Answer
(a) $P(0)=100$
(b)
$P(10)\approx482$
$P(20)\approx999$
$P(30)\approx1168$
(c) For example, $t=100$
$P(100)\approx1199.99997$
We can clearly see that as the value of $t$ approaches $\infty$ the value of $P(t)$ is getting infinitely close to $1200$.
Yes, the graph is perfect visualization of our calculations.
Work Step by Step
According to the information provided, we have:
$P(t) = \frac{1200}{1+11e^{-0.2t}}$
(a) To determine initial amount of the population we have to calculate $P(t)$ when $t=0$:
$P(0)=\frac{1200}{1+11e^{-0.2\times0}}=\frac{1200}{1+11e^{0}}=\frac{1200}{12}=100$
(b)
$t=10$
$P(10)=\frac{1200}{1+11e^{-0.2\times10}}=\frac{1200}{1+11e^{-2}}\approx482.18\approx482$
$t=20$
$P(20)=\frac{1200}{1+11e^{-0.2\times20}}=\frac{1200}{1+11e^{-4}}\approx998.77\approx999$
$t=30$
$P(30)=\frac{1200}{1+11e^{-0.2\times30}}=\frac{1200}{1+11e^{-6}}\approx1168.14\approx1168$
(c) Let's evaluate for $t=100$
$P(100)=\frac{1200}{1+11e^{-0.2\times100}}=\frac{1200}{1+11e^{-20}}\approx1199.99997$
We can clearly see that as the value of $t$ approaches $\infty$ the value of $P(t)$ is getting infinitely close to $1200$.
Yes, the graph is perfect visualization of our calculations.
