Answer:
Step-by-step explanation:
Let us denote probability of spoilage as follows
Transformer spoilage = P( T ) ; line spoilage P ( L )
Both P ( T ∩ L ) .
Given
P( T ) = .05
P ( L ) = .08
P ( T ∩ L ) = .03
a )
For independent events
P ( T ∩ L ) = P( T ) x P ( L )
But .03 ≠ .05 x .08
So they are not independent of each other .
b )
i )
Probability of line spoilage given that there is transformer spoilage
P L/ T ) = P ( T ∩ L ) / P( T )
= .03 / .05
= 3 / 5 .
ii )
Probability of transformer spoilage but not line spoilage.
P( T ) - P ( T ∩ L )
.05 - .03
= .02
iii )Probability of transformer spoilage given that there is no line spoilage
[ P( T ) - P ( T ∩ L ) ] / 1 - P ( L )
= .02 / 1 - .08
= .02 / .92
= 1 / 49.
i v )
Neither transformer spoilage nor there is no line spoilage
= 1 - P ( T ∪ L )
1 - [ P( T ) + P ( L ) - P ( T ∩ L ]
= 1 - ( .05 + .08 - .03 )
= 0 .9
Answer:
r = 3.7 cm
Step-by-step explanation:
19.02 = 1/2π(2radius) + diameter
expanding:
19.02 = 1/2(3.14)(2r) + 2r
19.02 = 3.14r + 2r = 5.14r
r = 19.02/5.14 = 3.7 cm
Answer:
the answer is 1
(-4-(-2)) divided by (-3-(-1))
These are the choices
y = 4 x
y = 220(4) x
y = 880(4) x
y = 880(0.25) x
Right..?
Answer:
Step-by-step explanation:
Hi there!
Expand the parentheses
I hope this helps!
