This is a combination problem.
Given:
12 students
3 groups consisting of 4 students.
Mark can't be in the first group.
The combination formula that I used is: n! / r!(n-r)!
where: n = number of choices ; r = number of people to be chosen.
This is the formula I used because the order is not important and repetition is not allowed.
Since Mark can't be considered in the first group, the value of n would be 11 instead of 12. value of r is 4.
numerator: n! = 11! = 39,916,800
denominator: r!(n-r)! = 4!(11-4)! = 4!*7! = 120,960
Combination = 39,916,800 / 120,960 = 330
There are 330 ways that the instructor can choose 4 students for the first group
30+18=30+6y
18=6y
y=3 so the answer is 3
Answer:
3x+2x=total number
Step-by-step explanation:
F(x) = 3 - 2x
g(x) = 1/(x + 5)
(f/g)(x) = (3 - 2x) / 1/(x + 5) = (3 - 2x)(x + 5)
(f/g)(8) = (3 - 2(8))(8 + 5) = -13 x 13 = -169
Answer:
11 quarters
Step-by-step explanation:
If Thomas ignores the extra 3 dimes, he can arrange his coins in groups of 3 dimes and 1 quarter, each group valued at 55¢. It takes
... 6.05/0.55 = 11
groups to bring the total value to the $6.05 remaining after ignoring the extra dimes.
Thomas has 11 quarters and 36 dimes.
