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2 years ago

The price of a pair of Jordans was reduced from $200 to $160. By percentage was the price reduced

Mathematics

1 answer:

hichkok12 [17]2 years ago

3 0

Answer:

20%

Step-by-step explanation:

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3 years ago

Orange juice is sold in cartons in three different size 350 ml selling price $4.20 450 ml selling price $5.35 500 ml selling pri

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Answer:

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Step-by-step explanation:

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2 years ago

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Read 2 more answers

One household is to be selected at random from a town. The probability that the household has a cat is 0.20.2 . The proba

dimulka [17.4K]

Answer:

There is a 50% probability that the household has a dog, given that the household has a cat.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a household has a cat.

B is the probability that a household has a dog.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a household has a cat but not a dog and $A \cap B$ is the probability that a household has both a cat and a dog.

By the same logic, we have that:

$B = b + (A \cap B)$

The probability that the household has a cat or a dog is 0.5

$a + b + (A \cap B) = 0.5$

The probability that the household has a dog is 0.4

$B = 0.4$

$B = b + (A \cap B)$

$b = 0.4 - (A \cap B)$

The probability that the household has a cat is 0.2.

$A = 0.2$

$A = a + (A \cap B)$

$a = 0.2 - (A \cap B)$

$a + b + (A \cap B) = 0.5$

$0.2 - (A \cap B) + 0.4 - (A \cap B) + (A \cap B) = 0.5$

$A \cap B = 0.1$

What is the probability that the household has a dog, given that the household has a cat?

20% of the households have a cat, and 10% have both a cat and a dog. So

$P = \frac{A \cap B}{A} = {0.1}{0.2} = 0.5$

There is a 50% probability that the household has a dog, given that the household has a cat.

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2 years ago