how do I solve this problem.
In 1920 the record for a certain race was 445.7 sec. In 1970, it was 44.2 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) =
b) What is the predicted record for 2003?
c) What is the predicted record for 2006?
d) In what year will the predicted record be 42.70 seconds?
To solve these problems, we'll need to find a linear equation (in the form of y = mx + b) that fits the given data. Let's go through each question one by one.
a) Find a linear function that fits the data R(t):
Step 1: Let's find the slope (m) of the linear function using the given data points (1920, 445.7) and (1970, 44.2).
Slope (m) = (change in y) / (change in x)
m = (44.2 - 445.7) / (1970 - 1920)
m = -401.5 / 50
m = -8.03
Step 2: Now, let's find the y-intercept (b) by substituting one of the data points (1920, 445.7) and the slope (m) into the equation y = mx + b.
445.7 = (-8.03)(1920) + b
445.7 = -15417.6 + b
b = 445.7 + 15417.6
b = 15863.3
So, the linear function that fits the data R(t) is: R(t) = -8.03t + 15863.3
b) What is the predicted record for 2003?
Step 1: Since the year 2003 is 83 years after 1920, we can calculate the predicted record (R) by substituting t = 83 into the linear function.
R(83) = -8.03(83) + 15863.3
R(83) = -669.49 + 15863.3
R(83) = 15193.81
The predicted record for 2003 is approximately 15193.81 seconds.
c) What is the predicted record for 2006?
Step 1: Since the year 2006 is 86 years after 1920, we can calculate the predicted record (R) by substituting t = 86 into the linear function.
R(86) = -8.03(86) + 15863.3
R(86) = -690.58 + 15863.3
R(86) = 15172.72
The predicted record for 2006 is approximately 15172.72 seconds.
d) In what year will the predicted record be 42.70 seconds?
Step 1: We can set up the linear equation R(t) = -8.03t + 15863.3 and solve for t when R(t) = 42.70
42.7 = -8.03t + 15863.3
-8.03t = 42.7 - 15863.3
-8.03t = -15820.6
t = -15820.6 / -8.03
t ≈ 1969.34
The predicted record of 42.70 seconds is likely to occur around the year 1969.
Note: Since t represents the number of years since 1920, we can calculate the actual year by adding t to 1920. In this case, the predicted record of 42.70 seconds may occur around the year 1969 or 1970.