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Algebra 2 - residual^2 or something ?

  1. Sep 11, 2008 #1
    i have a function [tex]y=f(x)[/tex] with three points;
    [tex]A (1, 40)[/tex]
    [tex]B (3, 80)[/tex]
    [tex]C (5, 90)[/tex]

    the [tex]x[/tex] model represents days. and the [tex]y[/tex] model represents failed parts on an assembly line.

    the question is: how many failed parts will have been identified [tex]1[/tex] week from now?

    I have no problem with doing the work, getting the slopes, figuring out all the algebraic gymnastics and getting the equations.

    heres what i have:
    [tex]A\rightarrow B, y=20x-20[/tex]

    [tex]A\rightarrow C, y=\frac{25}{2}x+\frac{55}{2}[/tex]

    [tex]B\rightarrow C, y=\frac{77}{5}x+26[/tex]


    I actually think my equation from [tex]B\rightarrow C [/tex] could be incorrect.

    now, the part that is getting me is im trying to get the residual? i need to get some medication because my ADHD is seriously out of control. its getting worse as i get older.

    i need to construct some table... this is what i have for the first one:

    real obverved measures residual
    [tex]x,... y,... y_{1}=20x-20,... (y-y_{1})^2[/tex]
    [tex]1,... 40, ...40, ..................0[/tex]
    [tex]3,... 80, ...80, ..................0[/tex]
    [tex]5,... 90, ...120, ................900[/tex]


    real obverved measures residual
    [tex]x,... y,... y_{2}=\frac{25}{2}x+\frac{55}{2},... (y-y_{1})^2[/tex]
    [tex]1,[/tex]
    [tex]2,[/tex]
    [tex]3,[/tex]

    not quite sure how to do all of this. :(
     
  2. jcsd
  3. Sep 11, 2008 #2
    You might want to check your equation for A to C as well as from B to C...

    I am assuming of course you want to write an equation that goes through both those points in y = mx +b form.

    Other than that I can't help you because I don't really understand the question. I know you want to predict what y is when x is 7, but there are many methods of doing this and I am not sure which one you are supposed to be doing.
     
  4. Sep 12, 2008 #3
    Here are my Equations:


    [tex]A\rightarrow B, y=20x-20[/tex]

    [tex]A\rightarrow C, y=\frac{25}{2}x+\frac{55}{2}[/tex]

    [tex]B\rightarrow C, y=5x+65[/tex]

    the first two are correct because we did them in class. the third i just corrected, there was a minor mistake in my arithmetic that knocked everything off the rocker.

    now, i have to do something with the [tex]y[/tex] points [tex] A(1,40) B(3,80) C(5,90) [/tex]

    what i have to do for this table

    I have each point
     
    Last edited: Sep 12, 2008
  5. Sep 12, 2008 #4
    I think A -> B is incorrect!

    y = 20x - 20

    If this is supposed to go through the point A(1,40) then when I put 1 in for x I should get y = 40.


    y = 20(1) - 20 = 20 - 20 = 0

    So y = 20x - 20 does not go through the point (1,40) it goes through (1, 0)



    It's probably just a typo from you notes from your class, but the correct equation should be y = 20x + 20
     
  6. Sep 12, 2008 #5
    YOURE RIGHT!! i actually got the answer early in the class which was [tex]y=20x+20[/tex] i dont know why its copied incorrectly thereafter

    "[tex]y = 20(1) - 20 = 20 - 20 = 0[/tex]"

    how would i construct a table just as that for the other two equations... but, there was three parts to each. one for the Y value for each point.
     
  7. Sep 12, 2008 #6
    I'm sorry can you elaborate on the table you are supposed to create, I don't really understand that part.
     
  8. Sep 12, 2008 #7
    real obverved measures residual
    [tex]x,... y,... y_{1}=20x+20,... (y-y_{1})^2[/tex]
    [tex]1,... 40, ...40, ..................0[/tex]
    [tex]3,... 80, ...80, ..................0[/tex]
    [tex]5,... 90, ...120, ................900[/tex]

    i think i finally understand it. because i had the equation copied wrong it all came out wrong, the numbers didnt match up and it threw me off.

    [tex]
    y_{1}=20(1)+20=40[/tex]

    and the residual thing. i take the real [tex]y[/tex] which is [tex]40[/tex] and subtract it from the new [tex]y[/tex] which is [tex]40[/tex]: [tex]
    y_{1}=20(1)+20=40[/tex]

    [tex](40-40)^2=0[/tex]

    let me try and set this up

    my line: [tex]y=20(1)+20=40, (40-40)^2=0[/tex]

    [tex]y=20(3)+20=80, (80-80)^2=0[/tex]

    [tex]y=20(5)+20=120, (120-90)^2=900[/tex]
     
  9. Sep 12, 2008 #8
    [tex]
    B\rightarrow C, y=5x+65[/tex]

    [tex]y=5(1)+65=70, (70-40)^2=900[/tex]

    [tex]y=5(3)+65=80, (80-80)^2=0[/tex]

    [tex]y=5(5)+65=90, (90-90)^2=0[/tex]


    now, i have to figure out where i went wrong with the second line.. from A\rightarrow C.

    i dont have my notebook with me at the moment though, ugh.
     
  10. Sep 13, 2008 #9
    just worked it out and the second equation is correct

    [tex]
    A\rightarrow C, y=\frac{25}{2}x+\frac{55}{2}[/tex]


    [tex]y=\frac{25}{2}(1)+\frac{55}{2}=40, (40-40)^2=0[/tex]

    [tex]y=\frac{25}{2}(3)+\frac{55}{2}=65, (80-65)^2=225[/tex]

    [tex]y=\frac{25}{2}(5)+\frac{55}{2}=90, (90-90)^2=0[/tex]


    thanks for all the help! i think i may be able to hold it down from here, until class on tuesday :p
     
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