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    I'm part way through question b but I'm stuck I've got the equation 17.5-1/3n all divided by the square root of 2/9n> (2.6 using normal tables) am I on the right track? Every time I try to do the quadratic formula I end up having to square root a negative which is obviously incorrect


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    (Original post by Abby5001)
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    I'm part way through question b but I'm stuck I've got the equation 17.5-1/3n all divided by the square root of 2/9n> (2.6 using normal tables) am I on the right track? Every time I try to do the quadratic formula I end up having to square root a negative which is obviously incorrect


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    Check your value of 2.6 - I think it should be something else.
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    (Original post by SeanFM)
    Check your value of 2.6 - I think it should be something else.
    I've checked and I can't see what else it would be equal to


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    (Original post by Abby5001)
    I've checked and I can't see what else it would be equal to


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    Sorry, I read 0.05 instead of 0.005. :facepalm: I'll check again.

    At what point did you get the square root of a negative?
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    (Original post by SeanFM)
    Sorry, I read 0.05 instead of 0.005. :facepalm: I'll check again.

    At what point did you get the square root of a negative?
    When I got -1/9n^2-1.502+306.25 which makes me think I multiplied it out wrong


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    (Original post by Abby5001)
    When I got -1/9n^2-1.502+306.25 which makes me think I multiplied it out wrong


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    The numbers you've given me are slightly confusing I've left it in terms of 2.5758, sqrt(2/9), 17.5 etc and turned it into a quadratic that has a positive discriminant, so I think you've multiplied it out incorrectly somewhere.
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    (Original post by SeanFM)
    The numbers you've given me are slightly confusing I've left it in terms of 2.5758, sqrt(2/9), 17.5 etc and turned it into a quadratic that has a positive discriminant, so I think you've multiplied it out incorrectly somewhere.
    Yes I probably have. Can you show me the quadratic equation you got


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    (Original post by Abby5001)
    Yes I probably have. Can you show me the quadratic equation you got


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    Since the probability of him passing must be less than 0.005, the probability of him getting less than 18 questions right must be less than 0.005, which means that the z value must be greater than 2.5758 (difficult to think about but important).

    So 17.5 - n/3 > sqrt(2n/9) * 2.5758. Moving everything over to the right and using a substitution yields a quadratic with a positive discriminant.
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    (Original post by SeanFM)
    Since the probability of him passing must be less than 0.005, the probability of him getting less than 18 questions right must be less than 0.005, which means that the z value must be greater than 2.5758 (difficult to think about but important).

    So 17.5 - n/3 > sqrt(2n/9) * 2.5758. Moving everything over to the right and using a substitution yields a quadratic with a positive discriminant.
    I thought in order to get a quadratic you would have to square everything, so you would get rid of the sqrt(2n/9)?


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    (Original post by Abby5001)
    I thought in order to get a quadratic you would have to square everything, so you would get rid of the sqrt(2n/9)?


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    If you did it'd leave you with n^(3/2), not very nice

    Or you could try substituting u = n^1/2.
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    (Original post by SeanFM)
    If you did it'd leave you with n^(3/2), not very nice

    Or you could try substituting u = n^1/2.
    How would it leave it n^3/2 ?? And if I substituted n^1/2=u then how would that help me


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    (Original post by Abby5001)
    How would it leave it n^3/2 ?? And if I substituted n^1/2=u then how would that help me


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    I see what you mean by squaring both sides, whereas I thought it was getting everything to one side before squaring everything.

    However, a>b doesn't mean that a^2 > b^2 (eg a = -3, b = -5) and using that method does get you a funny quadtraic that can't be solved.

    Substituting u = n^1/2 gives you 17.5 - u^2/3 > (sqrt2/9) * u * 2.5758.
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    (Original post by SeanFM)
    I see what you mean by squaring both sides, whereas I thought it was getting everything to one side before squaring everything.

    However, a>b doesn't mean that a^2 > b^2 (eg a = -3, b = -5) and using that method does get you a funny quadtraic that can't be solved.

    Substituting u = n^1/2 gives you 17.5 - u^2/3 > (sqrt2/9) * u * 2.5758.
    To be honest I'm getting more confused so do you mind just simply rearranging things and putting down the value of a,b and c


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    We both started with 17.5 - n/3 > sqrt(2n/9) * 2.5758.

    Putting in u = n^1/2 gives u^2 = n, so 17.5-u^2/3 > sqrt(2/9) * sqrt(n) * 2.5758 = sqrt(2/9) * u * 2.5758.

    So 17.5-(u^2)/3 > sqrt(2/9) * u * 2.5758, which means that (u^2)/3 + (sqrt(2/9)*2.5758)u - 17.5 < 0.

    Solve for u and then for x. (it may be 2.5756 instead of 2.5758 but :dontknow:)
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    (Original post by SeanFM)
    We both started with 17.5 - n/3 > sqrt(2n/9) * 2.5758.

    Putting in u = n^1/2 gives u^2 = n, so 17.5-u^2/3 > sqrt(2/9) * sqrt(n) * 2.5758 = sqrt(2/9) * u * 2.5758.

    So 17.5-(u^2)/3 > sqrt(2/9) * u * 2.5758, which means that (u^2)/3 + (sqrt(2/9)*2.5758)u - 17.5 < 0.

    Solve for u and then for x. (it may be 2.5756 instead of 2.5758 but :dontknow:)
    Got the right answer Thankyou so much


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    (Original post by Abby5001)
    Got the right answer Thankyou so much


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    As long as you understand why, I am happy
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    (Original post by SeanFM)
    As long as you understand why, I am happy
    So in future when the discriminant is negative I should replace the letter I am solving the equation for with random letter^2 ?


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    (Original post by Abby5001)
    So in future when the discriminant is negative I should replace the letter I am solving the equation for with random letter^2 ?


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    We didn't do it for that reason, only to turn it Into something that we can solve (or if we had ax^4 + bx^2 + c) etc.
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    (Original post by SeanFM)
    We didn't do it for that reason, only to turn it Into something that we can solve (or if we had ax^4 + bx^2 + c) etc.
    Really quickly could you just explain why they have use the power rule 2^x+y =2^x x 2^y when if fact they are doing 2^x-y ? Name:  ImageUploadedByStudent Room1446237515.283061.jpg
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    (Original post by Abby5001)
    Really quickly could you just explain why they have use the power rule 2^x+y =2^x x 2^y when if fact they are doing 2^x-y ? Name:  ImageUploadedByStudent Room1446237515.283061.jpg
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    Swap y for -y in the rule. In other words, it works for negative values of x and y too.
 
 
 
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