Answer :
Correction
The function is [tex]p(t)=361e^{kt[/tex]
Answer:
k=0.0259
Step-by-step explanation:
The population function is given as: [tex]p(t)=361e^{kt}[/tex]
Where t=0 corresponds to the year 2000.
In 1980, 1980-2000=-20, p(-20)=215
Therefore:
[tex]215=361e^{k*-20}\\$Divide both sides by 361$\\\frac{215}{361}= e^{-20k}\\$Take the natural logarithm of both sides$\\ln(\frac{215}{361})=-20k\\k=ln(\frac{215}{361})\div (-20)\\k=0.0259[/tex]
Answer:
k ≈ 0.44
Step-by-step explanation:
Given the equation
p(t) = 36e^(kt)
In the year 2000, when t = 0, the population
p(0) = 36e^k............(1)
In 1980, when t = -20, the population was 215000
This implies that
215000 = 36e^(-20k)
e^(-20k) = 215000/36
e^(-20k) = 5972.222
Taking natural logarithm of both sides
ln(e^(-20k)) = ln(5972.222)
-20k = 8.695
k = -8.695/20 = 0.435
k ≈ 0.44