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.
Answer:
They would be 44.6 miles apart
Step-by-step explanation:
Here, we want to calculate how far they will be in 1 hour
Mathematically;
Distance = speed * time
In 1 hour, Kiran would have traveled a distance of 3 * 1 = 3 miles
While Clare would have traveled a distance of 3.4 * 1 = 3.4 miles
Now, we want to calculate how far apart they would be
On the 24 miles trail, Kiran would be at a distance of 21 miles to the other end
On the 24 miles Claire would be at a distance of 20.6 miles
So how far apart they would be is
24 miles + 20.6 miles = 44.6 miles
Answer:
55
Step-by-step explanation:
m angle A and m angle C are equal sine two of those sides are equal. You would do 3x + 40 = x +50
You subtract x, leaving it to be 2x + 40 = 50
Subtract the 40, leaving it be 2x = 10
Divide both sides by 2, which leaves x = 5
and then you would do 3(5) + 40, which is 55
Answer:
P(3) is true since 2(3) - 1 = 5 < 3! = 6.
Step-by-step explanation:
Let P(n) be the proposition that 2n-1 ≤ n!. for n ≥ 3
Basis: P(3) is true since 2(3) - 1 = 5 < 3! = 6.
Inductive Step: Assume P(k) holds, i.e., 2k - 1 ≤ k! for an arbitrary integer k ≥ 3. To show that P(k + 1) holds:
2(k+1) - 1 = 2k + 2 - 1
≤ 2 + k! (by the inductive hypothesis)
= (k + 1)! Therefore,2n-1 ≤ n! holds, for every integer n ≥ 3.
Answer:
The distance is the absolute value of the difference
Step-by-step explanation:
