Search results

which book is it

You must have left something out... for example something along the lines of NML JKL add up to 90

Which university/course is this for? you should really look for more suitable forums than this one I assume your "A is true" means that the sentence A is true in the standard model that is N |= A where N denotes the standard model you can't have PA |-- A PA |-/- A in fact precisely one of...

It's difficult if you try to compare everything with english and try to translate everything

This guy knows what he is on about.

If sicmacao is who I think he is.. then he definitely knows his stuff. EDIT: confirmed..

You know you are doing pure maths when you don't see numbers more than 10.

as long as you know what you are on about they will give you the marks... math teachers at state schools were quite good, atleast when I did the HSC

Thats call drudgery.. not mathematics

that results in a simpler expression than using P(* <=n) since P(A and B) = P(A)P(B) but you can;t say the same with P(A or B) (when A, B are independent) with the other one you will have to interpret the meaning of "a geometric rv"

Don't drop if you have 12 or less units.. makes you lazy

The usual algorithm is: identify forces acting on the particle, this should be a complete list (I hope): 1. Gravity 2. Normal reaction force 3. Tension 4. Friction/resistance then fill in those you know and put an arrow and a letter in for those you don't. gravity is straight forward...

This my dear is just as indecipherable as a postmodernist text. the terms conceptual learning, integrity and analytical could mean almost anything. In most cases these terms are just flashed around without the user needing to know what they are on about. If I interpreted your post...

Stochastic analysis owns

You can go along the lines of: P(min(x1,x2)>n) = P(x1>n and x2 > n) = P(x1 > n)P(x2>n) = [(1-p)^2]^n or you can argue that a geometric random variable is the number of indepedent trials needed to record a success.. now with min(x1,x2) is the smaller of these so you can look at it as 1...

just stick with it..

you can't sum like that. If you condition, then you would need to write for, x1 <= x2 Sigma(x1=1 to infinity) Sigma(x2=x_1 to infinity) z^x1 p(1 - p)^(x1-1) . p(1 - p)^(x2-1) = ... for, x2 < x1 Sigma(x1=1 to infinity) Sigma(x2=1 to x_1-1) z^x2 p(1 - p)^(x1-1) . p(1 - p)^(x2-1) = ...

I think this is wrong, zp/(1-z+zp) = \sum_{n=1}^\infty z^n f(n) though

Oh this makes it different.. and easier. I knew it was something related.. If you consider 2 random variables X1 and X2 with geometric distribution given by the fs.. (that is with parameter p) then Y=min(X1,X2) is another geometric random variable with parameter q=[1-(1-p)^2]. (think why!) so...

Traditional