Answer:
4
Step-by-step explanation:
Step-by-step explanation:
step 1. P(0, 1) -> P'(0, -1)
step 2. Q(0, 0) -> Q'(0, 0)
step 3. R(-2, 0) -> R'(2, 0)
Answer:
see below
Step-by-step explanation:
7x-4 6x-4-x
Combine like terms on the second expression
7x -4 6x -x -4
5x -4
These are not equivalent expressions
7x -4 does not equal 5x-4
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.
