Recent content by HF08

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    Combinatorial Probmlem

    Let's work on the 2 weeks without replacement. There are C(9,3) ways to take out 3 people from a group of 9. This is where I get stuck. How can I calculate the odds of an individual being picked as 1 of 3 for the 2nd week? I have difficulty with understanding combinatorics/permutations. So...
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    Statistics: Confidence Intervals

    Hey LearnFrench, How do they determine the 0.93 statistic in the first place? Assuming I have no life, how could I derive a table on my own? Thanks, HF08
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    Combinatorial Probmlem

    Hi, I was reading a models book and read about a study where 9 analysts were chosen at random. Out of this group, 3 were selected for 2 week training, 3 were selected for 3 week training, and 3 were selected for 5 week training. Now I believe this models course had the idea that the first...
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    I have variance of response. How can I find it's MSE?

    To All, I did a study and my response is defined as y = b1x1 + b2x2 + e where e ~N(0,1). I have y~N(4,33). In my data results, I did an ordinary least squares regression model for y = b1x1+b2x2+ e. The ANOVA is telling me the mean of y is 4, but MSE is 1. So here is my question. If I...
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    Show multiplicative order is multiplicative (told this was easy but I can't see it)

    One more question I was trying to show this part rigorously. Do you have any ideas on how I might do this? I agree it is true, but I am wondering if I should shore it up exactly.
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    Show multiplicative order is multiplicative (told this was easy but I can't see it)

    Thanks Thank you very much for your help. I wonder if your interesting observation may give us an easier proof. lcm(x,y) = x*y/gcd(x,y) = x*y. I think this is the last proof of that. I guess we could call that a lemma or corollary? Seems to me something easier to prove after we...
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    Show multiplicative order is multiplicative (told this was easy but I can't see it)

    Thanks. Got it. (Almost got it?) Yes, you are correct. The notation was getting to me. I believe I am more clear on your presentation and comments now. Thank you. We know gcd(x,y) = 1 and x|k and y|k implies xy|k. Since ab^xy = 1, then k|xy. Hence, k= xy. Note: (a^y)^x = 1...
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    Show multiplicative order is multiplicative (told this was easy but I can't see it)

    So how can I justify just saying a^k=1 and b^k=1 for some k? I doubt I can just say, suppose there is some k>x and k>y such that.... Clearly, number 2 is just saying that the order of ab is k. I think I can show (ab)^k=a^k b^k=1 easily. So I get x|k and y|k. Since gcd(x,y) = 1, then xy|k...
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    Show multiplicative order is multiplicative (told this was easy but I can't see it)

    gcd(x, k ) = x, since x is the smallest integer such that a^x = = 1 mod n Right? Am I getting warmer? I have been stating x | k ?
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    Show multiplicative order is multiplicative (told this was easy but I can't see it)

    Thanks for the reply and some comments of my own... I found your post confusing. This is what I have been doing. Staying true to the notation in my original problem. Part I. Let x = ord_n(a). So a^x = = 1 mod n iff gcd(a,n) = 1 Now, let y = ord_n(b). (a^y)x = = 1 mod n. Hence...
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    Show multiplicative order is multiplicative (told this was easy but I can't see it)

    Thanks for your quick reply. Well, the way the problem is written is written with gcd ( ord_n(a), ord_n(b) ) = 1. So, does this condition imply gcd (a, b) = 1? Please tell me how to solve this with just (a , b) = 1. Perhaps there is something analogously that may be done. Thanks, HF08
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    Show multiplicative order is multiplicative (told this was easy but I can't see it)

    Please help me prove this. I worked hard to make my notation somewhat easy to follow: If gcd ( ord_n(a) , ord_n(b) ) = 1 , then ord_n(a*b) = ord_n(a) * ord_n(b) Attempted proof: I can't see how to use the gcd condition. Which is bad news. I do realize the following. Let k1 =...
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    Show integrable is uniformly continuous

    H = [a,b]\times[c,d] . f:H\rightarrowR is continuous, and g:[a,b]\rightarrowR is integrable. Prove that F(y) = \intg(x)f(x,y)dx from a to b is uniformly continuous. I initially ripped g(x) and f(x,y) apart and tried to show each was continuous. This failed. In short, I am...
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    Help me evaluate this limit

    Thanks Excellent post HallsofIvy. I have found both limits to be 0. This is more useful than the definition approach. If you agree with what the limit is, I will mark this solved. Thank You, HF08
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    Help me evaluate this limit

    [SOLVED] Help me evaluate this limit Part a: Show that f(x,y) = \frac{x^{4}+y^{4}}{x^{2}+y^{2}} as (x,y) -> (0,0) Part b: Similarily, show that f(x,y) = \frac{\sqrt{\left|xy\right|}}{\sqrt[3]{x^{2}+y^{2}}} as (x,y)->(0,0) Lets work with Part a, shall we? I seemed to be...
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