For three fair six-sided dice, the possible sum of the faces rolled can be any digit from 3 to 18.
For instance the minimum sum occurs when all three dices shows 1 (i.e. 1 + 1 + 1 = 3) and the maximum sum occurs when all three dces shows 6 (i.e. 6 + 6 + 6 = 18).
Thus, there are 16 possible sums when three six-sided dice are rolled.
Therefore, from the pigeonhole principle, <span>the minimum number of times you must throw three fair six-sided dice to ensure that the same sum is rolled twice is 16 + 1 = 17 times.
The pigeonhole principle states that </span><span>if n items are put into m containers, with n > m > 0, then at least one container must contain more than one item.
That is for our case, given that there are 16 possible sums when three six-sided dice is rolled, for there to be two same sums, the number of sums will be greater than 16 and the minimum number greater than 16 is 17.
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X + (x + 1) = 7
2x + 1 (-1) = 7 (-1)
2x = 6
2x/2 = 6/2
x = 3
x +1 = 3 + 1
x + 1 = 4
The two integers are 3, 4
hope this helps
Answer:
No... i'm sorry
Step-by-step explanation:
C = 3.5t....when t = 15
c = 3.5(15)
c = 52.50 <==
Answer:
Both of the equations are right because if you take s=4p, then reverse it, p=1/4 of s because 4 of p is equal to s so s/4 = p
