Three consecutive integers have a sum of 78. find the inte

2 answers:

5 0

Let the 3 numbers be x,x+1,x+2
Given sum of those 3 consecutive integers are 78
X+x+1+x+2=78
3x+3=78
3(x+1)=78
X+1=26
X=26-1
X=25

Therefore the numbers are 25,26,27

zalisa [80]3 years ago

3 0

Answer: 25, 26, and 27

Step-by-step explanation: Three consecutive integers can be represented as followed. Let's use "T" as our variable.

T → first integer

T + 1 → second integer

T + 2 → third integer

Since the sum of our three consecutive integers is 78, our equation will be as followed.

T + T + 1 + T + 2 = 78

We can simplify on the left.

3t + 3 = 78

-3 -3 ← subtract 3 on both sides

3t = 75

÷3 ÷3 ← divide by 3 on both sides

T = 25

T + 1 = 26

T + 2 = 27

Therefore, our three consecutive integers are 25, 26, and 27.

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Find the probability of being dealt a full house. (Round your answer to six decimal places.)?

vesna_86 [32]

I'm assuming a 5-card hand being dealt from a standard 52-card deck, and that there are no wild cards.

A full house is made up of a 3-of-a-kind and a 2-pair, both of different values since a 5-of-a-kind is impossible without wild cards.

Suppose we fix both card values, say aces and 2s. We get a full house if we are dealt 2 aces and 3 2s, or 3 aces and 2 2s.

The number of ways of drawing 2 aces and 3 2s is

$\dbinom42\dbinom43=24$

and the number of ways of drawing 3 aces and 2 2s is the same,

$\dbinom43\dbinom42=24$

so that for any two card values involved, there are 2*24 = 48 ways of getting a full house.

Now, count how many ways there are of doing this for any two choices of card value. Of 13 possible values, we are picking 2, so the total number of ways of getting a full house for any 2 values is

$2\dbinom{13}2\dbinom42\dbinom43=3744$