Question

Help please..The table lists data regarding the average salaries of several professional athletes in the years 1991 and 2001. A)use the date points to find a linear function that fits the date. B) use the function to predict the average salary in 2005 and 2010.
Year….Av erage Salary.
1991 $264,000
2001 $1,450,000
A linear function that fits the date is s(x)=___. (Let x=the number of years since 1990, and let s= the average salary years from 1990).
The predicted average salary for 2005 is $___
The predicted average salary for 2010 is $___ (round to the nearest whole number)

Answer 1

M(y) is what I would have named it:

Money (year)

Answer 2

you are given 2 points; subtract one point from the other and stack M/y to determine a slope

Answer 3

i got my first part, but stuck at how to get the 2005 prediction

Answer 4

y=1,444,000. i think i got that right, just need to determine the answer for 2005 and2010

Answer 5

2001 $1,450,000
-1991 - 264,000
-------------------
10 $1,186,000

slope = M/y: 1,186,000/10 = 118,600


now we can use this slope to determine an equation that will give us predictions for a given year:

M(y) = 1,186,000y - 2,361,062,000

this of course is given for the year format xxxx, as opposed to the "how far from" stuff

Answer 6

M(2005) = whatever the calculate makes of that :)

Answer 7

I get:

$16,868,000 hopefully

Answer 8

is that for 2005?

Answer 9

thats what I get when I input "2005" into my equation; if you have already calibrated it for "years after" you simply have to make:

x = 4 in your own equation and it should equal out

Answer 10

err, 1991 to 2005 aint 4 years, had a synapse explode

Answer 11

2,3,4,5,6,7,8,9,0,1,2,3,4,5

that would be 14 years after 1991

Answer 12

try x=14 in your first equation and see what you get

Answer 13

ok, hang on

Answer 14

i see a typo in mine thats persisting; il correct it and see if we agree

Answer 15

M(y) = 118,600y - 236,396,600

M(2005) = 118,600(2005) - 236,396,600
= 1,396,400

that should be better

Answer 16

im gonna have to do this the long way .....

Answer 17

how did you come up with the 236,396,600?

Answer 18

I used the slope intercept form of the line equation and simplified

Answer 19

or at least I tried to, but i think I got mixed up in it

Answer 20

If you know how to get an equation of a line using two points, then do that - that way your independent variable is just the year instead of a number like 14

Answer 21

\[\text{My}(y)=118600 y-2.35869\times 10^8 \]My(y) evaluated at 1991, 2001, 2005 and 2010 is:\[\{264000.,1.45\times 10^6,1.9244\times 10^6,2.5174\times 10^6\} \]

Answer 22

(2001,1450000) we have one of the points:

we have a slope:

M(y) = my-mPy+PM

M(y) = 118600y-118600Py+PM

M(y) = 118600y-118600(2001)+(1450000)

M(y) = 118600y-237318600+(1450000)

M(y) = 118,600y-235,868,600

should work

Answer 23

robtoebey - I'd pass that through N[#, 12] so it doesn't round

Answer 24

that does the job :) just had to remember how to add and multiply stuff

Answer 25

i'm confused...

Answer 26

the big numbers are prolly whats confusing you

Answer 27

yes.. just confusing how to figure for 2005 and 2010

Answer 28

so i subtract 118,600y-235,868,600 to get my answer for 2005?

Answer 29

given 2 points (x1,y1) and (x2,y2)

we can find the slope,m, by subtracting the points and ratio then as y/x

(x1,y1)
- (x2,y2)
-----------
x1-x2, y1-y2

m = \(\cfrac{y1-y2}{x1-x2}\)

knowing m we can convert the slope formula into the point slope form

\[\frac{y1-y2}{x1-x2}=m\]
\[{y1-y2}=m({x1-x2})\]
this in turn can be rearranged into slope intercept form:

\[{y1-y2}=m({x1-x2})\]
\[{y1}=m({x1-x2})+y2\]
\[{y1}=mx1-mx2+y2\]
\[{y1}=mx1+(-mx2+y2)\]
\[{y1}=mx1+b\]

make y1 and x1 generic

\[{y}=mx+b\]

Answer 30

you simpy use the formula you developed in part "a" to find the answer for part "b"

Answer 31

since you say you did "a" already, what formula did you develop?

Answer 32

OK, I'll save this formula and see if I can figure this out, I am just confused, first time taking algebra, doing online classes and now I have to figure how to do algebra 1, ugh..frustrating.

Answer 33

once i determine the slope; i use any given point to finish up the equation for part "a"

\[{y}=mx+(-mx2+y2)\]

\[{y}=118,600x+(-118,600(1991)+(264,000))\]

\[{y}=118,600x+(-235,868,600)\]

\[{y}=118,600x-235,868,600\]

when the year: x = 2005, we get

\[{y}=118,600(2005)-235,868,600\]
\[{y}=1,924,400\]

Answer 34

part a answer. i came up with y=144,000

m=264,000-1,450,000/10 years=1,186,000 divided by 10=118,600
y=264,000+(10*118,600)
y=144,000

Answer 35

m is good

Answer 36

the other parts are a bit odd to me

Answer 37

i am not use to using the graphing tools on here, sorry

Answer 38

oh ok..so i see what you are doing..

Answer 39

good, cause I got no idea what you were doing with this:

y=264,000+(10*118,600)

Answer 40

haha..my apologies..

Answer 41

i appreciate your help amistre, now i am going so see if i can figure for 2010 :) here's a medal for you help, thanks bunches..

Answer 42

youre welcome, and good luck ;)

Answer 43

is needed, thanks.

Answer 44

ktklown
"robtoebey - I'd pass that through N[#, 12] so it doesn't round"

Refer to the Mathematica attachment. To effect the result you suggested requires the "Round" function in Mathematica.