Question 510199
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Let the first year of your data (whether it is 1990 or 1991 you don't make clear) be represented by 0 as a value for your independent variable (x generally, but whatever you are using to represent your horizontal axis) then each subsequent year adds one.  So if 1990 is your first year 0 as an x value represnts 1990, 1 represents 1991, and so on.  Then the dependent variable, i.e., the value of the function at each of the two data points you are given is the actual salary value.  Since both salaries given end in 3 zeros, just drop the three zeros and label your graph values as $1000s.


Now each of your two data points has two values, an x value representing the year and a y value representing the value of the function (i.e. the salary) for that year.  You can now create two ordered pairs, *[tex \Large \left(x_1,y_1\right)] and *[tex \Large \left(x_2,y_2\right)], using the data from the table.


Knowing the coordinates of two points on a straight line, you can write an equation that represents all of the points on that line by using the two point form of an equation of a line:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ \left(\frac{y_1\ -\ y_2}{x_1\ -\ x_2}\right)(x\ -\ x_1) ]


where *[tex \Large \left(x_1,y_1\right)] and *[tex \Large \left(x_2,y_2\right)] are the coordinates of the given points.


So plug in your coordinates, do the arithmetic, and simplify.  Then determine the appropriate x-value for the two years you want to predict and plug that value in for x in the equation you just derived and do the indicated arithmetic to predict the salary value for that year.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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