Note that
n^m*n^r=n^(m+r)
2^5*2^x=2^(5+x)
the answer is D
Answer:
The fraction of the students who failed to went partying = 
Step-by-step explanation:
Let total number of students = 100
No. of students partied are twice the no. of students who not partied.
⇒ No. of students partied = 2 × the no. of students who are not partied
No. of students partied before the exam = 20 % of total students
⇒ No. of students partied before the exam = 
× 100
⇒ No. of students partied before the exam = 20
No. of students who not partied before the exam = 
Thus the fraction of the students who failed to went partying = 
15.6 in 13 minutes? are you asking how it got to 15.6 in 13 minutes?
So first you have to find the perfect square that matches up with x^2 + 6x
so half of 6, and square it. your perfect square is 9
x^2 + 6x + 9 = 7 + 9
then, condense the left side of the equation into a squared binomial:
(x + 3)^2 = 16
take the square root of both sides:
x + 3 = ± √16
therefore:
x + 3 = ± 4
x = - 3 ± 4
so your solution set is:
x = 1, -7
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)
