Answer:
The initial population is 200
The population after 10 years is 687
Step-by-step explanation:
Given
[tex]P(t) =\frac{800}{1+ 3e^{-0.29t.}}[/tex]
Solving (a): The initial population
Here:
[tex]t = 0[/tex]
Substitute 0 for t in P(t)
[tex]P(0) =\frac{800}{1+ 3e^{-0.29*0}}[/tex]
[tex]P(0) =\frac{800}{1+ 3e^{0}}[/tex]
[tex]P(0) =\frac{800}{1+ 3*1}[/tex]
[tex]P(0) =\frac{800}{1+ 3}[/tex]
[tex]P(0) =\frac{800}{4}[/tex]
[tex]P(0) =200[/tex]
The initial population is 200
Solving (b): Population after 10 years.
Here
[tex]t = 10[/tex]
Substitute 10 for t in P(t)
[tex]P(10) =\frac{800}{1+ 3e^{-0.29*10}}[/tex]
[tex]P(10) =\frac{800}{1+ 3e^{-2.9}}[/tex]
[tex]P(10) =\frac{800}{1+ 3*0.055}[/tex]
[tex]P(10) =\frac{800}{1+ 0.165}[/tex]
[tex]P(10) =\frac{800}{1.165}[/tex]
[tex]P(10) =687[/tex] -- approximated.
The population after 10 years is 687