Solve the problem.
-The logistic growth model $P ( t ) = \frac { 1,450 } { 1 + 28 e ^ { - 0.313 t } }$ represents the population of a bacterium in a culture tube after t hours. What was the initial amount of bacteria in the population?

Question

Quizplus · Accepted Answer

The initial amount of bacteria in the population is found by substituting $t = 0$ into the logistic growth model. So, $P(0) = \frac{1,450}{1 + 28e^{0}} = \frac{1,450}{1 + 28} = \frac{1,450}{29} = 50$.

Solve the Problem P(t)=1,4501+28e−0.313tP ( t ) = \frac { 1,450 } { 1 + 28 e ^ { - 0.313 t } }P(t)=1+28e−0.313t1,450​

Solve the Problem $P ( t ) = \frac { 1,450 } { 1 + 28 e ^ { - 0.313 t } }$