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A box contains nine $1 bills, eight $5 bills, two $10 bills, and four $20 bills. What is the expectation if one bill is selected? A.$6.48 B.$8.48 C.$5.48 D.$7.48

Probability
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answer: A Guy With More Money Than Me.
lol that's good!!^^^^^
There's a total of 23 bills, with the following probabilities of selecting a particular denomination: \[P($1)=\frac{9}{23}\\P($5)=\frac{8}{23}\\ P($10)=\frac{2}{23}\\P($20)=\frac{4}{23}\] If \(X\) denotes the random variable for bill denomination, then \(X\) can take on any value from \(\{1,5,10,20\}\). So, if \(x\) is one of the values of \(X\) and \(f(x)\) is the probability of a particular \(x\), the expected value is \[E(X)=\sum_{\text{all }x}x~f(x)=1\cdot\frac{9}{23}+5\cdot\frac{8}{23}+10\cdot\frac{2}{23}+20\cdot\frac{4}{23}\approx6.48\]

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