# math (am i correct)

posted by 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

• math (am i correct) - ,

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

• math (am i correct) - ,

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)

• math (am i correct) - ,

ok so what is the predicted record for 2006?

in what year will the predicted record be 44.98 seconds?

• math (am i correct) - ,

Scooby, we are not going to do everything for you.
From the way I did 2003 you should be able to do the rest.