The system of equations would be:
64x+132y = 1040
x = 2y
We multiply the number of adult tickets, 64, by the cost of the adult tickets, x. We multiply the number of student tickets, 132, by the cost of the student tickets, y. Together these add up to $1040.
We know that the cost of an adult ticket is twice as much as a student ticket, so x = 2y.
If i’m not dumb i think it’s the last one
The expected value formula is
We know that there are 125 tickets, so the probability of the ticket is
Then, we find the expected value
<h2>Hence, the expected value is $7.60.</h2>
Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).
Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2
