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liubo4ka [24]
3 years ago
5

Explain how the Quotient of Powers was used to simplify this expression. = 22 By finding the quotient of the bases to be , and c

ancelling common factors By finding the quotient of the bases to be , and simplifying the expression By simplifying 8 to 23 to make both powers base two, and subtracting the exponents By simplifying 8 to 23 to make both powers base two, and adding the exponents

Mathematics
1 answer:
9966 [12]3 years ago
5 0

the correct question in the attached figure

(2^5)/8=2^2

we have

8----------- >2^3

(2^5)/8----------- >  (2^5)/(2^3)=2^(5-3)=2^2

therefore 

2^2=2^2-- ------> is ok

 the answer is by simplifying 8 to 2^3 to make both powers base two and subtracting the exponents

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The answer for this is 4
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Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on
Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

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Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by =  \frac{^{3}C_2 }{^{47}C_2}   {∵ one king is already there}

              =  \frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}           {∵ ^{n}C_r = \frac{n!}{r! \times (n-r)!} }

              =  \frac{3}{1081}  =  0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by =  \frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}

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                                                                   =  \frac{6}{1081}  =  0.0055

Now, probability that he ends up with a full house = \frac{3}{1081} + \frac{6}{1081}

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3 years ago
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The game that is  used for the scenario above in terms of fair play is  using a balloon. Here, the player will hit the balloon.

<h3>What is the scenario under the balloon game?</h3>

The rule of play are:

This is a classic game with simple rules which are:

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A scenario of the worksheet game whose expected value is 0 is given below:

Assume that it costs about $1 for a player to play the billon game and as such, if  the player hits a balloon, they will be given $3. what can you say. Can you say that it this game is fair or not? and who has the biggest advantage.

Solution

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Thus, the expected value is $0.939 which tells that the game is fair.

Learn more about fair play from

brainly.com/question/24855677

#SPJ1

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