Q: 8 w p on a c b?

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a b c, t r w, z p t; any three variables

P = 2L + 2W P/2 = L + W 28/2 = 8 + W 14 = 8 + W 6 = W A = LW A = (8)(6) A = 48 unit2

it has three cases 1st E1- two balls are whiteE2- three balls are whiteE4-four balls are whiteE4/w= (p(E3)*4C2/4C2)/(P(E1)*2C2/4C2+P(E2)3C2/4C2+P(E3)*4C2/4C2)WHERE-P(E1)=P(E2)=P(E3)=1/3----------------------------------------------------------------------------------------------------2nd opinionLet's say we have 3 boxes with 4 balls each.Box A has 4 white balls.Box B has 3 white balls.Box C has 2 white balls.The probability of drawing 2 W balls from;Box A, P(2Wâ”‚A)=(4/4)âˆ™(3/3)=1Box B, P(2Wâ”‚B)=(3/4)âˆ™(2/3)=1/2Box C, P(2Wâ”‚C)=(2/4)âˆ™(1/3)=1/6Say the probability of picking any of the 3 boxes is the same, we have;P(A)=1/3P(B)=1/3P(C)=1/3Question is, given the event of drawing 2 W balls from a box taken blindlyfrom the 3 choices, what is the probability that the balls came from box A,P(Aâ”‚2W).Recurring to Bayes Theorem:P(Aâ”‚2W)=[P(A)P(2Wâ”‚A)]/[P(A)P(2Wâ”‚A)+P(B)P(2Wâ”‚B)+P(C)P(2Wâ”‚C)]=[(1/3)(1)]/[(1/3)(1)+(1/3)(1/2)+(1/3)(1/6)]=6/10=0.60=60%P(Aâ”‚2W)=0.60=60%Read more:Solution_to_a_bag_contains_4_balls_Two_balls_are_drawn_at_random_and_are_found_to_be_white_What_is_the_probability_that_all_balls_are_white

the odds theoretically are almost infinity to 1 as they could be any color. -------------------------------------------------------------------------------------------2nd opinionLet's say we have 3 boxes with 4 balls each.Box A has 4 white balls.Box B has 3 white balls.Box C has 2 white balls.The probability of drawing 2 W balls from;Box A, P(2Wâ”‚A)=(4/4)âˆ™(3/3)=1Box B, P(2Wâ”‚B)=(3/4)âˆ™(2/3)=1/2Box C, P(2Wâ”‚C)=(2/4)âˆ™(1/3)=1/6Say the probability of picking any of the 3 boxes is the same, we have;P(A)=1/3P(B)=1/3P(C)=1/3Question is, given the event of drawing 2 W balls from a box taken blindlyfrom the 3 choices, what is the probability that the balls came from box A,P(Aâ”‚2W).Recurring to Bayes Theorem:P(Aâ”‚2W)=[P(A)P(2Wâ”‚A)]/[P(A)P(2Wâ”‚A)+P(B)P(2Wâ”‚B)+P(C)P(2Wâ”‚C)]=[(1/3)(1)]/[(1/3)(1)+(1/3)(1/2)+(1/3)(1/6)]=6/10=0.60=60%P(Aâ”‚2W)=0.60=60%Read more:Solution_to_a_bag_contains_4_balls_Two_balls_are_drawn_at_random_and_are_found_to_be_white_What_is_the_probability_that_all_balls_are_white

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W=wrought p=pipe b=bend

It's been a while, but I'll take a stab at this one. Assuming this is a rectangular shape: width = w length = w + 12 perimeter = length + length + width + width As stated above, p = 7w, so: 7w (perimeter) = (w + 12) + (w + 12) + w + w 7w = 4w + 24 (simplifying the above) -4w -4w (getting all the w's on one side) ------------------ 3w = 24 w = 8 Checking our work. Substituting 8 for w: p = 7(8) p = 56 P = (8+12) + (8 + 12) + 8 + 8 p = 20 + 20 + 8 + 8 p = 56

P = 2L + 2W P/2 = L + W 28/2 = 8 + W 14 = 8 + W 6 = W A = LW A = (8)(6) A = 48 unit2

W=wrought p=pipe b=bend

2 W on a B P stands for "2 wheels on a bicycle pump," which is a common riddle or puzzle.