Question
5. The equation of the line of best fit of a scatter plot is f(x)=10.2217β0.0098x
π
(
π₯
)
=
10.2217
β
0.0098
π₯
, where π is the finishing time of the Olympic gold medalist in the menβs 100 m dash, and π₯ is the number of years since 1960.(1 point)
Part A: Interpret the π¦-intercept of the linear equation modeling the data.
The model suggests that the finishing time of the Olympic gold medalist in the menβs 100 m dash in 1960 was
seconds
Part B: Interpret the slope of the linear equation that models the data.
The finishing time since 1960 has decreased each year by
seconds
Part C: The Summer Olympic Games, where the 100 m dash is run, occur every 4 years. By how many seconds does the model predict the first-place-finishing time will decrease every 4 years?
seconds.
π
(
π₯
)
=
10.2217
β
0.0098
π₯
, where π is the finishing time of the Olympic gold medalist in the menβs 100 m dash, and π₯ is the number of years since 1960.(1 point)
Part A: Interpret the π¦-intercept of the linear equation modeling the data.
The model suggests that the finishing time of the Olympic gold medalist in the menβs 100 m dash in 1960 was
seconds
Part B: Interpret the slope of the linear equation that models the data.
The finishing time since 1960 has decreased each year by
seconds
Part C: The Summer Olympic Games, where the 100 m dash is run, occur every 4 years. By how many seconds does the model predict the first-place-finishing time will decrease every 4 years?
seconds.
Answers
GPT-4o mini
Let's break down the interpretations based on the given linear equation \( f(x) = 10.2217 - 0.0098x \):
### Part A: Interpret the \( y \)-intercept of the linear equation.
The \( y \)-intercept occurs when \( x = 0 \). In this context, \( x = 0 \) corresponds to the year 1960. To find the \( y \)-intercept, we substitute \( x = 0 \) into the equation:
\[
f(0) = 10.2217 - 0.0098(0) = 10.2217
\]
This means that the finishing time of the Olympic gold medalist in the men's 100 m dash in 1960 was **10.2217 seconds**.
### Part B: Interpret the slope of the linear equation that models the data.
The slope of the line, \( -0.0098 \), indicates the rate of change of the finishing time per year. Specifically, it suggests that for each additional year since 1960, the finishing time decreases by **0.0098 seconds**.
### Part C: The Summer Olympic Games, where the 100 m dash is run, occur every 4 years. By how many seconds does the model predict the first-place finishing time will decrease every 4 years?
Since the slope indicates the time decreases by \( 0.0098 \) seconds each year, over a span of 4 years, the total decrease can be calculated as:
\[
\text{Decrease in 4 years} = 0.0098 \text{ seconds/year} \times 4 \text{ years} = 0.0392 \text{ seconds}
\]
Thus, the model predicts that the first-place finishing time will decrease by **0.0392 seconds** every 4 years.
### Part A: Interpret the \( y \)-intercept of the linear equation.
The \( y \)-intercept occurs when \( x = 0 \). In this context, \( x = 0 \) corresponds to the year 1960. To find the \( y \)-intercept, we substitute \( x = 0 \) into the equation:
\[
f(0) = 10.2217 - 0.0098(0) = 10.2217
\]
This means that the finishing time of the Olympic gold medalist in the men's 100 m dash in 1960 was **10.2217 seconds**.
### Part B: Interpret the slope of the linear equation that models the data.
The slope of the line, \( -0.0098 \), indicates the rate of change of the finishing time per year. Specifically, it suggests that for each additional year since 1960, the finishing time decreases by **0.0098 seconds**.
### Part C: The Summer Olympic Games, where the 100 m dash is run, occur every 4 years. By how many seconds does the model predict the first-place finishing time will decrease every 4 years?
Since the slope indicates the time decreases by \( 0.0098 \) seconds each year, over a span of 4 years, the total decrease can be calculated as:
\[
\text{Decrease in 4 years} = 0.0098 \text{ seconds/year} \times 4 \text{ years} = 0.0392 \text{ seconds}
\]
Thus, the model predicts that the first-place finishing time will decrease by **0.0392 seconds** every 4 years.