Variables stand in the place of unknown numbers. For example, in the following equation, one number is unknown: 2+x=5. The x takes the place of the number that is unknown.
You could be learning algebra so the letter P and Q could be there filling in for another number. You can also use other letters.
p and q
The answer is negative four BECAUSE... 20/5 is POSITIVE four 20/-5 is a NEGATIVE four because a positive divided by a negative is a negative. Easy way to remember negs/pos: n * p = n p* n = n p * p = p n * n = p n / p = n p / n = n p / p = p n / n = p There are always two ways to get a positive, and two ways to get a negative. Very simple.
Multiplying by a negative changes the sign of the original number. P x N = N N x N = P Multiplying by a positive keeps the sign of the original number. P x P = P N x P = N
Proof: P{T>n+m/T>n}=P{T>n+m,T>n}/P{T>n} (Bayes theorem) =P{T>n+m}/P{T>n} =((1-p)^(n+m))/(1-p)^n = (1-p)^(n+m-n) = (1-p)^m (1-p)^m = {T>m} So T>m has the same probability as T>m+n given that T>n, which means it doesn't care (or don't remember) that n phases had passed.
In Algebra there are constants and variables. 'p' is a variable which means that its value can change with reference to the equation it is present in. In Physics, 'p' represents momentum.
YOU CAN USE ALGEBRA FOR A LOT OF THING. THE LETTERS JUST REPLACE NUMBERS THAT YOU DO NOT KNOW. SAY YOU DID 2P+5N( TAKE P=5 AND N=8) YOU WILL DO 2X5=10 AND 5X8=40, SO YOUR ANSWERS 50.
Ah ha! In algebra, a letter stands for an unknown value! Some of the most common letters used in algebra are: x, y, n, and p. Some might say 'n' is an uNknown number! haha
If X and Y are i.i.d Poisson variables with lambda1 and lambda2 then, P (X = x | X + Y = n) ~ Bin(n, p) where p = lambda1 / lambda1 + lambda2
No, but if you really want that, use type-cast, for example:printf ("main=%p=5*%lx\n", main, ((long)main)/5);Of course it is good for nothing at all.
You could be learning algebra so the letter P and Q could be there filling in for another number. You can also use other letters.
p and q
P! / q!(p-q)!
khara
Diodes.
P= positive N=negative P x N = N N x P = N P x P = P N x N = P Hope that helps!?!?!
We have to use the expression P(X=x) = nCx px (1--p)(n--x) Here n = n and p=p and x = 1 or x>1 P(X>/=1) = 1 -- P(X</=1) So, P(X<=1) = P(X=0) + P(X=1) This gives nC0 p0 (1--p)(n-0)+ nC1 p1 (1--p)(n--1) ie (1--p)n + n p (1--p)(n--1)