Answer:
1)
[tex]P(t)=1633e^{0.0087t}[/tex]Explanation:
Since we are taking t = 0 as 1900. The initial population is the population in 1900:
[tex]P_0=1633[/tex]We can start writting the exponential model:
[tex]P(t)=1633e^{kt}[/tex]:
We know that the population in 1950 (t = 50 in this model) is 2525. Then, we can find k by:
[tex]P(50)=2525=1633e^{k\cdot50}[/tex]And solve for k:
[tex]\frac{2525}{1633}=e^{50k}[/tex]Now, we can apply natural logarithm on both sides:
[tex]\ln(\frac{2525}{1633})=\ln(e^{50k})[/tex]Since natural log and the exponential functions are inverse functions:
[tex]\ln(\frac{2525}{1633})=50k[/tex][tex]k=\frac{\ln(\frac{2525}{1633})}{50}[/tex][tex]k\approx0.0087[/tex]Thus, the exponential model is:
[tex]P(t)=1633e^{0.0087t}[/tex] Question 2Now, we need to use the model we just create to find the predicted population by the model:
At year 1900, t = 0 (that's how the problem asked us to define t)
Then
[tex]P\left(0\right)=1633e^{0.0087\cdot0}=1633e^0=1633[/tex]At year 1920, t = 20
[tex]P(20)=1633e^{0.0087\cdot20}=1633e^{0.174}=1943[/tex]At year 1940, t = 40:
[tex]P(40)=1633e^{0.0087\cdot40}=1633e^{0.348}=2313[/tex]At year 1960, t = 60
[tex]P(60)=1633e^{0.0087\cdot60}=1633e^{0.522}=2752[/tex]At year 1980, t = 80
[tex]P(80)=1633e^{0.0087\cdot80}=1633e^{0.696}=3275[/tex]At year 2000, t = 100
[tex]P(100)=1633e^{0.0087\cdot100}=1633e^{0.87}=3898[/tex]At year 2020, t = 120
[tex]P(120)=1633e^{0.0087\cdot120}=1633e^{1.044}=4639[/tex]At year 2040, t = 140
[tex]P(140)=1633e^{0.0087\cdot140}=1633e^{1.218}=5520[/tex]