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dexar [7]
3 years ago
9

When randomly choosing two bills from a cup of bills that contains a dollar bill a two dollar bill a $5 bill a $10 bill and $20

bill and a $50 bill what is the probability of choosing a $20 bill and a $10 bill?
Mathematics
1 answer:
topjm [15]3 years ago
4 0

Answer:

The answer is 1/36

Step-by-step explanation:

Step one:

Given data

sample space= $1, $2,$5,$10,$20,$50

sample size= 6

Required

The probability of choosing 20 and 10 dollar bill

Step two

The probability is

1/6*1/6= 1/36

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serg [7]
132/2 = 66
66/2 = 33
33/3 = 11
11/11 = 1

132 = 2^2 * 3 * 11

55121/11 = 5011
5011/5011 = 1

55,121 = 11 * 5011

LCM = common factor with larger exponent * not common factors

LCM = 2^2 * 3 * 11 * 5011

LCM = 661,452
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How can you use a model to tell a percentage as a fraction with a denominator of 100?
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How do I solve for x?
notka56 [123]
Solving:
\frac{3+4x}{2} + \frac{5-x}{3} = \frac{29}{6}
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2,3,6\:|2
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<span>Replace denominators and resolve:
</span>\frac{3+4x}{2} + \frac{5-x}{3} = \frac{29}{6}
\frac{3(3+4x)}{6} + \frac{2(5-x)}{6} = \frac{29}{6}
Cancel the dominators
\frac{3(3+4x)}{\diagup\!\!\!\!6} + \frac{2(5-x)}{\diagup\!\!\!\!6} = \frac{29}{\diagup\!\!\!\!6}
3(3+4x) + 2(5-x) = 29
9 + 12x + 10 - 2x = 29
12x - 2x = 29 - 9 - 10
10x = 20 - 10
10x = 10
x =  \frac{10}{10}
\boxed{\boxed{x = 1}}\end{array}}\qquad\quad\checkmark


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Answer:

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Step-by-step explanation:

Hope this helps!

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