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what is the equation of the trend line in the scatter plot? use the two…

Question

what is the equation of the trend line in the scatter plot?
use the two yellow points to write the equation in slope - intercept form. write any coefficients as integers, proper fractions, or improper fractions in simplest form.

Explanation:

Step1: Identify the two yellow points

From the scatter plot, the two yellow points are \((4, 9)\) and \((8, 3)\).

Step2: Calculate the slope (\(m\))

The formula for slope is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let \((x_1,y_1)=(4,9)\) and \((x_2,y_2)=(8,3)\). Then \(m=\frac{3 - 9}{8 - 4}=\frac{-6}{4}=-\frac{3}{2}\).

Step3: Find the y - intercept (\(b\))

The slope - intercept form of a line is \(y = mx + b\). We know \(m = -\frac{3}{2}\) and we can use the point \((4,9)\) to find \(b\). Substitute \(x = 4\), \(y = 9\) and \(m=-\frac{3}{2}\) into \(y=mx + b\):
\(9=-\frac{3}{2}(4)+b\)
\(9=-6 + b\)
Add 6 to both sides: \(b=9 + 6=15\)? Wait, no, wait. Wait, let's recalculate. Wait, \(-\frac{3}{2}\times4=-6\), then \(9=-6 + b\), so \(b=9 + 6 = 15\)? Wait, that can't be right. Wait, no, maybe I mixed up the points. Wait, let's check the points again. Wait, the first yellow point: when \(x = 4\), \(y = 9\); the second yellow point: when \(x = 8\), \(y = 3\). Wait, let's recalculate the slope: \(\frac{3 - 9}{8 - 4}=\frac{-6}{4}=-\frac{3}{2}\). Then using the point \((4,9)\): \(y=-\frac{3}{2}x + b\). Substitute \(x = 4\), \(y = 9\): \(9=-\frac{3}{2}(4)+b\) \(9=-6 + b\), so \(b = 15\)? Wait, but when \(x = 10\), \(y = 0\), let's check with \(b = 15\) and \(m=-\frac{3}{2}\), \(y=-\frac{3}{2}(10)+15=-15 + 15 = 0\), which matches the end - point. Wait, but let's check the other point \((8,3)\): \(y=-\frac{3}{2}(8)+15=-12 + 15 = 3\), which also matches. So the slope - intercept form is \(y=-\frac{3}{2}x + 15\)? Wait, no, wait, maybe I made a mistake in the points. Wait, looking at the graph, the first yellow point is at \(x = 4\), \(y = 9\); the second is at \(x = 8\), \(y = 3\). Wait, but let's check the line. The line goes from \((4,9)\) to \((8,3)\) to \((10,0)\). So the slope is \(\frac{0 - 9}{10 - 4}=\frac{-9}{6}=-\frac{3}{2}\), and the y - intercept is 15? Wait, when \(x = 0\), \(y = 15\)? But the graph's y - axis only goes up to 10. Wait, maybe the points are misidentified. Wait, no, the yellow points are at \((4,9)\) and \((8,3)\). Let's recalculate the slope again: \(m=\frac{3 - 9}{8 - 4}=\frac{-6}{4}=-\frac{3}{2}\). Then using the point \((4,9)\): \(y=-\frac{3}{2}x + b\). \(9=-\frac{3}{2}\times4 + b\) \(9=-6 + b\), so \(b = 15\). But when \(x = 4\), \(y = 9\); \(x = 8\), \(y = 3\); \(x = 10\), \(y = 0\). Let's check \(x = 10\): \(y=-\frac{3}{2}\times10+15=-15 + 15 = 0\), which is correct. So the equation of the line is \(y=-\frac{3}{2}x + 15\)? Wait, no, that seems off because the y - intercept is 15, but the graph's y - axis is up to 10. Wait, maybe the first yellow point is at \(x = 3\), \(y = 10\)? No, the yellow point is at \(x = 4\), \(y = 9\). Wait, maybe I made a mistake in the slope calculation. Wait, let's take two points on the line: \((4,9)\) and \((8,3)\). The change in \(y\) is \(3 - 9=-6\), change in \(x\) is \(8 - 4 = 4\), so slope is \(\frac{-6}{4}=-\frac{3}{2}\). Then the equation is \(y=-\frac{3}{2}x + b\). Plugging in \((4,9)\): \(9=-\frac{3}{2}(4)+b\) \(9=-6 + b\), so \(b = 15\). So the equation is \(y=-\frac{3}{2}x + 15\).

Answer:

\(y = -\frac{3}{2}x + 15\)