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    Hey, too many questions.

    1.Is the reasoning in the first pic ok? Edit. No it wasn't.

    2. Is this correct?

    3. How is integral of Mx(t) not negative?

    4. Is my venn ok? I can get P(C n D) but then am not so sure what is going on, tried a few things. I should be aiming for the space to be 1 right?

    5. What is that

    \displaystyle \Phi^{-1}

    symbol about? How can I find these values using the z tables?
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    No, the reasoning is a bit faulty - I think you've mixed up where the "not" goes. You can tell it's faulty because it's true that if mn is even then m is even or n is even.
    You wanted "mn even => m even or n even" has contrapositive "m odd and n odd => mn odd".
    If you're asked to prove that mn is even implies m even or n is even, I think you need a special case of Bézout's Theorem,which states that if hcf(m,n) = k then am+bn = k has solutions. This adapts by:
    If m is not even, then hcf(2, m) = 1 and hence am + 2b = 1 has solutions, and hence amn + 2bn = n has solutions. But mn is assumed to be even, and hence amn is, so the LHS is even, and hence the RHS is even, so n is even. Hence at least one of m and n is even.
    Your answer on slide 2 is fine.
    Mx(t) is not negative because the integrand is positive (exponentials are positive).
    Your Venn diagram is fine. P(C U D) = P(we get one of b,e,d,f); you can look at this as P(we don't get a or c).
    The inverse-phi is exactly the inverse function of the phi function; that is, if phi(0) = 0.5, then phi^-1(0.5) = 0. You use the phi tables "in reverse" - that is, instead of looking at the row, and following along the columns to reach the right entry, you look for the entry in the table, then find which row/column it lies in.
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    (Original post by SubAtomic)
    Hey, too many questions.

    5. What is that

    \displaystyle \Phi^{-1}

    symbol about? How can I find these values using the z tables?
    It's the inverse of the cumulative normal distribution.

    You can either, for standard values, look it up in the small table that usually follows the standard normal distribution table,

    or, look for the value you're interested in in the body of the normal distribution table, and see what z values gives that.
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    (Original post by Smaug123)
    No, the reasoning is a bit faulty - I think you've mixed up where the "not" goes. You can tell it's faulty because it's true that if mn is even then m is even or n is even.
    Lol, after a second look it is indeed faulty, that pic is from a while ago,


    (Original post by Smaug123)
    You wanted "mn even => m even or n even" has contrapositive "m odd and n odd => mn odd".
    If you're asked to prove that mn is even implies m even or n is even, I think you need a special case of Bézout's Theorem,which states that if hcf(m,n) = k then am+bn = k has solutions. This adapts by:
    If m is not even, then hcf(2, m) = 1 and hence am + 2b = 1 has solutions, and hence amn + 2bn = n has solutions. But mn is assumed to be even, and hence amn is, so the LHS is even, and hence the RHS is even, so n is even. Hence at least one of m and n is even.
    Your answer on slide 2 is fine.
    :cool:

    (Original post by Smaug123)
    Mx(t) is not negative because the integrand is positive (exponentials are positive).
    Hmmm. I thought it should be lambda over t - lambda?

    (Original post by Smaug123)
    Your Venn diagram is fine. P(C U D) = P(we get one of b,e,d,f); you can look at this as P(we don't get a or c).
    So am I assuming that all events are contained in the sets A B C and D? P(C u D) = 0.5 ? Or am I making false assumptions that the sample space = 1? Maybe I have overcomplicated things :rolleyes:

    (Original post by Smaug123)
    The inverse-phi is exactly the inverse function of the phi function; that is, if phi(0) = 0.5, then phi^-1(0.5) = 0. You use the phi tables "in reverse" - that is, instead of looking at the row, and following along the columns to reach the right entry, you look for the entry in the table, then find which row/column it lies in.
    :cool:
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    (Original post by SubAtomic)
    Hmmm. I thought it should be lambda over t - lambda?
    Ah, I misunderstood you; I thought you meant Mx(t) was negative, which in hindsight was a silly thing for me to think given that you wrote "the integral of Mx(t)" :P I suspect it's only defined for t>lambda, but I have forgotten the circumstances as the picture seems to have vanished!

    (Original post by SubAtomic)
    So am I assuming that all events are contained in the sets A B C and D? P(C u D) = 0.5 ? Or am I making false assumptions that the sample space = 1? Maybe I have overcomplicated things :rolleyes:
    Yep, the fact that it's a sample space means every possible outcome is in the sample space, so there are six possible outcomes and their total probability is 1. C U D together rules out four of the six outcomes (if the outcome were e, then the outcome is in C U D, etc). To check, I made P(C) = 3/10. We can see that all possible outcomes are contained somewhere in A, B, C and D, because every one of the six possible letters is used somewhere.
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    (Original post by Smaug123)
    Ah, I misunderstood you; I thought you meant Mx(t) was negative, which in hindsight was a silly thing for me to think given that you wrote "the integral of Mx(t)" :P I suspect it's only defined for t>lambda, but I have forgotten the circumstances as the picture seems to have vanished!
    I edited my embarrassing proof out and must have taken that pic down, it is back now


    (Original post by Smaug123)
    Yep, the fact that it's a sample space means every possible outcome is in the sample space, so there are six possible outcomes and their total probability is 1. C U D together rules out four of the six outcomes (if the outcome were e, then the outcome is in C U D, etc). To check, I made P(C) = 3/10. We can see that all possible outcomes are contained somewhere in A, B, C and D, because every one of the six possible letters is used somewhere.
    Yep, I was just making it awkward for myself for some reason :cool:
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    Is this right?

    \displaystyle 3^{3^{3^{3}}}\equiv -72 \equiv 5\mod \ 11
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    (Original post by ghostwalker)
    It's the inverse of the cumulative normal distribution.

    You can either, for standard values, look it up in the small table that usually follows the standard normal distribution table,

    or, look for the value you're interested in in the body of the normal distribution table, and see what z values gives that.
    Hey, am still uncertain, oh, think it just clicked, so I was supposed to be looking for the value that gives 0.45 as the rhs of the distribution is full? I was looking at 0.475 :rolleyes:
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    (Original post by SubAtomic)
    I edited my embarrassing proof out and must have taken that pic down, it is back now
    Yep, I was right the first time - it says that it's only defined for t<lambda. It's like \cos^{-1}(t) is only defined for -1 \le t \le 1.
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    (Original post by SubAtomic)
    Is this right?

    \displaystyle 3^{3^{3^{3}}}\equiv -72 \equiv 5\mod \ 11
    I don't think so.

    3^3 = 27

    3^3^3 = 3^27 = 7625597484987

    So you are after 3^7625597484987

    Use FLT to make this less scary...
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    (Original post by SubAtomic)
    Hey, am still uncertain, oh, think it just clicked, so I was supposed to be looking for the value that gives 0.45 as the rhs of the distribution is full? I was looking at 0.475 :rolleyes:
    OK.

    It's more usual for tables to give the cumulative distribution from -infinity, rather than from 0, as the values will differ by 0.5 in the body of the table.

    As long as you're happy that such things are taken care of if/as necessary in your calculations.
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    D Franklin and ghostwalker in one thread :adore:

    So the mod one I must have done, not sure what I did looking at it now

    So would the correct solution be

    \displaystyle 3^{7625597484987} \equiv (3^{10})^x \cdot 3^7 \equiv 1 \cdot 3 \cdot 9 \cdot 4 \equiv 108 \equiv 9 \mod 11

    Maybe?
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    (Original post by ghostwalker)
    OK.

    It's more usual for tables to give the cumulative distribution from -infinity, rather than from 0, as the values will differ by 0.5 in the body of the table.

    As long as you're happy that such things are taken care of if/as necessary in your calculations.
    Thanks, I am not sure if I will ever be totally happy with my calculations, but I am as happy as I can be, thanks for the help :cool:
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    (Original post by SubAtomic)
    D Franklin and ghostwalker in one thread :adore:
    :rofl: I'll do the easy bit!
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    (Original post by ghostwalker)
    :rofl: I'll do the easy bit!
    all my questions are easy for you it is probably me that causes the most difficulty haha.
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    (Original post by SubAtomic)
    3. How is integral of Mx(t) not negative?
    What restrictions do you have on t? Look at the form of M_x(t) given to you after its been integrated. What do you notice?
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    (Original post by shamika)
    What restrictions do you have on t? Look at the form of M_x(t) given to you after its been integrated. What do you notice?
    t < lambda

    Confused, I will show what am doing then maybe it will be something ridiculous as usual :rolleyes:

    \displaystyle \lambda \int e^{(t- \lambda)x} \ dx = \frac{\lambda \cdot e^{(t- \lambda)x}}{t- \lambda}

    Where do I go from here to get that equation in the pic? Or is it just something I am supposed to take as given?
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    (Original post by SubAtomic)
    t < lambda

    Confused, I will show what am doing then maybe it will be something ridiculous as usual :rolleyes:

    \displaystyle \lambda \int e^{(t- \lambda)x} \ dx = \frac{\lambda \cdot e^{(t- \lambda)x}}{t- \lambda}

    Where do I go from here to get that equation in the pic? Or is it just something I am supposed to take as given?
    Evaluate it with the limits and then hopefully it'll make sense
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    (Original post by shamika)
    Evaluate it with the limits and then hopefully it'll make sense
    :emo: had to look at evaluating integrals with infinity, so I get this

    \displaystyle \frac{\lambda \cdot e^{-\infty}}{t-\lambda}- \frac{\lambda}{t- \lambda}}

    \displaystyle = \frac{\lambda \cdot e^{-\infty} - \lambda}{t- \lambda}

    Is this right? If so I think I may just about get it
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    (Original post by SubAtomic)
    :emo: had to look at evaluating integrals with infinity, so I get this

    \displaystyle \frac{\lambda \cdot e^{-\infty}}{t-\lambda}- \frac{\lambda}{t- \lambda}}

    \displaystyle = \frac{\lambda \cdot e^{-\infty} - \lambda}{t- \lambda}

    Is this right? If so I think I may just about get it
    Pretty much, yep - the infinite integral should technically be evaluated as a limit, but this works fine if you just use lim(e^-x) = 0 as x->inf.
 
 
 
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