ok i don't get this one
in 1920 the record for a certain race was 45.9 sec in 1980 it was 44.1 sec let r (T) = the record in the race and t= the number of years since 1920
(a) find a linear fuction that fits the data
(b) use the funcation in (a) to perdict the record for 2003 and 2006
(c) find t he year when the record will be 43.20 sec..
help please thatnks in advance
r = m (t) + b
when year = 1920, t = 0
where t = year - 1920
45.9 = m (0) + b
so b = 45.9
44.1 = m (1980-1920) + 45.9
44.1 = 60 m + 45.9
60 m = -1.8
m = -.03
so
r =-.03 t + 45.9
when year = 2003
t = 2003 - 1920 = 83
r = -.03 (83) + 45.9
r = 43.41
that should get you started
so what yr will it be 43.20
so r(t)=?
as I said
r =-.03 t + 45.9
then
43.2 = -.03 t + 45.9
.03 t = 2.7
t = 90
1920 + 90 = 2010
To find a linear function that fits the data, we need to find the equation of a straight line in the form of y = mx + b, where y represents the record time and x represents the number of years since 1920.
(a) Let's start by finding the slope (m) of the line. The formula for slope is:
m = (y2 - y1) / (x2 - x1)
Using the data points (1920, 45.9) and (1980, 44.1):
m = (44.1 - 45.9) / (1980 - 1920)
= -1.8 / 60
= -0.03
So, the slope of the line is -0.03.
Now, let's find the y-intercept (b) of the line. We can use the point-slope form of the equation:
y - y1 = m(x - x1)
Plugging in the coordinates (1920, 45.9):
y - 45.9 = -0.03(x - 1920)
Simplifying:
y - 45.9 = -0.03x + 57.6
Rearranging the equation to the slope-intercept form (y = mx + b):
y = -0.03x + 57.6 + 45.9
y = -0.03x + 103.5
Therefore, the linear function that fits the data is r(t) = -0.03t + 103.5.
(b) To predict the record for 2003 and 2006, we need to substitute the respective values of t into the function r(t).
For 2003:
t = 2003 - 1920 = 83
r(t) = -0.03(83) + 103.5
r(t) = 100.61 seconds
For 2006:
t = 2006 - 1920 = 86
r(t) = -0.03(86) + 103.5
r(t) = 100.22 seconds
Therefore, the predicted record for 2003 is 100.61 seconds, and for 2006 it is 100.22 seconds.
(c) To find the year when the record will be 43.20 seconds, we need to substitute this value into the function r(t) and solve for t.
43.20 = -0.03t + 103.5
Rearranging the equation:
-0.03t = 43.20 - 103.5
-0.03t = -60.3
Dividing both sides by -0.03:
t = -60.3 / -0.03
t = 2010
So, the year when the record will be 43.20 seconds is 2010.
I hope this helps! Let me know if you have any further questions.