Question

the scatter plot and line of best fit below show the length of 14 peoples femur (the long leg bone in the thigh) and their height in centimeters. based on the line of best fit, what would be the predicted femur length for someone with a height of 229 cm?
(scatter plot with x - axis labeled femur length (centimeters) and y - axis labeled height (centimeters), with points and a line of best fit passing through (60,201) and (69,208))

Explanation:

Step1: Find the slope of the line of best fit

We have two points on the line of best fit: \((60, 201)\) and \((69, 208)\). The slope \(m\) is calculated as \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{208 - 201}{69 - 60}=\frac{7}{9}\approx0.7778\).

Step2: Use the point - slope form to find the equation of the line

Using the point \((60, 201)\) and the slope \(m = \frac{7}{9}\), the point - slope form is \(y - y_1=m(x - x_1)\). So \(y-201=\frac{7}{9}(x - 60)\). Simplifying, we get \(y=\frac{7}{9}x- \frac{7\times60}{9}+201=\frac{7}{9}x-\frac{140}{3}+201=\frac{7}{9}x+\frac{- 140 + 603}{3}=\frac{7}{9}x+\frac{463}{3}\).

Step3: Predict the femur length for \(y = 229\)

We need to solve for \(x\) when \(y = 229\). So \(229=\frac{7}{9}x+\frac{463}{3}\). First, subtract \(\frac{463}{3}\) from both sides: \(229-\frac{463}{3}=\frac{7}{9}x\). \(229=\frac{687}{3}\), so \(\frac{687}{3}-\frac{463}{3}=\frac{224}{3}=\frac{7}{9}x\). Then, solve for \(x\): \(x=\frac{224}{3}\times\frac{9}{7}=\frac{224\times9}{3\times7}=\frac{224\times3}{7}=32\times3 = 96\)? Wait, no, wait. Wait, we mixed up \(x\) and \(y\). Wait, the \(x\) - axis is femur length (centimeters) and the \(y\) - axis is height (centimeters). So we should have the equation with \(x\) as femur length and \(y\) as height. Let's re - define: let \(x\) be femur length, \(y\) be height. So the two points are \((x_1,y_1)=(60,201)\) and \((x_2,y_2)=(69,208)\). The slope \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{208 - 201}{69 - 60}=\frac{7}{9}\) (height per femur length). The equation of the line is \(y=mx + b\). Using \((60,201)\): \(201=\frac{7}{9}\times60 + b\). \(201=\frac{140}{3}+b\), so \(b = 201-\frac{140}{3}=\frac{603 - 140}{3}=\frac{463}{3}\). So \(y=\frac{7}{9}x+\frac{463}{3}\). Now, we want to find \(x\) when \(y = 229\). So \(229=\frac{7}{9}x+\frac{463}{3}\). Multiply through by 9 to clear the fractions: \(229\times9 = 7x+463\times3\). \(2061=7x + 1389\). Subtract 1389 from both sides: \(2061 - 1389=7x\). \(672 = 7x\). Then \(x=\frac{672}{7}=96\)? Wait, no, that can't be. Wait, maybe we made a mistake in the axes. Wait, looking at the graph: the \(x\) - axis is femur length (centimeters) and the \(y\) - axis is height (centimeters). The two points are \((60,201)\) and \((69,208)\). So when height \(y = 229\), we solve for \(x\) (femur length). Let's use the slope formula correctly. The slope \(m=\frac{\Delta y}{\Delta x}=\frac{208 - 201}{69 - 60}=\frac{7}{9}\) (change in height over change in femur length). So the equation is \(y - 201=\frac{7}{9}(x - 60)\), where \(x\) is femur length and \(y\) is height. So when \(y = 229\), \(229-201=\frac{7}{9}(x - 60)\). \(28=\frac{7}{9}(x - 60)\). Multiply both sides by \(\frac{9}{7}\): \(28\times\frac{9}{7}=x - 60\). \(36=x - 60\). Then \(x=36 + 60=96\)? Wait, but let's check the other way. Wait, maybe the points are \((201,60)\) and \((208,69)\) (swapping \(x\) and \(y\)). Let's re - assign: let \(x\) be height, \(y\) be femur length. Then the two points are \((x_1,y_1)=(201,60)\) and \((x_2,y_2)=(208,69)\). The slope \(m=\frac{69 - 60}{208 - 201}=\frac{9}{7}\approx1.2857\). The equation of the line is \(y - 60=\frac{9}{7}(x - 201)\). So \(y=\frac{9}{7}x-\frac{9\times201}{7}+60=\frac{9}{7}x-\frac{1809}{7}+\frac{420}{7}=\frac{9}{7}x-\frac{1389}{7}\). Now, when \(x = 229\), \(y=\frac{9}{7}\times229-\frac{1389}{7}=\frac{2061-1389}{7}=\frac{672}{7}=96\). Wait, so the predicted femur length is 96 centimeters? Wait, no, wait, the \(x\) - axis is femur length (centimeters) and \(y\) - axis is height (centimeters). So if we have the line of best fit with points \((60,201)\) (femur length = 60, height = 201) and \((69,208)\) (femur length = 69, height = 208). We need to find femur length when height = 229. The difference in height from 201 to 229 is \(229 - 201=28\). Since the slope is \(\frac{7}{9}\) (height per femur length), the change in femur length \(\Delta x\) is \(\Delta x=\frac{\Delta y}{m}=\frac{28}{\frac{7}{9}}=28\times\frac{9}{7}=36\). So the femur length is \(60 + 36=96\).

Answer:

96