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Dvinal [7]
3 years ago
7

Roberta Arthur had the following reciepts for the week of April 12th: 50$ gift; 25$ rebate; April 13th, 2.45$ refund on aluminum

cans returned; april 14th, 299.89$ weekly pay from her regular job; april 16th, 7$ pay from her part time job. find Roberta's total reciepts for the week.​
Mathematics
1 answer:
oksian1 [2.3K]3 years ago
7 0

Answer:

$329.44

Step-by-step explanation:

 $50-25= 25

 25.00-2.45=22.55

 299.89+22.55=322.44

322.44+7.00= 329.44

I didnt see april 15 if its supposed to be added i there just add 299.89 to the total.

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Find the slope of the line passing through the points -2,7 and -6,3
Helga [31]

Answer: the slope is 1

Step-by-step explanation:

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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
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);

  • 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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3 0
4 years ago
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oee [108]
The answer to the question

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3 years ago
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