Scores that rise by about the same amount for each extra hour suggest a linear model, which predicts a steady change per hour. Plot the pairs in Desmos, then use a linear regression to find the slope and starting score. A best-fit line follows the whole trend, not necessarily every recorded score. One score jump alone can give a misleading slope.
Hints
- Hint 1
Compare the score increases over equal one-hour intervals. A linear model has about the same increase each time. Do the increases look close enough for a line to describe the overall pattern?
- Hint 2
Put hours in and scores in , with each student's values in the same row. A regression fits a line using all five pairs, rather than choosing a slope from one pair.
- Hint 3
In , the slope is the predicted score increase per hour, and the intercept is the prediction at zero hours. Compare both fitted numbers with the equations.
Step-by-step
Fit a line to all five students
Step 1Check the shape of the data
Enter the hours in Desmos's column and the scores in its column, keeping each student's pair together. The plotted points climb in an almost straight path, so a linear model, a line with a steady rise per hour, is a good fit.
- Step 2
Find the best-fit slope and intercept
Type below the table. The tilde tells Desmos to fit the line to all five pairs. Under PARAMETERS, it shows for the predicted score gain per hour and for the predicted score at zero hours. These are estimates, so the line needn't pass through every recorded score.
- Step 3
Match both parts of the equation
Choose the equation whose slope and intercept are both closest to Desmos's fitted values: . Don't choose a line only because its intercept matches the recorded score at zero hours; its slope must fit the other scores too. The best model predicts a score of about at zero hours and a gain of about points per hour. Choice B.