Answer:
Option (2)
Step-by-step explanation:
Given expression is, AX + B = C



AX + B = C
AX = C - B
C - B =
= 
C - B = 
Let 
AX =
= 
Since AX = C - B

Therefore, a = 1, b = 5
(-3a - 4c) = -35
3(1) + 4c = 35
3 + 4c = 35
4c = 32
c = 8
And (-3b - 4d) = -11
3(5) + 4d = 11
4d = -4
d = -1
Therefore, Option (2). X =
will be the answer.
Answer:
a)0.08 , b)0.4 , C) i)0.84 , ii)0.56
Step-by-step explanation:
Given data
P(A) = professor arrives on time
P(A) = 0.8
P(B) = Student aarive on time
P(B) = 0.6
According to the question A & B are Independent
P(A∩B) = P(A) . P(B)
Therefore
&
is also independent
= 1-0.8 = 0.2
= 1-0.6 = 0.4
part a)
Probability of both student and the professor are late
P(A'∩B') = P(A') . P(B') (only for independent cases)
= 0.2 x 0.4
= 0.08
Part b)
The probability that the student is late given that the professor is on time
=
=
= 0.4
Part c)
Assume the events are not independent
Given Data
P
= 0.4
=
= 0.4

= 0.4 x P
= 0.4 x 0.4 = 0.16
= 0.16
i)
The probability that at least one of them is on time
= 1-
= 1 - 0.16 = 0.84
ii)The probability that they are both on time
P
= 1 -
= 1 - ![[P({A}')+P({B}') - P({A}'\cap {B}')]](https://tex.z-dn.net/?f=%5BP%28%7BA%7D%27%29%2BP%28%7BB%7D%27%29%20-%20P%28%7BA%7D%27%5Ccap%20%7BB%7D%27%29%5D)
= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56
Answer:
The answer is 75
Step-by-step explanation:
3/8*200=75.
Answer:
That's incredible. I have that exact problem on my website.
Here is my explanation.
My website says probability = 0.131944444444445
https://www.1728.org/occupncy8.htm
Step-by-step explanation:
Answer:
Step-by-step explanation: