Solution :
Faulty electrical connects = P(A) = 0.80
Mechanical defects = P(B) = 0.92
Mechanical defects are related to loose keys = P(C/B) = 0.27
Improper assembly = P(D/B) = 0.73
Defective wires = P(E/A) = 0.35
Improper connections = P(F/A) = 0.13
Poorly welded wires = P(G/A) = 0.52
Now, the probability due to loose keys = 0.27 x 0.92 = 0.2484
Improperly connected = 0.13 x 0.80 = 0.1040
Poorly welded wires = 0.52 x 0.80 = 0.4160
So, the probability that a failure is due to improperly connected or poorly welded wires = 0.1040 + 0.4160
= 0.5200
Hello,
It 's difficult to translate that in french.
l=16 ^(1/5)
w=2^(1/5)
Area=32^(1/5)=(2^5)^(1/5)=2 (in²)
Answer:
RANGE=21
Step-by-step explanation:
Least to greatest is KEY!
Cot(x)sec(x) =
(cos(x)/sin(x))(1/cos(x))=
cos(x)/(sin(x)cos(x)) =
1/sin(x) =
csc(x)
Answer:50.97 or 51 rounded
Step-by-step explanation: hope this helps
