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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Answer:
The solutions are 1 and 5.
Step-by-step explanation:
x2 − 6x + 5 = 0
Step 1: Write the related function.
x2 − 6x + 5 = 0
y = x2 − 6x + 5
Step 2: Graph the function.
Use a graphing calculator.
Step 3: Find the zeros.
The zeros are 1 and 5.
Answer:
20 degrees
Step-by-step explanation:
to find the complement you subtract the angle by 90 degrees
The mode of this set of data is 12
