## delicious49 2 years ago 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

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1. asfdgl

answer: A Guy With More Money Than Me.

2. 221emily

lol that's good!!^^^^^

3. SithsAndGiggles

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 $E(X)=\sum_{\text{all }x}x~f(x)=1\cdot\frac{9}{23}+5\cdot\frac{8}{23}+10\cdot\frac{2}{23}+20\cdot\frac{4}{23}\approx6.48$