It’s the second one it’s right
Answer:
Step-by-step explanation:
Given that a parking lot has two entrances. Cars arrive at entrance I according to a Poisson distribution at an average of 3 per hour and at entrance II according to a Poisson distribution at an average of 2 per hour.
Assuming the number of cars arriving at the two parking lots are independent we have total number of cars arriving X is Poisson with parameter 3+2 = 5
X is Poisson with mean = 5
the probability that a total of 3 cars will arrive at the parking lot in a given hour
= P(X=3) = 0.1404
b) the probability that less than 3 cars will arrive at the parking lot in a given hour
= P(X<3)
= P(0)+P(1)+P(2)
= 0.1247
It is in standard form: y=mx+b
Answer:
Step-by-step explanation:
10d - 6d = 4
Take d common
d (10 - 6) = 4
d (4) = 4
4d = 4
Divide both sides by 4
d = 1
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
I believe the answer is C
