Question

The town of Pinedale, Wyoming, is experiencing a population boom. In 1990, the population was 860 and five years later it was 1210. (a) Find a linear model l(x) and an exponential model p(x) for the population of Pinedale in the year 1990+x. (b) What do these models estimate the population of Pinedale to be in the year 2000?

Answer 1

Answer:

a) $m = \frac{1210-860}{5-0} =70$

Now we can find the intercept using the first condition:

$860 = 70*0 +b$

$b =860$

And the model would be given by:

$y = 70 x +860$

$y= A e^{rt}$

The initial amount is A = 860 and we have:

$y = 860 e^{rt}$

Now we can use the second condition:

$1210= 860 e^{5r}$

And solving for r we got:

$\frac{121}{86} = e^{5r}$

$r= 0.0682886$

And the model would be:

$y = 860 e^{0.0682886 t}$

b) $y = 70*10 +860=1560$

Step-by-step explanation:

Part a

For this case we have the following info given:

$x_1 = 0 , y_1 = 860$

Where 0 represent the starting year in 1990.

And 5 years later we have 1210 people:

$x_2 = 5, y_2 = 1210$

And we want to create a model like this:

$y = mx +b$

And we can estimate the slope like this:

$m= \frac{y_2 -y_1}{x_2 -x_1}$

And replacing we got:

$m = \frac{1210-860}{5-0} =70$

Now we can find the intercept using the first condition:

$860 = 70*0 +b$

$b =860$

And the model would be given by:

$y = 70 x +860$

And if we want an exponential model like this:

$y= A e^{rt}$

The initial amount is A = 860 and we have:

$y = 860 e^{rt}$

Now we can use the second condition:

$1210= 860 e^{5r}$

And solving for r we got:

$\frac{121}{86} = e^{5r}$

$r= 0.0682886$

And the model would be:

$y = 860 e^{0.0682886 t}$

Part b

For this case we want to estimate the population in the year 2000. And that represent 10 years from 1990 so then x =10 and replacing we got:

$y = 70*10 +860=1560$

And for the exponential model we have:

$y = 860 e^{0.0682886*10} =1702.441$