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maks197457 [2]
3 years ago
12

A carnival game has 18 balloons. Amy popped 2/3 the first game and 5/6 the second game. How many more balloons did she pop the s

econd game?
​
Mathematics
1 answer:
faust18 [17]3 years ago
8 0

Answer:

3 more balloons she has popped in the second game.

Step-by-step explanation:

Given that,

Total no. of balloons = 18

Amy popped 2/3 the first game and 5/6 the second game.

Balloons popped in first game = \dfrac{2}{3}\times 18=12\ \text{balloons}

Balloons popped in second game = \dfrac{5}{6}\times 18=15\ \text{balloons}

Difference of balloons popped in first and second game,

= 15-12

= 3 balloons

Hence, 3 more balloons she has popped in the second game.

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Let AA and BB be events such that P(A∩B)=1/73P(A∩B)=1/73, P(A~)=68/73P(A~)=68/73, P(B)=21/73P(B)=21/73. What is the probability
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Answer:

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Step-by-step explanation:

Let A and B events. We have defined the probabilities for some events:

P(A') =\frac{68}{73} , P(B) =\frac{21}{73} , P(A \cap B) =\frac{1}{73}

Where A' represent the complement for the event A

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: P(A)+P(A') =1

So for this case we can solve for P(A) like this:

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P(A \cup B) = P(A)+P(B) -P(A \cap B)

And if we replace the values given we got:

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