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3 years ago

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The probability that a randomly-selected U.S. household owns a pet is 0.39. Given that a household owns a pet, the probability t

hat it owns a dog is 0.67 and the probability that it owns a cat is 0.24. a) Given that a household owns a pet, what is the probability that the pet is not a cat or dog?

1 answer:

xeze [42]3 years ago

5 0

Answer:

0.2508

Step-by-step explanation:

The question asks for a conditional probability.

P(a household owns a dog if it owns a pet) = 0.67

Then P(a household does not own a dog if it owns a pet) = 1 - 0.67 = 0.33

P(a household owns a cat if it owns a pet) = 0.24

Then P(a household does not own a cat if it owns a pet) = 1 - 0.24 = 0.76

P(if a household owns a pet but the pet is not a cat or dog) = P(if a household owns a pet, it does not own a cat) × P(if a household owns a pet, it does not own a dog) = 0.33 × 0.76 = 0.2508

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Answer:

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Step-by-step explanation:

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3 years ago

The coordinates of ∆ABC are A(-3, 2), B(5, 8) & C(11, 0). Which type of triangle is ∆ABC? Select All that apply.

stepan [7]

Given:

The coordinates of ∆ABC are A(-3, 2), B(5, 8) & C(11, 0).

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The type of the given triangle.

Solution:

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using the distance formula, we get

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Similarly,

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And,

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$AC=\sqrt{(14)^2+(-2)^2}$

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garri49 [273]

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4 years ago

Use the data from problem:

navik [9.2K]

Answer:

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Sample variance: 5.41

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Step-by-step explanation:

a) Sample mean:

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Sample median: M=48.3

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Sample variance:

$s^2=\frac{1}{N-1}\sum (X_i-\bar X)^2 =(\frac{1}{80-1})*427.74=5.41$

Sample standard deviation

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b)

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