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Homework Help: Calculation of Ratio

  1. Aug 21, 2006 #1
    I have data arranged like this:


    So (18,0)--> 5.44 and (15,3)--> 7.63. Now different values of X and Y affect "Value". So it's two indepent variables affecting the depend variable, "Value".

    I need to figure out what the exchange rate of X and Y is. In other words, i need the Ratio X=kY where k is a constant of some sort.

    My first strategy was to generate 2 random pairs of values for (X,Y) that both give the same "Value". Then, i could deduce the ratio. E.g. If (10,0) and (15,10) give the same Value, then the ratio is 5X=10Y or X =2Y. But this has proven difficult because the values are difficult to generate!

    How could i determine the ratio strictly from the 2 columns above (keeping the Value(s) different)?! Is this possible?!
  2. jcsd
  3. Aug 21, 2006 #2


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    Staff: Mentor

    If x=18 and y=0, then k=infinity....
  4. Aug 21, 2006 #3
    I know. That's why there must always be a change in both X and Y.

    For example, if (15,0) and (10,2) give the same Value, then the ratio is 5X = 2y. And hence, k = 2/5.
  5. Aug 22, 2006 #4


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    What berkeman is saying is that the function of two variables value(X,Y)=z is not such that given a fixed value of z, the solutions {(X,Y)} are related by a proportionality relation X=kY, because we know that (18,0) is a solution of z(X,Y)=5.44, but there is no k such that 18=k0.

    There is however, a k such that k18=0; it is k=0.

    In general, the function that will give a relation btw X and Y of the form kX=Y when you fix z is of the form


    You can find the values of c_1 and c_2 using the values in your table. Now just generate a 3rd set of value (X,Y,z(X,Y)). If they do not satisfy the above equation, then z(X,Y) is not of this form and givena fixed z, X and Y are not related by a simple proportionality constant. They are related by something more complex.

    A simpler way still would be to verify if z(X=0,Y=anything) exists (i.e. is not infinity). If it does, then it is sufficient to conclude that z is not of the above form.
    Last edited: Aug 22, 2006
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