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    OK when you intergrate to find the cdf from the pdf
    wen do you have to find c and do you not need to find c
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    you don't have to do the +c as there should always be limits to the integration. If that makes sense/helps. We just went through this today in class.
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    There is no + c with s2 integration. Have no idea why, but that's the way it is. Not much help i know :o: Oh. Maybe there is according to Rachie. But i guess that's only when you don't have the limits.
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    If you're finding the cdf from the pdf, you'll end up with an integrated function - in this case, you do need to find C because it won't always be zero and the value for C will complete the function for the cdf.
    When you use limits to work out probability, you don't need to include C because you are taking away the two sets of values so they will cancel each other out.
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    (Original post by *Rachie*)
    If you're finding the cdf from the pdf, you'll end up with an integrated function - in this case, you do need to find C because it won't always be zero and the value for C will complete the function for the cdf.
    When you use limits to work out probability, you don't need to include C because you are taking away the two sets of values so they will cancel each other out.
    so how do you find the mode of

    fx= 1/17(-x^3+8x^2-7)
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    Try drawing the graph and it should be clearer. Then you will probably have to differentiate to find the max. point.
    What are the limits?
    I'm not sure about this one though, cba to do it myself but try that!
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    (Original post by Raow)
    Try drawing the graph and it should be clearer. Then you will probably have to differentiate to find the max. point.
    What are the limits?
    I'm not sure about this one though, cba to do it myself but try that!

    erm i knida did try but it goes over the limits as the limits is 1 to 2
    donna hw to draw the graph!!
    help
    ok wat about this one
    find cdf of 2/11 and intervals are bewteen 2<x<5
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    To draw the graph: Draw the graph as normal. Ignore parts of the graph where x doesnt equal 1 to 2 (in this case). Where x is any other value the graph is just 0. They can look a bit weird though. So you should end up with a straight line along the x axis and then a random bit of the graph fx= 1/17(-x^3+8x^2-7) between x=1 and x=2.

    What do you have to do for your other question? Explain more!
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    (Original post by Raow)
    To draw the graph: Draw the graph as normal. Ignore parts of the graph where x doesnt equal 1 to 2 (in this case). Where x is any other value the graph is just 0. They can look a bit weird though. So you should end up with a straight line along the x axis and then a random bit of the graph fx= 1/17(-x^3+8x^2-7) between x=1 and x=2.

    What do you have to do for your other question? Explain more!
    for the other one
    fx= 2/11 and you have to find the cdf
    interval is 2<x<5
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    I think it's 6/11
    As if the pdf is f(x)=2/11 then you have to integrate to find the cdf.
    So F(x)=[2x/11] between 2 and 5.
    =10/11-4/11
    =6/11
    voila (:
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    Where does the -2/11 come from?
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    (Original post by Raow)
    Where does the -2/11 come from?

    i dont kno
    goin to bed now nite nite
    wen i kno i post it to u
    nite then
 
 
 
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