A population of bacteria is introduced into a nutrient solution in a dish.The nutrient solution is replenished by the exact same amount each day.At first the bacteria population grows very quickly.Soon, population growth must slow down, as the nutrients become more scarce given the number of bacteria.Eventually the bacteria population stabilizes.A model for the bacteria population growth is given by $P(t)=\frac{5000}{6+e^{-5 t}}$
thousand bacteria.What is $\lim _{t \rightarrow \infty} P(t)$
?
A)6000
B)1000
C)833
D)5000
E)None of the above

Question

A population of bacteria is introduced into a nutrient solution in a dish.The nutrient solution is replenished by the exact same amount each day.At first the bacteria population grows very quickly.Soon, population growth must slow down, as the nutrients become more scarce given the number of bacteria.Eventually the bacteria population stabilizes.A model for the bacteria population growth is given by $P(t)=\frac{5000}{6+e^{-5 t}}$
 thousand bacteria.What is $\lim _{t \rightarrow \infty} P(t)$
 ?
A)6000
B)1000
C)833
D)5000
E)None of the above

Quizplus · Accepted Answer

The Answer of A population of bacteria is introduced into a nutrient solution in a dish.The nutrient solution is replenished by the exact same amount each day.At first the bacteria population grows very quickly.Soon, population growth must slow down, as the nutrients become more scarce given the number of bacteria.Eventually the bacteria population stabilizes.A model for the bacteria population growth is given by $P(t)=\frac{5000}{6+e^{-5 t}}$
 thousand bacteria.What is $\lim _{t \rightarrow \infty} P(t)$
 ?
A)6000
B)1000
C)833
D)5000
E)None of the above

A Population of Bacteria Is Introduced into a Nutrient Solution P(t)=50006+e−5tP(t)=\frac{5000}{6+e^{-5 t}}P(t)=6+e−5t5000​

A Population of Bacteria Is Introduced into a Nutrient Solution $P(t)=\frac{5000}{6+e^{-5 t}}$