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.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
is the number of different combinatios of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
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.
Answer:
3
Step-by-step explanation:
Answer:
A would have to be the best answered all the all ones dont seen right
So first we assume that half is adult and have is student.
So
215 adult 215 student
Now since it said they were 70 less student tickets than adult tickets we subtract 70.
215 - 70 = 145
Now we add the 70 we subtracted from student tickets to the adult tickets.
215 + 70 = 285
So there were
285 adult tickets sold and 145 student tickets sold
I hope this helped. Have a great day!