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Vadim26 [7]
3 years ago
9

Every​ morning, my neighbor goes out walking. I observe that​ 20% of the time she walks with her​ beagle, 70% of the time she wa

lks with her golden​ retriever, and​ 30% of the time she walks alone. If these events are all​ disjoint, is this an example of a valid probability​ distribution?
Mathematics
1 answer:
Oduvanchick [21]3 years ago
5 0

Answer:

No, This is not a valid Probability Distribution

Step-by-step explanation:

Probability [Neighbour walks with Beagle] = 20%

Probability [Neighbour walks with Golden Retriever] = 70%

Probability [Neighbour walks alone] = 30%

Disjoint Events are the events that have zero probability of occurring together. If all the three above items are disjoint, it means that it can never happen that two of them happen together.

The total (summed) probability of a valid probability distribution, with disjoint sets = 1 . In given case, total probability = 0.2 + 0.7 + 0.3 = 1.2 ; i.e > 1.

So, this probability distribution with stated disjoint events is not a Valid Probability distribution.

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ANSWER FAST!!!!!!!
Lemur [1.5K]

Answer:

The angle between the 190 ft. side and the 330 ft. side is 57.9^o

Step-by-step explanation:

<u>The Law of Cosines</u>

When we know the value of all sides of a triangle, we can compute all of its interior angles by using the Law of Cosines, which is a generalization of the Pythagoras's theorem. If a,b, and c are the known sides of a triangle and \alpha is the angle formed by sides a and b (opposite to c), then

\displaystyle c^2=a^2+b^2-2ab\ cos\alpha

We'll use the values a=190, b=330, c=280 because we want to compute the angle opposite to c

\displaystyle cos\alpha =\frac{a^2+b^2-c^2}{2ab}

\displaystyle cos\alpha =\frac{190^2+330^2-280^2}{2(190)(330)}

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