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oksian1 [2.3K]
3 years ago
13

Lisa lives in Atlanta. She is going to a baseball game with her two sisters and her parents. They have $100 to spend. After they

buy their tickets, how much money will they have left to buy refreshments?
Mathematics
1 answer:
9966 [12]3 years ago
4 0

Not enough information. How much do the tickets cost?

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Select the expression that is equivalent to (x-3)^2
Elan Coil [88]

Answer:

B.

{ \tt{ {(x - 3)}^{2} }} \\  \\  = { \tt{(x - 3)(x - 3)}} \\  \\  = { \tt{ {x}^{2} - 6x + 9 }}

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Please help ill give brainlest
UkoKoshka [18]

Answer:

A

Step-by-step explanation:

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Help me on number 12 13 14 and 15
aleksandr82 [10.1K]

12. 1.625 [terminating]; 13. 0.83 [bar notation over 3 (repeating)]; 14. 900 cm = 9 m; 15. 0.23 cm = 2.3 mm

Repeating decimals are parts of decimals that have repetitive digits; terminating decimals are decimals whose digits end.

Whether you are using Metric or Imperial, you have to determine whether you are going from a small unit to a big unit or vice versa. Then perform your operation. So, in exercise 14, the smaller unit is centimeters, so you would be going from big to small. Exercise 15 has you going from small to big.

There are centimeters in one meter, so multiply 9 by to get 900 centimeters.

There are 10 millimeters in one centimeter, so divide 2.3 by 10 simply by moving the decimal point ONCE to the left [Power of 10].

small to BIG → Division

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5 0
3 years ago
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olga2289 [7]
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4 0
3 years ago
Choose 5 cards from a full deck of 52 cards with 13values (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A) and 4 kinds(spade, diamond, h
Delvig [45]

Answer:

a) 182 possible ways.

b) 5148 possible ways.

c) 1378 possible ways.

d) 2899 possible ways.

Step-by-step explanation:

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

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

In this question, we have that:

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.

4 0
3 years ago
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