Answer:
0.8340
Step-by-step explanation:
The aim of this question is to find the probability that the student makes it to the second class prior to when the lecture commences.
Provided that A, B, and C are independent of each other from the full complete question.
Then; we want P(A + C < B)
So A
N (2, 1.5)
B
N (10, 1)
C
N (6, 1)
P(A + C - B < 0)
Since they are normally distributed( i.e. A + C - B)
Then;
E(A + C -B) = E(A) + E(C) - (B)
E(A + C -B) = 2 - 10 + 6
E(A + C -B) = -2
Var(A+C -B) = Var(A) + Var (B) + Var (C)
Var(A + C -B) = (1.5)² + (1)² + (1)²
Var(A + C -B) = 4.25
The standard deviation = 
The standard deviation = 2.06155



Answer:
Choices C and D
Step-by-step explanation:
The most simplest way is just to solve each one and see which two have the same answer as the first expression.
But you could also see that both of the
are going to cancel each other out so you are left with
which is also choice C.
For choice D I distributed the negative one to the insides of the parenthesis that left me with
which is the same expression as the first one.
#1:(A)
#2:(E)
#3:(C)
Hope it helps!