Question

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

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

yes sir!

Answer 6

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).

Answer 7

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.

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

ok

Answer 13

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.

Answer 14

And then post what you get for slope

Answer 15

11,210,000 over 19

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

No, that's the slope of your line.

Answer 20

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.

Answer 21

I am lost!!!! SRY

Answer 22

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

Answer 23

I am not good at this type of question

Answer 24

Which point do you want to use?

Answer 25

269,000

Answer 26

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

Answer 27

1991

Answer 28

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.

Answer 29

so what is s(x)= then

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

No.

Answer 35

1,120,000/19=58,947

Answer 36

I am so confused!!!!

Answer 37

You picked your point, so you should have

\[y - 269,000 = \frac{1,121,000}{19}(x - 1991)\]

Answer 38

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

Answer 39

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

Answer 40

ok

Answer 41

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

Answer 42

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

Answer 43

or s(x), whichever you prefer.

Answer 44

2,030,530,000

Answer 45

26,410,000 to many zero when working

Answer 46

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

Answer 47

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