Answer: 19/7 ,3.14, 3.2
Step-by-step explanation: Hope this helps Have a brilliant day of learning!
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:
i cant see the question to ypur test that well but wish you the best
Complimentary angles add up to 90 degrees
x + y = 90
x = y + 8
y + 8 + y = 90
2y + 8 = 90
2y = 90 - 8
2y = 82
y = 82/2
y = 41
x = y + 8
x = 41 + 8
x = 49
so ur angles are : 41 and 49
Answer:
No
Step-by-step explanation:
They are similar
