Answer:
a. p = the population proportion of UF students who would support making the Tuesday before Thanksgiving break a holiday.
Step-by-step explanation:
For each student, there are only two possible outcomes. Either they are in favor of making the Tuesday before Thanksgiving a holiday, or they are against. This means that we can solve this problem using concepts of the binomial probability distribution.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which
is the number of different combinatios of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.
So, the binomial probability distribution has two parameters, n and p.
In this problem, we have that
and
. So the parameter is
a. p = the population proportion of UF students who would support making the Tuesday before Thanksgiving break a holiday.
If u raise a power to another power, then u multiply the powers. In this case, u can simplify to 6^2p =6^10
Now you have like bases, so you can set the powers equal to each other. So now, 2p=10. Now p=5
Your equation should be
$4(40 tickets) + $5(X) = $400
160 +5x = 400
-160 -160
0 5x = 340
divide both sides by 5
5x/5 = 340/5
x = 68
so they need to sell 68 tickets at the door.
4(40) + 5(68) = 400
160 + 340 + 400
400 + 400