Answer:
option A and B are correct.
Step-by-step explanation:
Given: [tex]p(t) = \frac{64}{1 + 11e^{-.08t} }[/tex]
Option A: [tex]\lim_{t \to \infty} \frac{64}{1 + 11e^{-.08(\infty)} } = \frac{64}{1 + 0} = 64[/tex]
Option A is true,
Option B: [tex]P(0) = \frac{64}{1 + 11} = 5.33[/tex]
Option B is also true.
Option C:
[tex]P(t + 1) = \frac{64}{1 + 11e^{-.08(t + 1)} } = \frac{64}{1 + 10.15e^{-.08t} } \\1.08 · P(t) = \frac{64 · 1.08 }{1 + 11e^{-.08t} } = \frac{69.12}{1 + 11e^{-.08t} }[/tex]
P(t + 1) ≠ 1.08 · P(t)
Option C is incorrect.
Option D: It is also incorrect, because according to option 2 earth's population will not grow exponentially without Bound.
