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.
</span>
Answer:
Always
Step-by-step explanation:
If the irrational parts of the numbers have zero sum, the sum is rational. If not, the sum is irrational. It is true, for example, for 2 7, 3. It is false, for example, for3+2 5,−2 5 .
Ill say, C) Place the compass' point on the other end of the segment.
Answer:
P(x < 5) = 0.70
Step-by-step explanation:
Note: The area under a probability "curve" must be = to 1.
Finding the sub-area representing x < 5 immediately yields the desired probability.
Draw a dashed, vertical line through x = 5. The resulting area, on the left, is a trapezoid. The area of a trapezoid is equal to:
(average length)·(width, which here is:
2 + 5
----------- · 0.02 = (7/2)(0.2) = 0.70
2
Thus, P(x < 5) = 0.70
Answer:
Step-by-step explanation:
5(y+4) = 4(y+5)
5y+20 = 4y + 20
5y = 4y
y=0
