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2 years ago

6

A bug was sitting on the tip of the propeller blade when the propeller started to rotate. The bug held on for 5 rotations before

flying away. How far did the bug travel before it flew off?

1 answer:

Andrej [43]2 years ago

8 0

Answer:

around 1800 full rotations

Step-by-step explanation:

a circle is 360 degrees so if the bug was on it for 5 rotations then it would have traveled 360x5

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Which type of function is shown in the table below?

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3 years ago

A= h(b+c) solve for b

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A=h(b+c)
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Partners A, B and C share the profit of a business deal. Partner A gets 1/6, Partner B gets 3/4 and Partner C gets €25. How much

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2 years ago

What is the relationship between 9.125×10−3 and 9.125×102 ? Select from the drop-down menus to correctly complete the statement.

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Given $9.125\times10^{-3}$ and $9.125\times10^{2}$

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3 years ago

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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.

$A = 0.2$

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

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

So

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

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

$A \cap B = 0.1$

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.

4 0

2 years ago