Answer:
A) 0 , B) 0.0199 , C) 200
Step-by-step explanation:
Let X denotes no. of computers fail on day
X ~ Bin (n = 4 , p = 0.05)
A] P (X = x) = (n c x) p^x q^n-x
P (X = 4) = (4 c 4) (0.005)^4 (0.995)^0
(0.05)^4 ~ 0
B] P (X <u>></u> 1) = 1 - P (X < 1) = 1 - P (X = 0)
1 - { ( 4 c 0) (0.005)^0 (0.995)^ 4 }
1 - (0.995)^ 4 ~ 0.0199
C] Let Y depict mean number of days until a specific computer fails
Y ~ Geom (p = 0.005)
Mean = E (Y) = 1/p = 1/0.05 = 200
Answer:
D
Step-by-step explanation:
Answer:

Step-by-step explanation:
First, use distribution to simplify

Then, simplify further by combining like terms

Solve

Answer:
-9(7+3)x
Step-by-step explanation:
7+3=10×9=90 so that'll be 90x
Question:
Sam is delivering two refrigerators that each weigh 105 kilograms. There is an
elevator with a weight limit of 1,000 pounds. Can he take both refrigerators on the
elevator in one trip? (1 kilogram 2.2 pounds)
Answer: yes he can
Step-by-step explanation:
Yes he can.
He can if the total weight of the 2 refrigerator didn't exceed the weight limit of the elevator.
Therefore:
Since 1 refrigerator weigh 105kg
2 refrigerator = 2 x 105kg = 210kg
The weight limit of the elevator 1000pounds convert to kg.
Since
2.2 pounds = 1kg
1000 pounds = 1000/2.2 kg
=454.55kg
Weight limit of the elevator = 454.55kg.
Since Weight of the 2 refrigerator is less than
weight limit of the elevator, he can take both refrigerators on the elevator in one trip