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nydimaria [60]
3 years ago
14

What is 1 over 3 x plus 1 over 4 equals 5 over 12

Mathematics
1 answer:
guajiro [1.7K]3 years ago
5 0
Simplify 1/3y to y/3

subtract 1/4 from both sides

simplify 5/12 - 1/4 to 1/6

multiply both sides by 3

simplify 1/6 x 3 to 3/6

simplify 3/6 to 1/2

Answer: y = 1/2
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HELPPPP‼️‼️
Dmitriy789 [7]

Answer:

The answer is 75 and here's why.

Let's figure out the amount of remaining votes to prove this.

250 total votes   -    150 votes for candidate a = 100 votes.

               minus ↑ sign

Candidate B received 25% of the votes, so let's find 25% of 100

100 * 0.25 = 25 votes

Candidate B only got 25 votes (that's kinda sad, poor guy)

100 - 25 votes = the # of votes candidate C got

Candidate C got 75 votes!

4 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
Which additional congruence statement could you use to prove that CAB CAD by HL?
icang [17]

Answer:

#2, AB is congruent to AD

Step-by-step explanation:

AB is congruent to AD.

since the bottom sides are corresponding, and the middle sides are corresponding, then if the outside sides are corresponding, then CAB is congruent to CAD

8 0
3 years ago
What is the value of the expression below
natulia [17]

Answer:

a=2,b=4. (3a+b)=3*2+4=10.

5 0
2 years ago
Sabrina reads 15 pages per hour. She needs to finish a book with 120 pages. Which equation would you use to determine how many h
frozen [14]

Answer:

15x=120

Step-by-step explanation:

8 0
3 years ago
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