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Sagot :
Answer:
The answer is "19.71 million".
Step-by-step explanation:
[tex]\to P(t) = \frac{19.71}{1+61.22 e^{(-0.03513t)}}[/tex]
where
[tex]\to p= \text{population in million} \\\\\to t= \text{time in years}\\[/tex]
when t is increases, then [tex]61.22 \ e^{-0.035 t}=\frac{61.22}{e^{0.035t}}[/tex] when decrease so that p(t) continually grows in size to reach the optimum population which you find if t is quite high [tex]t \to \infty[/tex]
[tex]\to P_{max} = \lim_{ \ t \to \infty} \frac{19.71}{1+61.22 e^{(-0.03513t)}}[/tex]
[tex]when \ t \to \infty \ , 0.035 t \to \infty \ , e^{(0.03513t)} \to e^{\infty} \to \infty\ \ e^{\frac{1}{0.035t} = \ e^{-0.035t}} \to 0[/tex]
[tex]= \frac{19.71}{1+0}\\\\= \frac{19.71}{1}\\\\ = 19.71 \ million[/tex]
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