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...
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
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...
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...
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.
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...
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...
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...
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...
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
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 =...
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...
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
[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...