Question

show what you know
a. drag the points to fit the line to the data.
b. use your line of fit to predict the ( y )-value of a new data point whose ( x )-value is 4.

Explanation:

Step1: Determine the equation of the line of fit

Looking at the line, when \( x = 0 \), \( y = 8 \) (y-intercept \( b = 8 \)). The slope \( m \) can be calculated using two points, e.g., \( (0, 8) \) and \( (9, 3) \). The slope formula is \( m=\frac{y_2 - y_1}{x_2 - x_1} \), so \( m=\frac{3 - 8}{9 - 0}=\frac{-5}{9}\approx - 0.56 \). The equation of the line is \( y=mx + b \), so \( y = - \frac{5}{9}x+8 \).

Step2: Predict the y - value when \( x = 4 \)

Substitute \( x = 4 \) into the equation \( y=-\frac{5}{9}(4)+8 \). First, calculate \( -\frac{5}{9}(4)=-\frac{20}{9}\approx - 2.22 \). Then, \( y=-2.22 + 8 = 5.78\approx6 \) (or using a more approximate slope, if we consider the line's trend, when \( x = 4 \), from the graph's line, we can also estimate by looking at the line: at \( x = 0 \), \( y = 8 \); at \( x = 3 \), \( y = 7 \); at \( x = 6 \), \( y = 5 \); so the rate of change is about \( - 1 \) every 3 units of \( x \). For \( x = 4 \), which is 4 units from \( x = 0 \), the change in \( y \) is \( -\frac{4}{3}\approx - 1.33 \), so \( y = 8-1.33 = 6.67\approx7 \)? Wait, maybe a better way: looking at the line, when \( x = 4 \), the line passes near \( y = 6 \) or \( y = 7 \)? Wait, let's re - examine the line. The leftmost point is \( (0,8) \) and the rightmost is \( (9,3) \). The slope is \( \frac{3 - 8}{9 - 0}=\frac{-5}{9}\approx - 0.555 \). So for \( x = 4 \), \( y=8-\frac{5}{9}(4)=8-\frac{20}{9}=\frac{72 - 20}{9}=\frac{52}{9}\approx5.78 \), which is approximately 6. But maybe from the graph's line, when \( x = 4 \), the line is at \( y = 6 \) (since at \( x = 3 \), \( y = 7 \); at \( x = 4 \), a bit less than 7, maybe 6 or 7. But using the equation, it's approximately 6.

Answer:

6 (or a value around 6, depending on the line - fitting precision)