Respuesta :
Answer:
a) [tex] m = \frac{1210-860}{5-0} =70[/tex]
Now we can find the intercept using the first condition:
[tex] 860 = 70*0 +b[/tex]
[tex] b =860[/tex]
And the model would be given by:
[tex] y = 70 x +860[/tex]
[tex] y= A e^{rt}[/tex]
The initial amount is A = 860 and we have:
[tex] y = 860 e^{rt}[/tex]
Now we can use the second condition:
[tex] 1210= 860 e^{5r}[/tex]
And solving for r we got:
[tex] \frac{121}{86} = e^{5r}[/tex]
[tex]r= 0.0682886[/tex]
And the model would be:
[tex] y = 860 e^{0.0682886 t}[/tex]
b) [tex] y = 70*10 +860=1560[/tex]
Step-by-step explanation:
Part a
For this case we have the following info given:
[tex] x_1 = 0 , y_1 = 860[/tex]
Where 0 represent the starting year in 1990.
And 5 years later we have 1210 people:
[tex] x_2 = 5, y_2 = 1210[/tex]
And we want to create a model like this:
[tex] y = mx +b[/tex]
And we can estimate the slope like this:
[tex] m= \frac{y_2 -y_1}{x_2 -x_1}[/tex]
And replacing we got:
[tex] m = \frac{1210-860}{5-0} =70[/tex]
Now we can find the intercept using the first condition:
[tex] 860 = 70*0 +b[/tex]
[tex] b =860[/tex]
And the model would be given by:
[tex] y = 70 x +860[/tex]
And if we want an exponential model like this:
[tex] y= A e^{rt}[/tex]
The initial amount is A = 860 and we have:
[tex] y = 860 e^{rt}[/tex]
Now we can use the second condition:
[tex] 1210= 860 e^{5r}[/tex]
And solving for r we got:
[tex] \frac{121}{86} = e^{5r}[/tex]
[tex]r= 0.0682886[/tex]
And the model would be:
[tex] y = 860 e^{0.0682886 t}[/tex]
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:
[tex] y = 70*10 +860=1560[/tex]
And for the exponential model we have:
[tex] y = 860 e^{0.0682886*10} =1702.441[/tex]
The linear model estimates the population to be 1560, while the exponential model estimates the population to be 1692 in 2000.
x represents the number of years after 1990
So, the given parameters are:
[tex]\mathbf{(x,y) = (0,860)\ (5,1210)}[/tex]
(a) The linear and exponential functions
The linear function
Start by calculating the slope (m)
[tex]\mathbf{m = \frac{y_2 - y_1}{x_2 - x_1}}[/tex]
So, we have:
[tex]\mathbf{m = \frac{1210 - 860}{5-0}}[/tex]
[tex]\mathbf{m = \frac{350}{5}}[/tex]
[tex]\mathbf{m = 70}[/tex]
So, the linear function is:
[tex]\mathbf{l(x) = m(x - x_1) + y_1}[/tex]
This gives
[tex]\mathbf{l(x) = 70(x - 0) + 860}[/tex]
[tex]\mathbf{l(x) = 70x + 860}[/tex]
The exponential function
An exponential function is represented as:
[tex]\mathbf{y = ab^x}[/tex]
When x = 0, we have:
[tex]\mathbf{860 = ab^0}[/tex]
[tex]\mathbf{860 = a}[/tex]
So, we have:
[tex]\mathbf{a = 860 }[/tex]
When x = 5, we have:
[tex]\mathbf{1210 = ab^5}[/tex]
Substitute [tex]\mathbf{a = 860 }[/tex]
[tex]\mathbf{1210 = 860b^5}[/tex]
Divide both sides by 860
[tex]\mathbf{1.40697674419 = b^5}[/tex]
Take 5th roots of both sides
[tex]\mathbf{1.07067431388 = b}[/tex]
Rewrite as:
[tex]\mathbf{b = 1.07067431388 }[/tex]
Approximate
[tex]\mathbf{b = 1.07}[/tex]
So, the exponential model is:
[tex]\mathbf{p(x) = 860 \times 1.07^x}[/tex]
(b) The value of the models in the year 2000
Here, x = 10
So, we have:
[tex]\mathbf{l(x) = 70x + 860}[/tex]
[tex]\mathbf{l(10) = 70 \times 10 + 860}[/tex]
[tex]\mathbf{l(10) = 1560}[/tex]
Similarly,
[tex]\mathbf{p(x) = 860 \times 1.07^x}[/tex]
[tex]\mathbf{p(10) = 860 \times 1.07^{10}}[/tex]
[tex]\mathbf{p(10) = 1692}[/tex]
Hence, the linear model estimates the population to be 1560, while the exponential model estimates the population to be 1692 in 2000.
Read more about population models at:
https://brainly.com/question/24172878
