math (am i correct)
posted by scooby9132002 on .
in 1920 the record for a certain race was 46.9 sec. In 1990 it was 45.5 sec. let R(t) = the record in the race and t = the number of years since 1920
1.)find a linear function that fits this data?
2.) use a function to predict the record in 2003 and 2006.
3.) find the year when the record will be 44.98 sec.
I think number 1 is 46.9

Record = m (year 1920) + b
where year 1920 = t so
R = m t + b
46.9 = m (0) + b
so b = 46.9
in 1990, t = 1990  1920 = 70
so
45.5 = m (70) + 46.9
m = .02
so
R = .02 t + 46.9
for 2003 for example
t = 2003  1920 = 83
so
R = .02(83)+46.9
R = 45.24 
1 is not 46.9.
Question 1 asks you to find a linear function that fits the data.
Find a function f(t) where t is the number of years since 1920. There are two points given.
1920, 46.9s
1990, 45.5s
Or, for t expressed as years since 1920...
0 years, 46.9s
70 years, 45.5s
for a linear function, y = mx + b
Where m is the slope and b is the y axis intercept.
The slope, m, is (y2  y1)/(x2  x1)
or
(45.5  46.9)/(70  0)
Calculate this value for the slope.
The y intercept is at x=0 which is already given as one of the points (0, 46.9) 
ok so what is the predicted record for 2006?
in what year will the predicted record be 44.98 seconds? 
Scooby, we are not going to do everything for you.
From the way I did 2003 you should be able to do the rest.