Answer: 25/7
Decimal form 3.57143.....
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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Even numbers end with 0,2,4,6,8
7 in hundreds place
odd in thousands place, odd numbers are 1,3,5,7,9
multiplue of 10 end with 0
so therfor the number has 7 in hundreds place and ends with 0 with odd in thousands and undefined hundreds place
50 options
some examples are
1730
9790
any number, x7y0 such that x is odd and y can be any number
0.08*x+0.03*(29000-x)<span> </span>
Answer:
b 5/12
Step-by-step explanation:
