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photoshop1234 [79]
3 years ago
15

Please answer this question and you will receive points.

Mathematics
1 answer:
VashaNatasha [74]3 years ago
8 0

Answer:

(-2,-4)

Step-by-step explanation:

The given system is

x+y=-6\\x-y=2

Add both equations to eliminate y

x+x=-6+2

Simplify both sides to get:

2x=-4

Divide both sides by 2

x=\frac{-4}{2}\\ x=-2

Put x=-2 into one of the equations say, x+y=-6 and solve for y

y+-2=-6\\y=-6+2\\y=-4

The solution is (-2,-4)

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20

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x = 4

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180 = 102 + 24x - 18

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In a manufactory, there are two types of spoilages. It is found that 5% of spoilages are due to transformer spoilage, 8% are due
jok3333 [9.3K]

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

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3 years ago
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