A) Plan A requires for a percentage increase of a number of students. This means that year after year the number of new students will increase. Plan B requires for a constant number of new students each year. This means that year after year the percentage increase would get smaller.
B) To solve this problem we will use formula for a growth of population:

Where:
final = final number of students
initial = initial number of students
percentage = requested percentage increase
t = number of years
We can insert numbers and solve for t:

For Plan B we can use simple formula
increase = 120
increase per year = 20
number of years = increase / (increase per year) = 120 / 20 = 6 years
Plan B is better to double the <span>enrollment.
C)We use same steps as in B) to solve this.
</span>

For Plan B we can use simple formula
increase = 240
increase per year = 20
number of years = increase / (increase per year) = 240 / 20 = 12 years
Plan A is better to triple the enrollment.
18+4=27\qo/89 so that answer must me 18+4=27\qo but carry the y
Answer:
=(k−1)*P(X>k−1) or (k−1)365k(365k−1)(k−1)!
Step-by-step explanation:
First of all, we need to find PMF
Let X = k represent the case in which there is no birthday match within (k-1) people
However, there is a birthday match when kth person arrives
Hence, there is 365^k possibilities in birthday arrangements
Supposing (k-1) dates are placed on specific days in a year
Pick one of k-1 of them & make it the date of the kth person that arrives, then:
The CDF is P(X≤k)=(1−(365k)k)/!365k, so the can obtain the PMF by
P(X=k) =P (X≤k) − P(X≤k−1)=(1−(365k)k!/365^k)−(1−(365k−1)(k−1)!/365^(k−1))=
(k−1)/365^k * (365k−1) * (k−1)!
=(k−1)*(1−P(X≤k−1))
=(k−1)*P(X>k−1)
Step-by-step explanation:
Put the value of x = -6 to all expressions:
