When you have 3 choices for each of 6 spins, the number of possible "words" is
3^6 = 729
The number of permutations of 6 things that are 3 groups of 2 is
6!/(2!×2!×2!) = 720/8 = 90
A) The probability of a word containing two of each of the letters is 90/729 = 10/81
The number of permutations of 6 things from two groups of different sizes is
(2 and 4) : 6!/(2!×4!) = 15
(3 and 3) : 6!/(3!×3!) = 20
(4 and 2) : 15
(5 and 1) : 6
(6 and 0) : 1
B) The number of ways there can be at least 2 "a"s and no "b"s is
15 + 20 + 15 + 6 + 1 = 57
The probability of a word containing at least 2 "a"s and no "b"s is 57/729 = 19/243.
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These numbers were verified by listing all possibilities and actually counting the ones that met your requirements.
I would certainly be willing to do that for you. But in order to write the composite function, I would first need to know the elemental functions ... f (x) and g(x).
Answer:
C
Step-by-step explanation:
"at most" meaning less than or equal to
Volume of a cone: (1/3)πr²h
Data:
V=31.5 in³,
r=3 in
Then:
V=(1/3)πr²h
31.5 in³=(1/3)π(3 in)²h
<span>(1/3)π(3 in)²h=31.5 in</span>³
(1/3)π(9 in²)h=31.5 in³
3πin² h=31.5 in³
h=(31.5 in³) / (3π in²)
h≈3.34 in
answer: the height of the cone is 3.34 in.
Answer A
In^2 this multiplies the number used by itself therefore in your case 283 x 283. You are using the measurement for inches.
