Use the image and table to answer the question.

Question

An illustration shows a scatterplot titled Number of Laps the Swim Team Completes Per Week. The x-axis shows weeks ranging from 1 through 10 in increments of one. The y-axis shows number of laps and ranges from 100 to 1,000 in 100 unit increments.

Week Number of Laps
1 50
2 150
3 200
4 300
5 500
6 600
7 700
8 800
Which of the following is an appropriate regression model, rounded to the nearest tenth, given the bivariate dataset?

(1 point)
Responses

f(x)∼111.9x
f left parenthesis x right parenthesis is similar to 111.9 x

f(x)∼−111.9x−91.1
f left parenthesis x right parenthesis is similar to negative 111.9 x minus 91.1

f(x)∼111.9x+50
f left parenthesis x right parenthesis is similar to 111.9 x plus 50

f(x)∼111.9x−91.1
f left parenthesis x right parenthesis is similar to 111.9 x minus 91.1
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Answer 1

To determine the appropriate regression model for the given data, we can analyze the trend in the number of laps completed by the swim team as the weeks progress.

From the data provided in the table:

| Week | Number of Laps | |------|-----------------| | 1 | 50 | | 2 | 150 | | 3 | 200 | | 4 | 300 | | 5 | 500 | | 6 | 600 | | 7 | 700 | | 8 | 800 |

Looking at the values in the table:

1. Growth Pattern: The number of laps increases as the weeks increase. The increase is not linear, especially in weeks 5 to 8, where more laps are completed compared to previous increases.

2. Finding the Model:

   • We can first calculate the average change in the number of laps per week to estimate the slope.
   • The total change from week 1 (50 laps) to week 8 (800 laps) is 800 - 50 = 750 laps over 7 weeks, which yields an average rate of change of about 107.1 laps/week.

3. Linear Regression Estimation:

   • A linear regression might take a form of f(x) = mx + b, where m is the slope (average growth per week) and b is the y-intercept.
   • Since the values seem to start from a base of approximately 50 and rise with a slope of around 111.9 would be reasonable given the changes.

Now we can analyze the model options given in the question:

1. f(x)∼111.9x has no intercept; it doesn’t align with the data starting at 50.
2. f(x)∼−111.9x−91.1 has a negative slope, which isn't suitable since we see an increasing trend.
3. f(x)∼111.9x+50 has both the positive slope and a sensible starting point that aligns with the data, starting around 50.
4. f(x)∼111.9x−91.1 has the same issue with a negative slope.

Based on this analysis, the most appropriate regression model is:

f(x)∼111.9x+50.