A=b*h area = base (times) height P=2b + 2h perimeter = 2 (times) base (plus) 2 ( times) height A= b*h 12 = b*h 12/b = h let x = b h = 12/x since x = b and h = 12/x therefore P = 2x + 2(12/x)
If B is between P and Q, then: P<B<Q
The probability of A is denoted P(A) and the probability of B is denoted P(B). P(A or B) = P(A) + P(B) - P(A and B). Say P(A) = Probability of drawing a heart, which is 13/52. Say P(B) = Probability of drawing a three, which is 4/52. We now have to determine P(A and B) which is the probability of a heart and a three, which is 1/52. We now can determine the probability of drawing a heart or a three which is 13/52 + 4/52 - 1/52 = 16/52 = 4/13.
A triangle with side a: 5, side b: 8, and side c: 11 units has an area of 18.33 square units.This is not a right triangle, so it has to be calculated as two smaller right triangles, or using Heron's Formula since the side lengths are known.Here p= half-perimeter = (5 + 8 +11)/2 =12A = sqrt {p [(p - a) * (p - b) * (p - c)]}A = sqrt [ 12 (7 x 4 x 1)]A = sqrt [12 x 28]A = sqrt 336 = about 18.33
P = 5 A = 6 B = 11 Or.... P = 1 A = 30 B = 55
12 buttons on a phone
P. B. S. Pinchback died on 1921-12-21.
12 strikes for a perfect game
There are symbols missing from your question which I cam struggling to guess and re-insert. p(a) = 2/3 p(b ??? a) = 1/2 p(a ∪ b) = 4/5 p(b) = ? Why use the set notation of Union on the third given probability whereas the second probability has something missing but the "sets" are in the other order, and the order wouldn't matter in sets. There are two possibilities: 1) The second probability is: p(b ∩ a) = p(a ∩ b) = 1/2 → p(a) + p(b) = p(a ∪ b) + p(a ∩ b) → p(b) = p(a ∪ b) + p(a ∩ b) - p(a) = 4/5 + 1/2 - 2/3 = 24/30 + 15/30 - 20/30 = 19/30 2) The second and third probabilities are probabilities of "given that", ie: p(b|a) = 1/2 p(a|b) = 4/5 → Use Bayes theorem: p(b)p(a|b) = p(a)p(b|a) → p(b) = (p(a)p(b|a))/p(a|b) = (2/3 × 1/2) / (4/5) = 2/3 × 1/2 × 5/4 = 5/12
12 scored in a perfect game of bridge.
if P(A)>0 then P(B'|A)=1-P(B|A) so P(A intersect B')=P(A)P(B'|A)=P(A)[1-P(B|A)] =P(A)[1-P(B)] =P(A)P(B') the definition of independent events is if P(A intersect B')=P(A)P(B') that is the proof
A=b*h area = base (times) height P=2b + 2h perimeter = 2 (times) base (plus) 2 ( times) height A= b*h 12 = b*h 12/b = h let x = b h = 12/x since x = b and h = 12/x therefore P = 2x + 2(12/x)
Sum Rule: P(A) = \sum_{B} P(A,B) Product Rule: P(A , B) = P(A) P(B|A) or P(A, B)=P(B) P(A|B) [P(A|B) means probability of A given that B has occurred] P(A, B) = P(A) P(B) , if A and B are independent events.
It means the answer is -11p.
The probability of inclusive events A or B occurring is given by P(A or B) = P(A) + P(B) - P(A and B), where P(A) and P(B) represent the probabilities of events A and B occurring, respectively.
P(A|B)= P(A n B) / P(B) P(A n B) = probability of both A and B happening to check for independence you see if P(A|B) = P(B)
If they're disjoint events: P(A and B) = P(A) + P(B) Generally: P(A and B) = P(A) + P(B) - P(A|B)