Question 1208291
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Part (a)


B(t) = B<sub>0</sub>*e^(-at)
B(t) = 200*e^(-at)
B(1) = 200*e^(-a*1)
198 = 200*e^(-a)
e^(-a) = 198/200
e^(-a) = 0.99
a = -Ln(0.99)
a = 0.01005 approximately when rounding to 5 decimal places.


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Part (b)


B(t) = B<sub>0</sub>*e^(-at)
B(t) = 200*e^(-0.01005t)
100 = 200*e^(-0.01005t)
e^(-0.01005t) = 100/200
e^(-0.01005t) = 0.5
-0.01005t = Ln(0.5)
t = -Ln(0.5)/0.01005
t = 68.96986871243 approximately
t = 69 years when rounding to the nearest whole number.
Adding this duration to the year 2021 lands on the year 2021+69 = 2090.


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Part (c)


B(t) = B<sub>0</sub>*e^(-at)
B(30) = B<sub>0</sub>*e^(-0.01005*30)
100 = B<sub>0</sub>*e^(-0.01005*30)
100 = B<sub>0</sub>*0.73971
B<sub>0</sub> = 100/0.73971
B<sub>0</sub> = 135.18811
B<sub>0</sub> = 135
If 100 species will be observed in 2021+30 = 2051, then we estimate roughly 135 species were present in 2021.


Side note: I interpret the "now", from "If thirty years from now", to refer to when the problem was written (assuming 2021) rather than this current year (2024). The math professor who wrote this problem should be more careful in her/his wording. 
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