Drawing Line of Best Fit Scatter Plot Algebra Statistics

The problem asks us to draw a line of best fit for the given scatter plot. The scatter plot shows the relationship between theater revenue (on the x-axis, in millions of dollars) and rental revenue (on the y-axis, in millions of dollars).

A line of best fit is a straight line that best represents the data on a scatter plot. It should pass through the “middle” of the data points, with roughly an equal number of points above and below the line. The line of best fit helps us to see the trend in the data.

Looking at the scatter plot, we can observe a positive correlation between theater revenue and rental revenue, meaning that as theater revenue increases, rental revenue also tends to increase. The points are generally clustered, and they form an upward trend from left to right.

To draw the line of best fit, we need to choose a line that visually minimizes the distance from the line to all the data points. We can try to approximate this by picking two points that seem to lie on the trend line, or by visually balancing the points above and below the line.

Let’s examine some of the data points to get a sense of the trend:

We can try to estimate two points that the line of best fit might pass through.
Let’s consider a point near the beginning of the trend, say when theater revenue is around 10 million dollars, the rental revenue is around 3 million dollars. So, a point could be (10, 3).
Let’s consider a point near the end of the trend, say when theater revenue is around 100 million dollars, the rental revenue is around 18 million dollars. So, a point could be (100, 18).

Now, let’s use these two points to define the line.
The slope of the line passing through and is:

The equation of the line in point-slope form is . Using the point (10, 3):

Let’s check if this line is a good approximation.
If x = 0, y = 4/3 ≈ 1.33. This is a reasonable starting point for the line, as some points are close to the y-axis with low y values.
If x = 10, y = 10/6 + 4/3 = 5/3 + 4/3 = 9/3 = 3. This matches our chosen point.
If x = 100, y = 100/6 + 4/3 = 50/3 + 4/3 = 54/3 = 18. This matches our chosen point.

Now, let’s visualize this line on the graph. The line starts at approximately (0, 1.33) and goes up to (100, 18).

Alternatively, we can try to pick two points directly from the graph that seem to be on the line of best fit.
Let’s pick a point where the line might pass near (0, 2).
Let’s pick another point where the line might pass near (80, 14).

Using these points (0, 2) and (80, 14):
Slope
Equation:

Let’s check some points with this line:
If x = 0, y = 2.
If x = 20, y = 3/20 * 20 + 2 = 3 + 2 = 5. Looking at the graph, at x=20, y is around 4-6. This is reasonable.
If x = 40, y = 3/20 * 40 + 2 = 6 + 2 = 8. At x=40, y is around 8-10. This is reasonable.
If x = 60, y = 3/20 * 60 + 2 = 9 + 2 = 11. At x=60, y is around 12-14. This is a bit low.
If x = 80, y = 3/20 * 80 + 2 = 12 + 2 = 14. This matches our chosen point.
If x = 100, y = 3/20 * 100 + 2 = 15 + 2 = 17. At x=100, y is around 18-20. This is a bit low.

Let’s try to adjust the points to get a better fit. We want the line to be a bit steeper to catch the higher points.

Let’s try to select two points that appear to be well-placed relative to the distribution of the points.
Consider the point (20, 6) and (90, 17).
Slope .
Equation:

Let’s check this line:
If x = 0, y ≈ 2.86.
If x = 20, y = 11/70 * 20 + 20/7 = 22/7 + 20/7 = 42/7 = 6. (Matches)
If x = 40, y = 11/70 * 40 + 20/7 = 44/7 + 20/7 = 64/7 ≈ 9.14. (At x=40, y is around 8-10. Good.)
If x = 60, y = 11/70 * 60 + 20/7 = 66/7 + 20/7 = 86/7 ≈ 12.29. (At x=60, y is around 12-14. Good.)
If x = 80, y = 11/70 * 80 + 20/7 = 88/7 + 20/7 = 108/7 ≈ 15.43. (At x=80, y is around 14-16. Good.)
If x = 100, y = 11/70 * 100 + 20/7 = 110/7 + 20/7 = 130/7 ≈ 18.57. (At x=100, y is around 18-20. Good.)

This line seems to be a good approximation. The equation of the line of best fit is approximately .

Since we are asked to draw the line of best fit, we need to represent this visually.
The line passes through approximately (0, 2.86) and (100, 18.57).
So, we can draw a line from approximately (0, 3) to (100, 18.5).
Another way to draw it is by picking two points on the grid that appear to be on the line.
For example, we can draw a line passing through the point (0, 3) and another point around (60, 12).
Let’s check the slope between (0, 3) and (60, 12).
.
Equation: .
Let’s check points:
x=0, y=3.
x=20, y = 0.1520 + 3 = 3 + 3 = 6. (Good, points around (20, 6))
x=40, y = 0.1540 + 3 = 6 + 3 = 9. (Good, points around (40, 8-10))
x=60, y = 0.1560 + 3 = 9 + 3 = 12. (Good, points around (60, 12-14))
x=80, y = 0.1580 + 3 = 12 + 3 = 15. (Good, points around (80, 14-16))
x=100, y = 0.15*100 + 3 = 15 + 3 = 18. (Good, points around (100, 18-20))

This line appears to be a very good visual approximation.

To draw this line on the graph, we would start at y=3 when x=0 and draw a line upwards. We can use the points (0, 3) and (100, 18) to define the line.

Final Answer: The line of best fit should pass through the data points such that it approximates the trend. A visually reasonable line can be drawn using points such as (0, 3) and (100, 18). This corresponds to an approximate equation of .

The final answer is $\boxed{The line of best fit can be approximated by the equation .}$.