To find an appropriate regression model for the given dataset, we can look at the number of laps completed by the swim team per week. The data appears to show a pattern where the number of laps increases as the weeks progress.
Here is the data provided:
\[ \begin{array}{|c|c|} \hline \text{Week} & \text{Number of Laps} \ \hline 1 & 50 \ 2 & 150 \ 3 & 200 \ 4 & 300 \ 5 & 500 \ 6 & 600 \ 7 & 700 \ 8 & 800 \ \hline \end{array} \]
From this data, we can see that for most weeks, the number of laps increases consistently.
We will use basic linear regression to find an equation of the form \( f(x) = mx + b \).
- Calculate the slope (m) and the y-intercept (b).
- Fit the data based on these calculations.
Given the options:
- \( f(x) \approx 111.9x - 91.1 \)
- \( f(x) \approx 111.9x + 50 \)
- \( f(x) \approx 111.9x \)
- \( f(x) \approx -111.9x - 91.1 \)
Since the number of laps increases linearly over the weeks, we can eliminate any model with a negative slope (option 4).
Also, given that the data starts at 50 laps for week 1 and continues upward, the most fitting linear options would be options 1, 2, and 3.
To determine the right model, note that:
- The first slope options match a general linear increase, but we need the intercept in relation to the data.
This means we’ll likely use either 1 or 2.
Finally, if we were to plot this data or calculate further, the slope of 111.9 corresponds more realistically to the data range, as the absolute value is close to increments observed.
In conclusion, the best-fit regression model, rounded to the nearest tenth and based on the data’s trend, would be:
\[ f(x) \sim 111.9x - 91.1 \]
This model fits as per the increasing trend of the data, while providing a reasonable intercept.