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DedPeter [7]
3 years ago
12

The function h is defined by h(y) = 2y - 7 VWhat is the value of h (8) ? АВАА

Mathematics
1 answer:
MaRussiya [10]3 years ago
3 0
H(y) = 2y - 7
h(8) = 2(8) - 7
h(8) = 16 - 7
h(8) = 9
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Jon has to pay $7.50 admission for the skating rink and $1.50 per hour to rent rollerblades. Write an equation for the cost (y)
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a) 182 possible ways.

b) 5148 possible ways.

c) 1378 possible ways.

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The order in which the cards are chosen is not important, which means that we use the combinations formula to solve this question.

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C_{n,x} = \frac{n!}{x!(n-x)!}

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There are 52 total cards, of which:

13 are spades.

13 are diamonds.

13 are hearts.

13 are clubs.

(a)Two-pairs: Two pairs plus another card of a different value, for example:

2 pairs of 2 from sets os 13.

1 other card, from a set of 26(whichever two cards were not chosen above). So

T = 2C_{13,2} + C_{26,1} = 2*\frac{13!}{2!11!} + \frac{26!}{1!25!} = 182

So 182 possible ways.

(b)Flush: five cards of the same suit but different values, for example:

4 combinations of 5 from a set of 13(can be all spades, all diamonds, and hearts or all clubs). So

T = 4*C_{13,5} = 4*\frac{13!}{5!8!} = 5148

So 5148 possible ways.

(c)Full house: A three of a kind and a pair, for example:

4 combinations of 3 from a set of 13(three of a kind ,c an be all possible kinds).

3 combinations of 2 from a set of 13(the pair, cant be the kind chosen for the trio, so 3 combinations). So

T = 4*C_{13,3} + 3*C_{13.2} = 4*\frac{13!}{3!10!} + 3*\frac{13!}{2!11!} = 1378

So 1378 possible ways.

(d)Four of a kind: Four cards of the same value, for example:

4 combinations of 4 from a set of 13(four of a kind, can be all spades, all diamonds, and hearts or all clubs).

1 from the remaining 39(do not involve the kind chosen above). So

T = 4*C_{13,4} + C_{39,1} = 4*\frac{13!}{4!9!} + \frac{39!}{1!38!} = 2899

So 2899 possible ways.

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