# Thread: Develop algebraically an equation...

1. ## Develop algebraically an equation...

that related the height of students arm span to their height. I have collected the arm span and height in cm of 15 students in my class and have graphed them in a scatterplot. What does it mean to develop an algebraic equation relating arm span to height? How do I go about this? Is it like y=ax+b??? I have no idea! Please help

2. Originally Posted by bemypenguinxx
that related the height of students arm span to their height. I have collected the arm span and height in cm of 15 students in my class and have graphed them in a scatterplot. What does it mean to develop an algebraic equation relating arm span to height? How do I go about this? Is it like y=ax+b??? I have no idea! Please help
I believe that you have a positive correlation. Now, draw a line of best fit and find your equation of the line.

It will be in the form, $\displaystyle y=mx+c$, where:
• $\displaystyle y$ and $\displaystyle x$ will be your changing variable.
• $\displaystyle m$ is your gradient.
• $\displaystyle c$ is your $\displaystyle y$-intercept.
When you have drawn your line of best fit, draw a triangle and find the gradient which is calculated by:
$\displaystyle m= \frac{y_1 - y_2}{x_1-x_2}$ where $\displaystyle y_1,\ y_2, \ x_1, \ x_2$ are coordinates of the triangle.

$\displaystyle c$ will be the value at which the line of best fit crosses the $\displaystyle y$-axis.

3. How do I know where to draw the triangle to find the gradient? I don't know where on the line of best fit to draw it????

4. Originally Posted by bemypenguinxx
How do I know where to draw the triangle to find the gradient? I don't know where on the line of best fit to draw it????
It doesn't matter as long as the triangle touches the line of best fit. Be sure to draw a large triangle to get an accurate value of the gradient.

5. OK, I think I've gotten the gradient by using that rule. I have it so y2=190, y1=140, x2=173 and x1=159.5

I did this, but I don't think it's right :S

= y2-y1
x2-x1
= 190-140
173-159.5
≈ 3.57

c (y-intercept)
Using y-y1=m(x-x1)
y-140=3.57(x-159.5)
y-140=3.57x-569.42
+140 +140
y=3.57x-429.42

6. I then need to use that equation (presuming it's right) to predict my own arm span. Then I have to say whether it was accurate. I then need to figure out if that model is a direct variation relationship, and I have no idea what that is

THEN it says does your model have limitations in its domain and range?

Ahhh I'm really confused this is due soon and I don't know what to do

Thanks for helping.

7. Originally Posted by bemypenguinxx
OK, I think I've gotten the gradient by using that rule. I have it so y2=190, y1=140, x2=173 and x1=159.5

I did this, but I don't think it's right :S

= y2-y1
x2-x1
= 190-140
173-159.5
≈ 3.57

c (y-intercept)
Using y-y1=m(x-x1)
y-140=3.57(x-159.5)
y-140=3.57x-569.42
+140 +140
y=3.57x-429.42
My answer came slightly different to yours. I got:

$\displaystyle m = \frac{y_2 - y_1}{x_2 - x_2} = \frac{190 - 140}{173-159.5} = \frac{50}{13.5} = \frac{100}{27}$

$\displaystyle y-140 = \frac{100}{27}\left(x-\frac{319}{2}\right)$
$\displaystyle y-140 = \frac{100x}{27} - \frac{15950}{27}$
$\displaystyle y = \frac{100x}{27} - \frac{12170}{27}$
$\displaystyle \therefore \ \approx y = 3.70x-450.74$

8. OK that makes much more sense to me now. I've worked through the equation you gave me and I understand it a lot better now. The question asks me if the model represents a direct variation relationship? I don't know what direct variation is?

Thank you for all your help so far, it's really helped me!

EDIT: Is it the same as direct proportion? If so, I have some work I've done on it but it's how to solve proportion questions, not how to identify them which I don't know how to do?