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DaniilM [7]
3 years ago
10

three consecutive integers such that 3 times the sum of the first and second is one greater than 4 times the third.??

Mathematics
1 answer:
weeeeeb [17]3 years ago
8 0

Remark

I take it you want the integers to this problem.

Givens

The smallest integer = x - 1

The  middle integer = x

The largest integer = x + 1

Equation

3( x - 1 + x) = 4(x + 1) + 1    Remove the brackets on both sides.

3(2x - 1) = 4x + 4 + 1

6x - 3 = 4x + 5                        Subtract 5 from both sides.

-3 - 5 + 6x = 4x

- 8 + 6x = 4x                          Subtract 6x from both sides                              

- 8 = 4x - 6x

- 8 = -2x                                 Divide by -2

-8/-2  = x

x = 4  

Check

smallest integer = 3

middle integer = 4

largest integer = 5

3(3 + 4) = 4(5) + 1

3(7) = 4(5) + 1                            

21 = 20 + 1

21 = 21 and it checks.

Answer

the smallest number is 3

the middle number is 4

The largest number = 5

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probability that the card drawn is:

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<u>Step-by-step explanation:</u>

Here we have , A card is drawn at random from a deck of 52 playing cards. We need to find that Find the probability that the card drawn is:

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A deck of standard 52 cards contain four aces. There are four kings in a standard deck of playing cards. So ,we need to find probability that card drawn is either a ace or king i.e.

probability = \frac{Number-of-(ace+king)-cards}{total-number-of-cards}

⇒ probability = \frac{Number-of-(ace+king)-cards}{total-number-of-cards}

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A deck of standard 52 cards contain four kings. There are 13 Diamonds in a standard deck of playing cards. So ,we need to find probability that card drawn is either a ace or king i.e.

probability = \frac{Number-of-(ace+king)-cards}{total-number-of-cards}

⇒ probability = \frac{Number-of-(ace+king)-cards}{total-number-of-cards}

⇒ probability = \frac{(13+4=17)}{52}

⇒ probability = \frac{(17)}{52}

⇒ probability =0.3269

Therefore, probability that the card drawn is:

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