someone plz help me....
In 1920 the record for a certain race was 46.6 sec. In 1950 it was 46.0 sec. let r(t)= the record in the race and t= the number of years since 1920.
(a) find a linear function that fits the data r(t)= (round to the nearest hundredth)
(b)use the function in (a)to predict the record in 2003 and in 2006.What is the predited record for 2003 and 2006 roun to the neareat tenth.
(c) In what year will the predicted record be 44.7 seconds?
(a) To find a linear function that fits the data, we need to calculate the slope (m) and the y-intercept (b) of the linear equation.
We have two data points:
(1920, 46.6) and (1950, 46.0)
Let's start by finding the slope:
slope (m) = (change in y) / (change in x)
= (46.0 - 46.6) / (1950 - 1920)
= -0.6 / 30
= -0.02
Now let's find the y-intercept (b) by substituting one of the data points into the slope-intercept form of a linear equation, y = mx + b:
46.0 = (-0.02)(1950) + b
46.0 = -39 + b
b = 46.0 + 39
b = 85.0
So the linear function that fits the data is:
r(t) = -0.02t + 85.0
(b) To predict the record in 2003 and 2006, we need to substitute the corresponding values of t into the linear function r(t).
For 2003, t = 2003 - 1920 = 83
r(2003) = -0.02(83) + 85.0
= -1.66 + 85.0
= 83.34 (rounded to the nearest tenth)
For 2006, t = 2006 - 1920 = 86
r(2006) = -0.02(86) + 85.0
= -1.72 + 85.0
= 83.28 (rounded to the nearest tenth)
Therefore, the predicted record for 2003 is 83.3 seconds and for 2006 is 83.3 seconds.
(c) To find the year when the predicted record will be 44.7 seconds, we need to solve the linear equation for t.
44.7 = -0.02t + 85.0
-0.02t = 44.7 - 85.0
-0.02t = -40.3
t = (-40.3) / (-0.02)
t = 2015
Therefore, the predicted record will be 44.7 seconds in the year 2015.