Question

The population of an aquatic species in a certain body of water is approximated by the logistic\\ 22,500\\ function $G(t) = \frac{22,500}{1 + 10e^{-0.5t}}$, where $t$ is measured in years.\\ Calculate the growth rate after 5 years.\\ The growth rate in 5 years is $oxed{}$\\(Do not round until the final answer. Then round to the nearest whole number as needed.)

          The population of an aquatic species in a certain body of water is approximated by the logistic\\
22,500\\
function $G(t) = \frac{22,500}{1 + 10e^{-0.5t}}$, where $t$ is measured in years.\\
Calculate the growth rate after 5 years.\\
The growth rate in 5 years is $oxed{}$\\(Do not round until the final answer. Then round to the nearest whole number as needed.)

The population of an aquatic species in a certain body of water is approximated by the logistic

22,500

function G(t) = (22,500)/(1 + 10e^-0.5t), where t is measured in years.

Calculate the growth rate after 5 years.

The growth rate in 5 years is oxed
(Do not round until the final answer. Then round to the nearest whole number as needed.)

Added by Jennifer M.

Calculus: Early Transcendentals

James Stewart 8th Edition

Thanks for your feedback!

The population of an aquatic species in a certain body of water is approximated by the logistic 22,500 functionG(t)= Ay 30,000 20,000 Calculate the growth rate after 5 years. G(I) 10.000 121620 The growth rate in 5years is[ Do not round until the final answer. Then round to the nearest whole number as needed