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    Mean and s.d of normal distribution? (The time taken for a randomly chosen student to complete the test is...)

    I will see if i can get hold of more details.
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    Mean and s.d of normal distribution? (The time taken for a randomly chosen student to complete the test is...)

    Thank you, please explain more! in particular the uniqueness of the solutiion..
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    Mean and s.d of normal distribution? (The time taken for a randomly chosen student to complete the test is...)

    Anyway, thanks for your help. I wanted some reassurance that i wasn't going mad. No need to ponder much more on it.
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    Mean and s.d of normal distribution? (The time taken for a randomly chosen student to complete the test is...)

    The question appeared in a book that was sent to me online from a teacher friend of a friend. Don't know much more about it i am afraid. But i can't see a way of doing it except maybe interpreting the statement so we can ignore then X>41 bit because sigma might be small enough. The topic is...
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    Mean and s.d of normal distribution? (The time taken for a randomly chosen student to complete the test is...)

    Maybe but not sure how you would use that in this question. What z value would you use?
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    Mean and s.d of normal distribution? (The time taken for a randomly chosen student to complete the test is...)

    Any stats experts out there who can help me with this question? For anyone familiar it looks like a fairly common question where you are given two bits of information and use it to find the mean and s.d. So its easy enough to interpret the second bit ( 1 in 8 take less than 35 minutes) and...
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    Nice problem. Is there an elegant solution? (Find curve for set of chords' midpoints)

    It turns out the resulting quadratics is y=2x^2 This is how i did it but it feels like there might be a neater solution so i'd be interested if anyone has the time/inclination for alternative approaches! SPOILER BELOW... ( I have missed out some steps for brevity) Let P be a point on the...
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    Inverse Trig Problem: Prove arcsin(1/x) = -arccos((sqrt{x^2 - 1})/x)

    Ok, i have had a go: If anyone can check it over i would be grateful. The tricky bit is knowing which cos y to pick and i think i can justify it, but would prob fail under intense interrogation. I think we would pick the negative sqrt because we need to cos y to be positive?
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    Inverse Trig Problem: Prove arcsin(1/x) = -arccos((sqrt{x^2 - 1})/x)

    Thanks for your thoughts so far. I will have a play.
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    Inverse Trig Problem: Prove arcsin(1/x) = -arccos((sqrt{x^2 - 1})/x)

    What i have tried so far: siny =1/x and cos y = - sqrt(1-1/x^2) Then i am thinking there are conditions on siny and cosy which would lead to inequalities: -1<=siny<=1 which leads to -1<=1/x<=1 and similarly for the cos y expression And this is where i get stuck I am also thinking of using...
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    Inverse Trig Problem: Prove arcsin(1/x) = -arccos((sqrt{x^2 - 1})/x)

    Can anyone offer any thoughts on this problem please?
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    Gradient at a point on a quadratic: gradient at (3, 21) for y = x^2 + 4x

    Yes i had assumed every realised that was a typo, given the original problem!
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    Gradient at a point on a quadratic: gradient at (3, 21) for y = x^2 + 4x

    That makes a lot of sense thank you. Its not what i was thinking but explains why the correct result stumbles out. Thank you again, you have changed my life!
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    Gradient at a point on a quadratic: gradient at (3, 21) for y = x^2 + 4x

    This was where i was heading but i don't understand why this shortcut(above) seems to give the same answer:
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    Gradient at a point on a quadratic: gradient at (3, 21) for y = x^2 + 4x

    Yes, i can do this by calculus and work it out. But i was exploring an alternative method to find the gradient using the idea of discriminant and points of intersection
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    Gradient at a point on a quadratic: gradient at (3, 21) for y = x^2 + 4x

    I can't explain it. This is what i am really asking. It started off as an attempt to use the intersection points of a general line that passes through the point with the given quadratic. I actually wanted to form a quadratic in terms of x with 'm's and use the discriminant. But i didn't get that...
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    Gradient at a point on a quadratic: gradient at (3, 21) for y = x^2 + 4x

    Can anyone offer an explanation as to why the following works. It feels like there is some dodgy maths going on ...
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    Integration by Parts: Integrate e^{ax} * sin(bx) with respect to x

    I think you have done all the hard work already? Just tidy up step 5 what is 1/b^2 divided by (1+ a^2/b^2) ?
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    Stuck on this (Find the value of x from drawing of angles inside a circle)

    So you can still retain the 68 and 31 for different x?
  20. A

    Stuck on this (Find the value of x from drawing of angles inside a circle)

    This suggests its is not a well formed question?
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