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2 years ago

13

Need answer ASAP! Please see screen shot.

1 answer:

vivado [14]2 years ago

6 0

Answer:

0.4 or $\frac{2}{5}$

Step-by-step explanation:

James spun the arrow 30 times which means 30 is the total. In other words, it's your denominator ( $\frac{x}{30}$ ). It landed on the blue section 12 times which means 12 is the part of the whole. In other words, it's your numerator ( $\frac{12}{y}$ ). Plug in the entire fraction and you get $\frac{12}{30}$ . Divide 12 by 30 and you get 0.4 or if you divide both the numerator & denominator by 6, you get $\frac{2}{5}$ . Either one works, depending on what your question asks for.

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Since this polynomial has 4 terms, factoring by grouping should be the first thing we try here.

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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.

$A = 0.2$

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

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

So

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

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$A \cap B = 0.1$

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