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Novosadov [1.4K]
3 years ago
10

A season has96 home games. If you purchase a season ticket in the bleacher section, what is the cost per game,in dollars, for th

is ticket
Mathematics
1 answer:
sattari [20]3 years ago
5 0

Answer:

$8.4375 per game

Step-by-step explanation:

From. The picture attached :

Season long Ticket cost for the bleachers sitting area is : $810

Given that the number of home games in a season is 96

The cost per unit of game could be calculated as :

Cost of season long Ticket / number of home games

= $810 / 96

= $8.4375

$8.4375 per game

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The perimeter will be
.. P = 2√(16^2 +30^2) = 2*34 = 68
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If your truck gets 22 highway miles per gallon and has a 25 gallon gas tank, how many miles should you be able to drive on a ful
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550 should be correct

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do 22 times 25

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What is P(2 or divisor of 9)?<br> please explain!
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The probability that a divisor of 2 or 9 is chosen when the spinner is spun once is 2/3.

<h3>What is the probability?</h3>

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P(2 or divisor of 9) = section that has a divisor of 2/ total number of sections + section that has a divisor of 2/ total number of sections

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2 years ago
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
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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);

  • If he is given with two kings.
  • If he is given one king and one ace.

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}

                                                                   =  \frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}

                                                                   =  \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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Answer:

Step-by-step explanation:

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