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.
Answer:
or 1.5
Step-by-step explanation:
Now we know it's a dilation. So we can use any two points. I chose V and V'.
V: (3,6)
V': (2,4)
So how we get from 6 to 4?
![4(x)=6\\x=\frac{6}{4} \\x=1.5](https://tex.z-dn.net/?f=4%28x%29%3D6%5C%5Cx%3D%5Cfrac%7B6%7D%7B4%7D%20%5C%5Cx%3D1.5)
You can tell it's 1.5 aswell because 4 times 1 is 4 and 4 times 0.5 is 2. Add them to get 6.
Therefore, the dilation factor is 1.5.
Step-by-step explanation:
step 1 = Divide 675 by 8 keeping notice of the quotient and the remainder.
step 2 =Continue dividing the quotient by 8 until you get a quotient of zero.
step 3 = Then just write out the remainders in the reverse order to get octal equivalent of decimal number 675.
Using the above steps, here is the work involved in the solution for converting 675 to octal number:
675 / 8 = 84 with remainder 3
84 / 8 = 10 with remainder 4
10 / 8 = 1 with remainder 2
1 / 8 = 0 with remainder 1
Then just write down the remainders in the reverse order to get the answer, The decimal number 675 converted to octal is therefore equal to :
1243
1: 1.6 2: 28 Step by step something whatever
Answer:
4(6x-7)
Step-by-step explanation: