Answer:
I'd say that is an "occupancy problem".
I ran a spreadsheet simulation of that and I'd say the probability is approximately .13
Those problems are rather complex to solve. What I think you would have to do is calculate the probability of
A) ZERO sixes appearing in 4 rolls.
B) exactly 1 six appears in 4 rolls.
C) exactly 2 sixes appear in 4 rolls.
D) exactly 3 sixes appear in 4 rolls. and
E) exactly 4 sixes appear in 4 rolls.
4 rolls of a die can produce 6^4 or 1,296 combinations.
A) is rather easy to calculate: The probability of NOT rolling a six in one roll is 5/6. In 4 rolls it would be (5/6)^4 = 0.4822530864
E) is fairly easy to calculate: The probability of rolling one six is (1/6). The probability of rolling 4 sixes is (1/6)^4 = 0.0007716049
Then we need to:
D) calculate how many ways can we place 3 objects into 4 bins
C) calculate how many ways can we place 2 objects into 4 bins
B) calculate how many ways can we place 1 objects into 4 bins
I don't know how to calculate D C and B
Step-by-step explanation:
The answer is 42 because 42 times 1 equals 42, 2 times 21 equals 42 and 3 times 14 equals 42.
Y = abˣ
(a ≠ 0, b ≠ 0)
(1, 7.5)
x = 1
y = 7.5
↓
7.5 = ab
(3, 16.875)
x = 3
y = 16.875
↓
16.875 = ab³
7.5 = ab
b b
7.5/b = a
16.875 = ab³
b³ b³
16.875/b³ = a
7.5/b = 16.875/b³
7.5/b(b³) = 16.875/b³(b³)
7.5b² = 16.875
7.5 7.5
b² = 2.25
√(b²) = √(2.25)
b = 1.5
a = 7.5/b
a = 7.5/1.5
a = 5
y = 5(1.5)ˣ
Answer:
The right answer is, -420
Step-by-step explanation:
Because we use common factor:
60 (x - 15)
We multiply for each term:
60 (x) - 60 (15)
60 (8) - 60 (15)
We solve:
480 - 900 = -420
Answer:
Step-by-step explanation:
base*hight
