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:
Step-by-step explanation:
45
Answer:
rjogir
grggreijgojrejow
g
Step-by-step explanation:
Answer:
A
Step-by-step explanation:
x=
−b±√b2−4ac
2a
x=
−(2)±√(2)2−4(2)(15)
2(2)
x=
−2±√−116
4
and there is really no solution
Answer:
SA ≈ 1134 cm²
General Formulas and Concepts:
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Geometry</u>
- Radius: r = d/2
- Surface Area of a Sphere: SA = 4πr²
Step-by-step explanation:
<u>Step 1: Define</u>
d = 19 cm
<u>Step 2: Find </u><em><u>SA</u></em>
- Substitute [R]: r = 19 cm/2
- Divide: r = 9.5 cm
- Substitute [SAS]: SA = 4π(9.5 cm)²
- Exponents: SA = 4π(90.25 cm²)
- Multiply: SA = 361π cm²
- Multiply: SA = 1134.11 cm²
- Round: SA ≈ 1134 cm²
