Answer:
[tex] m =\frac{y_2 -y_1}{x_2-x_1}= \frac{5425-3733}{3-0}= 564[/tex]
Now we can use the value for 1994 and we can find the intercept like this:
[tex] 3733 = 564*0 +b[/tex]
And solving for b we got:
[tex] b = 3733 -0= 3733[/tex]
So then oir model would be given by:
[tex]y= 564x +3733[/tex]
Option C
Step-by-step explanation:
For this case we want to create a linear function for the the value of the investment in the vear x, where x =0 represents 1994.
We know that for 1994 (x= 0) the value is y = $3733 and for 1997 (x=3) the value of y = $5425
We want to find a model given by:
[tex] y = mx +b[/tex]
Where m is the slope and b the intercept. We can find the slope with this formula:
[tex] m =\frac{y_2 -y_1}{x_2-x_1}= \frac{5425-3733}{3-0}= 564[/tex]
Now we can use the value for 1994 and we can find the intercept like this:
[tex] 3733 = 564*0 +b[/tex]
And solving for b we got:
[tex] b = 3733 -0= 3733[/tex]
So then oir model would be given by:
[tex]y= 564x +3733[/tex]
Option C
