The table lists data regarding the average salaries of several professional athletes in the years 1991 and 2001.
a) Use the data points to find a linear function that fits the data .
b) Use the function to predict the average salary in 2005 and 2010.
Year 1991 the average salary is 269,000
year 2010 the average salary is 1,390,000

- anonymous

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- anonymous

Treat this as a linear equation (y=mx+b). The year is your x value, and the salary is your y-value, so you really have two ordered pairs - (1991, 269000) and (2010, 1390000). Use these to find the slope (m) of your line.

- anonymous

Once you have the slope, then use the values from one of the ordered pairs to find the y-intercept (b).

- anonymous

I need to solve the linear function that fits data S(x)=
And then predicted the average salary for 2005 and 2010?

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## More answers

- anonymous

Is the data you listed above all of the data that is given?

- anonymous

yes sir!

- anonymous

Then what I said above should work, though your equation will take the form S(x)=mx+b, since it is supposed to be a function. Use the ordered pairs to find the slope (change in y value over change in x value).

- anonymous

Plug the two points into the slope formula to find the slope of the linear function.
Then plug the slope and one of the points into the point slope formula:
\[y-y_1 = m(x-x_1)\]
Where \((x_1,y_1)\) is one of your points and m is the slope.

- anonymous

That make no scence to me at all.....!

- anonymous

Ok, do you know the formula for the slope if you have 2 points?

- anonymous

x1 and y2 over x2 and y2
but it is hard for to get the answer

- anonymous

That's not quite right.
\[\frac{y_2-y_1}{x_2-x1} = Slope\]

- anonymous

ok

- anonymous

So plug in the salaries as y2 and y1 and the years as x2 and x1. Make sure you keep them consistent, so the salary for one year is y2 then that year must be x2.

- anonymous

And then post what you get for slope

- anonymous

11,210,000 over 19

- anonymous

You have an extra 0 there. Should be 1,121,000 over 19

- anonymous

Ok, so now plug in that slope, along with one of your two points into the point slope formula.

- anonymous

yea sry so the answer is 1,121,000 over 19 is s(x)?

- anonymous

No, that's the slope of your line.

- anonymous

Now you plug in that slope along with one of your points into the point slope formula:
\[y-y_1 = m(x-x_1)\]
Where \(x_1,y_1\) are the x and y values of your point, and m is the slope.

- anonymous

I am lost!!!! SRY

- anonymous

You have the slope. The slope is m. you have 2 points. Pick one.

- anonymous

I am not good at this type of question

- anonymous

Which point do you want to use?

- anonymous

269,000

- anonymous

That's a salary, the point would be (1991,269,000)

- anonymous

1991

- anonymous

Where 1991 is your \(x_1\) and 269,000 is your \(y_1\). Plug them into the equation and plug in the slope you found for m. That will give you the equation for the line which represents the salary over time.

- anonymous

so what is s(x)= then

- anonymous

After you plug it in, s(x) = y

- anonymous

I also have to predict the salary for 2005 and 2010?

- anonymous

Which you can do easily once you have the formula. Have you plugged it in yet?

- anonymous

so you have 1,120,000/9=124,444

- anonymous

No.

- anonymous

1,120,000/19=58,947

- anonymous

I am so confused!!!!

- anonymous

You picked your point, so you should have
\[y - 269,000 = \frac{1,121,000}{19}(x - 1991)\]

- anonymous

well 1,121,000 / 19=59,000

- anonymous

Solving for s(x) = y we have:
\[ s(x) = y = \frac{1}{19}(1,121,000x -2226800000)\]

- anonymous

ok

- anonymous

So that's the equation for the salary at a given year x.

- anonymous

You want to know the salary at 2010, plug that in for x, and see what y is.

- anonymous

or s(x), whichever you prefer.

- anonymous

2,030,530,000

- anonymous

26,410,000 to many zero when working

- anonymous

2010 would be 26,410,000
and 2005 would be 20,805,000

- anonymous

I am sorry if I am causing you grief I am just bad at these problem.

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