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2 B on a P of S?

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Freefromchaos

Lvl 1
16y ago
Updated: 8/17/2019

2 blades on a pair of scissors.

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16y ago

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What 2 numbers added together to equal -6 and multiplied to equal -432?

The two numbers are 18 and -24.If you're having trouble with the factoring, you can always use the quadratic formula:Let the two numbers be a & b. a*b = P and a+b = S {for Product and Sum}So substitute b = S-a {from the Sum formula}, and you have a*(S-a) = P, or:a*S - a² = P ----> a² - S*a + P = 0.So with the quadratic formula:a1 = (-S + sqrt(S^2 - 4*1*P)/(2*1) Anda2 = (-S - sqrt(S^2 - 4*1*P)/(2*1)Substituting -432 for P and -6 for S, we get a1 = 18, and a2 = -24. Note that substituting a1 into the original formulas, gives b = a2, or if you use a2, then you get b=a1. So the two numbers are a1 and a2.


What is the geometrical proof of a-b whole square?

Given the graphic capability of this site, you are going to have to use some imagination! <---------a---------> <---a-b---><--b--> +-----------+-------+ |...............|..........| |.......P......|....Q...| |...............|..........| +-----------+-------+ |.......R......|....S....| |...............|..........| +-----------+-------+ In the above graphic, P, S and the whole figure are meant to be squares. The total area is P+Q+R+S = a2 P = (a-b)2 Q = b*(a-b) = (a-b)*b = a*b - b2 R = (a-b)*b = a*b = a*b - b2 and S = b2 Now, P = {P+Q+R+S} - Q - R - S = a2 - ab + b2 - ab + b2 - b2 = a2 - 2ab + b2


What is the probability of getting exactly 5 boys and 5 girls if you have 10 kids?

Here the sample space is(s)=10, =>mod(s)=10 a=be the event of getting exactly 5 boys =>mod(a)=5 b=be the event of getting exactly 5 girls =>mod(b)=5 thus, p(a)=mod(a)/mod(s)=5/10=1/2 p(b)=mod(b)/mod(s)=5/10=1/2 p(5 boys and 5 girls)=p(a)*p(b)=1/2*1/2=1/4


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


What is the geometrical proof of a plus b whole square?

In view of the graphic capabilities of this site, you will need to use a fair amount of imagination! Here goes: <------a+b------> <----a----><-b-> +-----------+-----+ |. . . . . . . .|. . . .| |. . . P. . . .|. .Q. | |. . . . . . . .|. . . .| +-----------+-----+ |. . . R. . . .|. .S. | |. . . . . . . .|. . . .| +----------+-----+ Where P = a*a = a2 Q = a*b R = b*a = a*b S = b*b = b2 (a + b)2 = P + Q + R + S = a2 + ab + ab + b2 = a2 + 2ab + b2


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