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mr_godi [17]
3 years ago
13

If p+q+r = 1 and pq+qr+pr = -1, and pqr = -1, find the value of 

D%2B%20r%5E%7B3%7D%20%20%20" id="TexFormula1" title=" p^{3}+ q^{3}+ r^{3} " alt=" p^{3}+ q^{3}+ r^{3} " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Ugo [173]3 years ago
6 0
Pq+qr+pr= pqr (because they are both equal to -1)
then you divide both sides by p to get
q+ qr/p + r = qr. then you divide both sides by q and get
r/p + r/q = r. so I divided both sides by r to get
P + q =0 therefore p= -q. putting that into the first equation allows you to get that r=1 because p and q cancel each other out. then I put that into pq+qr+pr=-1 so pq+q+p=-1 since p=-q the q+p gets canceled out and you're left with pq=-1 and since it is the product of the two numbers and q is the negative of p. q=-1 while p=1. therefore p=1 r=1 and q=-1
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Using conditional probability as above:

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Cases for denominator when:

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Cases for denominator when:

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Cases for numerator when:

P (B = 12 & A + B + C = 21) = P(5,12,4) + P(6,12,3)

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