Results 1 to 8 of 8

Math Help - Develop algebraically an equation...

  1. #1
    Newbie
    Joined
    Aug 2008
    Posts
    14

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by bemypenguinxx View Post
    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, y=mx+c, where:
    • y and x will be your changing variable.
    • m is your gradient.
    • c is your y-intercept.
    When you have drawn your line of best fit, draw a triangle and find the gradient which is calculated by:
    m= \frac{y_1 - y_2}{x_1-x_2} where y_1,\ y_2, \ x_1, \ x_2 are coordinates of the triangle.

    c will be the value at which the line of best fit crosses the y-axis.
    Attached Thumbnails Attached Thumbnails Develop algebraically an equation...-scatter-graph-analysis.jpg  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2008
    Posts
    14
    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????
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by bemypenguinxx View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2008
    Posts
    14
    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

    m (gradient)
    = 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
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Aug 2008
    Posts
    14
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by bemypenguinxx View Post
    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

    m (gradient)
    = 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:

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

    y-140 = \frac{100}{27}\left(x-\frac{319}{2}\right)
    y-140 = \frac{100x}{27} - \frac{15950}{27}
    y = \frac{100x}{27} - \frac{12170}{27}
    \therefore \ \approx y = 3.70x-450.74
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Aug 2008
    Posts
    14
    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?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Help solving an exponential equation algebraically!
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: August 11th 2010, 10:13 AM
  2. solving an equation algebraically
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: February 21st 2010, 01:55 PM
  3. Solve the exponential equation algebraically?
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: October 12th 2009, 12:24 PM
  4. How Do I Develop This Equation?
    Posted in the Calculus Forum
    Replies: 0
    Last Post: April 28th 2009, 12:13 PM
  5. Develop a formula for cosine
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: July 8th 2008, 07:35 AM

Search Tags


/mathhelpforum @mathhelpforum