25-3=22
22/2 = 11
John made 14 free throws while Jose made 11.
Answer:
m < -
Step-by-step explanation:
8m + 95 < -87m + 5
8m + 87m < 5 - 95
95m < -90
m < -<u>
</u>
Simplifying gives;
m < -
Answer:
Step-by-step explanation:
Hello, please find below my work.
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
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.
6(2x + 5y)
6[2(3) + 5(4)]
6[6 + 20]
6(26)
156
The answer is 156
Hope this helps :)
