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Orlov [11]
3 years ago
7

Suppose P(C|D)=0.59, P(D)=0.44, and P( D|C)=0.38. What is P(C) rounded to two decimal places?

Mathematics
1 answer:
Galina-37 [17]3 years ago
8 0
From the information given.
P(CΙD) = P(C∩D)/P(D)
P(DΙC) = P(C∩D)/P(C)
Therefore;
P(C∩D) = P(CΙD) × P(D)= P(DΙC) ×P(C)
P(C∩D)= 0.59×0.44
             = 0.2596
P(C) = P(C∩D)/P(DΙC)
        =  0.2596/0.38
        = 0.68316
        ≈ 0.68
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Answer:

See explanation

Step-by-step explanation:

There are three possible cases:

1. Point N lies between M and P, then MN + NP = MP. Consider needed difference:

\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MN}{NP}-\dfrac{MN}{MN+NP}=\dfrac{MN(MN+NP)-MN\cdot NP}{NP(MN+NP)}=\\ \\=\dfrac{MN^2+MN\cdot NP-MN\cdot NP}{NP(MN+NP)}=\dfrac{MN^2}{NP(MN+NP)}

2. Point N lies to the right from point P, then MP + PN = MN.  Consider needed difference:

\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MP+PN}{NP}-\dfrac{MP+PN}{MP}=\dfrac{MP}{NP}+1-1-\dfrac{NP}{MP}=\dfrac{MP^2-NP^2}{NP\cdot MP}

3. Point N lies to the left from point M, then NM + MP = NP. Consider needed difference:

\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MN}{MN+MP}-\dfrac{MN}{MP}=\dfrac{MN\cdot MP-MN(MN+MP)}{MP(MN+MP)}=\\ \\=\dfrac{MN\cdot MP-MN^2-MN\cdot MP}{MP(MN+MP)}=\dfrac{-MN^2}{MP(MN+MP)}

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Answer:

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Step-by-step explanation:

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