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.
_____
These numbers were verified by listing all possibilities and actually counting the ones that met your requirements.
XY = 17.
Because YW is a perpendicular bisector, we can say that TW and WZ are both equal to 3. It tells us that XZ is 12, so that means that XW must be 12+3 which equals 15.
It also tells us that YW is 8. So we can use the Pythagorean Theorem to find the hypotenuse of Triangle XWY. Thus,

To solve this equation, square and add both of the terms on the right like this:

And then take the square root of both sides. Your final answer should be:
XY = 17.
Answer:
A.$3,600
Step-by-step explanation:
24,000*15% = 3600
or
24,000*0.15 = 3600
Answer:
A
Step-by-step explanation: