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In Calculus what is the formula for the described function A if a rectangle has area 12 m2 and the perimeter P of the rectangle is expressed as a function of the length x of one of its sides?
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?
If B is between P and Q, then: P<B<Q
What is the probability of drawing either a 3 or a heart from a regular bridge deck of cards?
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.
What is the area of a triangle with sides a 5 b 8 and c 11?
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
What is PA and B for PA equals 30 and PB equals 55?
P = 5 A = 6 B = 11 Or.... P = 1 A = 30 B = 55
12 B on a P?
12 buttons on a phone
When did P. B. S. Pinchback die?
P. B. S. Pinchback died on 1921-12-21.
How do you find p(b) when p(a) is 23 p(ba) is 12 and p(a U b) is 45 and is a dependent event?
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 S for a P G in B?
12 strikes for a perfect game
12 s in a p g of b?
12 scored in a perfect game of bridge.
If A and B are independent events then are A and B' independent?
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
In Calculus what is the formula for the described function A if a rectangle has area 12 m2 and the perimeter P of the rectangle is expressed as a function of the length x of one of its sides?
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)
What is the product rule and the sum rule of probability?
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.
What does 12 equals b on a p-bp mean?
It means the answer is -11p.
What is the formula for inclusive events?
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.
How do you find P A given B?
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)
Addition rule for probability of events A and 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)
