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.
"31-inch piece of steel is cut into three pieces"
Call the pieces a,b,c
a + b + c = 31
"second piece is twice as long as the first piece"
b = 2a
"third piece is one inch more than seven times the length of the first piece"
c = 1 + 7a
Ok, we have our equations, let's substitute the second and third into the first.
a + 2a + 1 + 7a = 31
10a + 1 = 31
10a = 30
a = 30/10 = 3
b = 2a = 6
c = 1 + 7a = 22
Check: 3 + 6 + 22 = 31, good
Answer: 3 inches, 6 inches, 22 inches
<span>n = 5
The formula for the confidence interval (CI) is
CI = m ± z*d/sqrt(n)
where
CI = confidence interval
m = mean
z = z value in standard normal table for desired confidence
n = number of samples
Since we want a 95% confidence interval, we need to divide that in half to get
95/2 = 47.5
Looking up 0.475 in a standard normal table gives us a z value of 1.96
Since we want the margin of error to be ± 0.0001, we want the expression ± z*d/sqrt(n) to also be ± 0.0001. And to simplify things, we can omit the ± and use the formula
0.0001 = z*d/sqrt(n)
Substitute the value z that we looked up, and get
0.0001 = 1.96*d/sqrt(n)
Substitute the standard deviation that we were given and
0.0001 = 1.96*0.001/sqrt(n)
0.0001 = 0.00196/sqrt(n)
Solve for n
0.0001*sqrt(n) = 0.00196
sqrt(n) = 19.6
n = 4.427188724
Since you can't have a fractional value for n, then n should be at least 5 for a 95% confidence interval that the measured mean is within 0.0001 grams of the correct mass.</span>
If you put brackets around (2+1), your method of working is:
1) 15-4*(2+1)=3
2) 15-4*3=3
3) 15-12=3
You don't need any more brackets, as the BIDMAS (brackets, Indices, division, multiplication, addition, subtraction) rule does the rest of the job for you.
The answer is therefore: 15-4*(2+1)=3