No. A can be independent of both B and C and this doesn't give use any information about the relationship between B and C.
A=B , A-B=B-B , A-B =0 B=C , B-B=C-B, 0=C-B So A-B=0 but also C-B=0 A-B=C-B ...add +b ...A-B+B=C-B+B , A=C
a/b = c/d Multiply both sides by b/c Thus (a/b)*(b/c) = (c/d)*(b/c) ie a/c = b/d
b + b + b + c + c + c + c = 3b + 4c
a - b = c -(a - b) = -c b - a = -c
A or B or C = A + B + C - A and B - A and C - B and C - 2 (A and B and C) I'm not sure by the way;
an independent variable
c
Forest fire B. Drought C. Hurricane
C. B. Mason has written: 'Aelink - an independent program linking facility for the IBM 360' -- subject(s): IBM 360 (Computer), Programming
If x depends on a, b and c, then x is the dependent variable, and a, b, and c are the independent variables - you can vary them at will, and x depends on them. Often it appears on the right hand side of an equation, such as x = a +b + 2/c, showing how x depends on the independent variables.
p(A and B) = p(A) x p(B) for 2 independent events p(A and B and ...N) = p(A) x p(B) x p(C) x ...x p(N) In words, if these are all independent events, find the individual probabilities if each and multiply them all together.
(b b b)( b b b )(b d g a)(b....)(c c c c)(c b b b)(a a a b)(a...d)(b b b)(b b b)(b d g a)(b....)(c c c c)(c b b b)(d d c a)(g.....)
C is hardware independent
Yes.
a b c a c a d a c c c a c b b b a a c b b b a c c b
a b c c c c b a g g a b g a b c c c c b a g b a a b c c c c b a g g a b a b c d b c e c b a b a g g
a b c c c c b a g g a b g a b c c c c b a g b a a b c c c c b a g g a b a b c d b c e c b a b a g g