Question 249990
1990 E(t) = 64.7
1995 E(t) = 67.8


your equation is:


E(t) = m*t + b


m = slope of your line (I'm presuming they are asking for a straight line projection.


b = the y-intercept which is the value of E(t) when t = 0


In 1990, t = 0 because 1990 is your base year.
In 1995, t = 5 because 1995 minus 1990 = 5
In 2005, t = 15 because 2005 minus 1990 = 15


You have 2 data points to work with.


They are:


E(0) = 64.7
E(5) = 67.8


E(0) means the value of E(t) when t = 0.
E(5) means the value of E(t) when t = 5.


you can use these points to find the slope of your line.


slope of your line is equal to ((E(5) - E(0))/(5-0).  


This becomes:


slope of your line is equal to ((67.8) - (64.7)) / 5 = 3.1/5 = .62


replace m with .62 and your equation becomes:


E(t) = .62 * t + b


Take either of your data points and substitute in this equation to solve for b.


Use E(t) = 64.7 when t = 0


64.7 = .62 * 0 + b which becomes 64.7 which equals b in your equation.


your equation becomes:


E(t) = .62 * t + 64.7


In 2005, t will equal 15.


Substitute for t in the equation to get:


E(15) = .62 * 15 + 64.7


Solve for E(15) to get:


E(15) = 74


To graph this equation, replace E(t) with y and replace t with x to get:


y = .62 * x + 64.7


There's a horizontal line at y = 74 to show you that the graph of the equation intersect this line when x = 15.  x = 15 in the graph is the same as t = 15 in the original equation.


graph of your equation looks like this:


{{{graph(600,600,-5,20,-10,100,.62*x + 64.7, 74)}}}