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OverLord2011 [107]
2 years ago
5

A motion sensor has a range of 65 meters as shown. The ultrasonic waves sweeping out from the sensor span an angle of 135°.

Mathematics
1 answer:
VMariaS [17]2 years ago
4 0

The approximate area that is within range of the motion sensor is 4975 square meters

<h3>How to determine the area of the range?</h3>

The figure is not given, however the area of the range can be calculated without the figure.

The given parameters are:

Angle, ⊕ = 135°

Radius, r = 65 meters

The area of the range is calculated using the following sector area

A = \frac{\theta}{360} * \pi r^2

This gives

A = \frac{135}{360} * 3.14 *  65^2

Evaluate

A = 4974.9375

Approximate

A = 4975

Hence, the approximate area that is within range of the motion sensor is 4975 square meters

Read more about sector areas at:

brainly.com/question/16736105

#SPJ1

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Which of the following products is negative? Select all that apply. *
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Answer:

  1. (-4)(9)
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Step-by-step explanation:

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8 0
3 years ago
An alarming number of U.S. adults are either overweight or obese. The distinction between overweight and obese is made on the ba
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Answer:

(A) The probability that a randomly selected adult is either overweight or obese is 0.688.

(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C) The events "overweight" and "obese" exhaustive.

(D) The events "overweight" and "obese" mutually exclusive.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = a person is overweight

<em>Y</em> = a person is obese.

The information provided is:

A person is overweight if they have BMI 25 or more but below 30.

A person is obese if they have BMI 30 or more.

P (X) = 0.331

P (Y) = 0.357

(A)

The events of a person being overweight or obese cannot occur together.

Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.

So, P (X ∩ Y) = 0.

Compute the probability that a randomly selected adult is either overweight or obese as follows:

P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.331+0.357-0\\=0.688

Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.

(B)

Commute the probability that a randomly selected adult is neither overweight nor obese as follows:

P(X^{c}\cup Y^{c})=1-P(X\cup Y)\\=1-0.688\\=0.312

Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C)

If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.

For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.

In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.

Thus, the events "overweight" and "obese" exhaustive.

(D)

Mutually exclusive events are those events that cannot occur at the same time.

The events of a person being overweight and obese are mutually exclusive.

5 0
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