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marusya05 [52]
3 years ago
15

On circle O below, the measure of SV is 120°.

Mathematics
1 answer:
kumpel [21]3 years ago
4 0
<h2>Answer:</h2><h2 />

<em>In this context:</em>

<h2 /><h2>\boxed{\angle VU=60^{\circ}}</h2>

<h2>Explanation:</h2><h2 />

Hello! Remember you have to write complete questions in order to get good and exact answers. Here I'll assume the diagram shown below. Both ∠SV and ∠VU are central angles. This is an angle between two radii in a circle. Another fact we know is that ∠SU measures 180 degrees so it is true that:

\angle SV + \angle VU=\angle SU \ldots eq1 \\ \\ \\ But: \\ \\ \angle SV=120^{\circ} \\ \\  \angle SU=180^{\circ} \\ \\ \\ Substituting \ into \ eq1: \\ \\ 120^{\circ}  + \angle VU=180^{\circ} \\ \\ \\ Isolating \  \angle VU: \\ \\ \angle VU=180^{\circ}-120^{\circ} \\ \\ \boxed{\angle VU=60^{\circ}}

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An insurance company examines its pool of auto insurance customers and gathers the following information: (i) All customers insu
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The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

An insurance company examines its pool of auto insurance customers and gathers the following information: (i) All customers insure at least one car. (ii) 70% of the customers insure more than one car. (iii) 20% of the customers insure a sports car. (iv) Of those customers who insure more than one car, 15% insure a sports car. Calculate the probability that a randomly selected customer insures exactly one car, and that car is not a sports car?

Answer:

P( X' ∩ Y' ) = 0.205

Step-by-step explanation:

Let X is the event that the customer insures more than one car.

Let X' is the event that the customer insures exactly one car.

Let Y is the event that customer insures a sport car.

Let Y' is the event that customer insures not a sport car.

From the given information we have

70% of customers insure more than one car.

P(X) = 0.70

20% of customers insure a sports car.

P(Y) = 0.20

Of those customers who insure more than one car, 15% insure a sports car.

P(Y | X) = 0.15

We want to find out the probability that a randomly selected customer insures exactly one car, and that car is not a sports car.

P( X' ∩ Y' ) = ?

Which can be found by

P( X' ∩ Y' ) = 1 - P( X ∪ Y )

From the rules of probability we know that,

P( X ∪ Y ) = P(X) + P(Y) - P( X ∩ Y )    (Additive Law)

First, we have to find out P( X ∩ Y )

From the rules of probability we know that,

P( X ∩ Y ) = P(Y | X) × P(X)       (Multiplicative law)

P( X ∩ Y ) = 0.15 × 0.70

P( X ∩ Y ) = 0.105

So,

P( X ∪ Y ) = P(X) + P(Y) - P( X ∩ Y )

P( X ∪ Y ) = 0.70 + 0.20 - 0.105

P( X ∪ Y ) = 0.795

Finally,

P( X' ∩ Y' ) = 1 - P( X ∪ Y )

P( X' ∩ Y' ) = 1 - 0.795

P( X' ∩ Y' ) = 0.205

Therefore, there is 0.205 probability that a randomly selected customer insures exactly one car, and that car is not a sports car.

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