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Bogdan [553]
4 years ago
12

A DJ for a school dance has a CD with 6 slow songs and 5 fast songs on it. As he plays each song he removes it from the play lis

t. What is the probability that the first two songs he plays are slow
Mathematics
2 answers:
mrs_skeptik [129]4 years ago
7 0
2:11 that`s the ratio of the first 2 being slow

Dima020 [189]4 years ago
7 0
The answer is 6/11 because there are 11 total songs and 6 slow songs are being played or 54%


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A box of Munchkins contains chocolate and glazed donut holes. If Jacob ate 2 chocolate
neonofarm [45]

Answer:

24 munchkins.

Step-by-step explanation:  

Let C be the number of chocolate and D be number of glazed donut holes in the original box.

We are told if Jacob ate 2 chocolate  munchkins, then 1/11 of the remaining Munchkins would be chocolate. We can represent this information as:

C-2=\frac{1}{11}*(C+D-2)...(1)

We are also told if he instead added 4  glazed Munchkins to the original box, 1/7 of the Munchkins would be chocolate. We can represent this information as:

C=\frac{1}{7}*(C+D+4)...(2)

Upon substituting C's value from equation (2) in equation (1) we will get,

\frac{1}{7}*(C+D+4)-2=\frac{1}{11}*(C+D-2)

Let us have a common denominator on right side of equation.

\frac{1}{7}*(C+D+4)-\frac{7*2}{7}=\frac{1}{11}*(C+D-2)

\frac{C+D+4-14}{7}=\frac{1}{11}*(C+D-2)

Multiplying both sides of our equation by 7, we will get,

7*\frac{C+D-10}{7}=7*\frac{1}{11}*(C+D-2)

C+D-10=\frac{7}{11}*(C+D-2)  

Multiplying both sides of our equation by 11, we will get,

11*(C+D-10)=11*\frac{7}{11}*(C+D-2)  

11*(C+D-10)=7*(C+D-2)

11C+11D-110=7C+7D-14

11C-7C+11D-7D=-14+110  

4C+4D=96

4(C+D)=96  

(C+D)=\frac{96}{4}

(C+D)=24

Therefore, the total number of Munchkins in original box is 24.

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