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san4es73 [151]
2 years ago
13

Which is the solution to this system of equations?

Mathematics
2 answers:
DENIUS [597]2 years ago
7 0
The 1st one is the one that is the Answer
vivado [14]2 years ago
5 0
A graphing calculator shows the only solution to be ...
  (0, 3)

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Please show your work!<br> 25% of 80
spin [16.1K]

Answer:

20

Step-by-step explanation:

25% change to .25

.25 x 80

Just solve 25 x 80 which equals 2000

then move decimal over twice giving you 20.00 or 20

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2 years ago
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A company charges a shipping fee that is 4% of the purchase price for all items
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4 percent of 66
0.04 = 4 percent
66 * 0.04 = 2.64
The shipping fee is $2.64
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3 years ago
How do you write 9.59 x 102 in a standard form?
Nataly [62]

Answer:

978.18

Step-by-step explanation:

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3 years ago
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If CE and FH are parallel lines and m&lt;HGI are parallel lines and m&lt;HGI = 126º, what is m&lt;FGI? HELP ASAP​
lisov135 [29]

Answer: <FGI = 54

Step-by-step explanation:

180-126=54

u go to agora cyber charter school?

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6 0
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}

                                                                                    =  \frac{9}{1081} = <u>0.0083</u>.

3 0
3 years ago
Read 2 more answers
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