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.
To find a linear function that fits the given data, we need to determine the equation of a straight line that passes through the two data points.
We are given two data points: (1920, 46.9) and (1990, 45.5).
The equation of a straight line can be written in the form y = mx + b, where m is the slope of the line and b is the y-intercept.
To find the slope, we can use the formula:
m = (y2 - y1) / (x2 - x1)
Using the values of our two data points:
m = (45.5 - 46.9) / (1990 - 1920)
m = -1.4 / 70
m = - 0.02
Now, let's substitute the values of one of the data points into the equation to solve for the y-intercept (b).
We'll use the point (1920, 46.9):
46.9 = (- 0.02) * 1920 + b
46.9 = - 38.4 + b
b = 85.3
Therefore, the linear function that fits the data is given by:
R(t) = -0.02t + 85.3, where t is the number of years since 1920.
Now, let's proceed to the next questions using this linear function:
2.) To predict the record in 2003 and 2006, we need to substitute the respective values of t into the equation R(t) = -0.02t + 85.3:
For 2003 (t = 2003 - 1920 = 83):
R(83) = -0.02 * 83 + 85.3
R(83) = 83.4 seconds
For 2006 (t = 2006 - 1920 = 86):
R(86) = -0.02 * 86 + 85.3
R(86) = 82.5 seconds
Therefore, the predicted record in 2003 is 83.4 seconds and in 2006 is 82.5 seconds.
3.) To find the year when the record will be 44.98 seconds, we can set up the equation R(t) = 44.98 and solve for t:
44.98 = - 0.02t + 85.3
- 0.02t = 44.98 - 85.3
- 0.02t = -40.32
t = -40.32 / -0.02
t = 2016
Therefore, the year when the record will be 44.98 seconds is 2016.