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stealth61 [152]
2 years ago
12

In a particular city, 82% of the residents have a desktop computer, 47% have a desktop computer and a laptop computer, and 3% ha

ve neither
a desktop nor a laptop computer. Given that a randomly selected home has a desktop computer, the probability this home also has a laptop
computer is
A)0.28
B)0.57
C)0.43
D)0.53
Mathematics
1 answer:
vagabundo [1.1K]2 years ago
3 0

Answer:

B) 0.57

Step-by-step explanation:

Let's define:

  • A: the home has a laptop
  • B: the home has a desktop computer.

Data

  • P(B) = 0.82
  • P(A∩B) = 0.47

We want to find the probability that a home has a laptop given that it has a  desktop computer, that is, P(A|B).

P(A|B) = P(A∩B)/P(B) = 0.47/0.82 = 0.57

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Answer:

The quantity of frosted donuts = 2

The quantity of glazed donuts = 9

The quantity of custard filled donuts = 1

Step-by-step explanation:

Let the quantity of frosted donuts = x

Let the quantity of glazed donuts = y

Let the quantity of custard filled donuts = z

As per question statements, following equations can be made:

x+y+z=12......(1)\\2x+y+5z=18.....(2)\\x=2z ...... (3)

Putting x=2x in (1) and (2):

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Subtracting (4) from (5):

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By equation (3):

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By equation (1):

y=\bold{9}

Therefore, the answers are:

The quantity of frosted donuts = 2

The quantity of glazed donuts = 9

The quantity of custard filled donuts = 1

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Step-by-step explanation:

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