Question

This is regarding salary? confused::

1991 - 268,000

2001-1,420,000
These are the avg in years!!

use data plots to find linier function that fits the data?

Let x=number of years since 1990 and s=avg salary x years from 1990

Find a linier function that fits the data S(X)=

Predict the avg for
2005 (round to whole numb)

2010(round to whole Numb)

Answer 1

I find word probleems hard never mind with this!! why

Answer 2

its the formulas to get all the information i get all confused.

Answer 3

formula for slope is
\[m=\frac{y_2-y_1}{x_2-x_1}\]

Answer 4

who has an average salary of $268,000 in 1991?

Answer 5

investment bankers?

Answer 6

athelete

Answer 7

oooooooooooooh ok

Answer 8

so you want the change in y over the change in x.
\[\frac{1,420,000 -268,000}{2001-1991}\]

Answer 9

\[m=\frac{1152000}{10}=115,200\] i.e. salaries increased by $115,200 per year since 1991

Answer 10

1152000/10

Answer 11

i got that part
but then im lost

Answer 12

got it

Answer 13

oh that was the hard part. now comes the easy part. you are turning 1991 in to year 0, so your y - intercept is what you started with in 1991, namely 268,000 and your linear equation is
\[S(x)=115200 x + 268,000\] where x is in years since 1991

Answer 14

looks just like
\[y=mx+b\] where "m" is the slope and "b" is the y - intercept

Answer 15

so you have to use two formulas

Answer 16

is this one to get the yeas 2005 and 2010 prediction

Answer 17

first find the slope.
then write it in "point-slope" form

Answer 18

yes but don't forget you are using x = years since 1991, so the year 2005 corresponds to
\[x=2005-1991= 14\] in simple english 14 years since 1991

Answer 19

likewise 2010 is x = 19

Answer 20

ok so since im looking for since years 1990 it would be 2005-1990 and 2010-1990 giving me 15 and 10

Answer 21

unless i read it wrong it says years since 1991, not 1990

Answer 22

i mean if it tells you what you start with in 1991, then that is where you start counting