I think that this is a combination problem. From the given, the 8 students are taken 3 at a time. This can be solved through using the formula of combination which is C(n,r) = n!/(n-r)!r!. In this case, n is 8 while r is 3. Hence, upon substitution of the values, we have
C(8,3) = 8!/(8-3)!3!
C(8,3) = 56
There are 56 3-person teams that can be formed from the 8 students.
Let x represent the amount of games played.
When x is used, it need to be multiplied by the price ($35).
x(35) = cost in $
Hope this helped :)
Answer:
the pay one
Step-by-step explanation:
i thinks sorry if i get it wrong
Answer:
yes it is
Step-by-step explanation:
anything took to zerk is zero, 1 is higher thsn zero