions. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.