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Identify the probability to the nearest hundredth that a point chosen randomly inside the rectangle is either in the triangle or in the circle.

Answer Options:
0.73
0.35
0.33
0.26

Identify the probability to the nearest hundredth that a point chosen randomly inside the rectangle is either in the triangle or in the circle Answer Options 07 class=

Respuesta :

simon2131

Answer:

0.26

Step-by-step explanation:

In geometric probability, the probability of an event is based on a ratio of geometric measures such as length or area.

The area model is used here. The measures of various dimensions of the sample space and the included figures are given.

The sample space is the area of the large rectangle.

Find the area of the sample space.

Substitute the known values for the length, l=16

m, and width, w=11.8 m, into the formula for the area of a rectangle, A=lw

.

A=(16)(11.8)=188.8

m2

The probability that a randomly chosen point will lie either inside the triangle or inside the circle is the sum of the individual probabilities.

The probability that a randomly chosen point will lie inside the triangle is equal to the ratio of the area of the triangle to the area of the large rectangle.

Find the area of the triangle.

The height of the triangle equals 11.8−7.3=4.5

m

.

Substitute the known values for the base, b=8

m, and height, h=4.5 m, into the formula for the area of a triangle, A=12bh

.

A=12(8)(4.5)=18

m2

Find the probability that a point chosen randomly inside the rectangle is in the triangle.

P1=18188.8

The probability that a randomly chosen point will lie inside the circle is equal to the ratio of the area of the circle to the area of the large rectangle.

Find the area of the circle.

The radius is half of the diameter. So, r=6.22=3.1

m

.

Substitute the known value for the radius, r=3.1

m, into the formula for the area of a circle, A=πr2

.

A=π(3.12)=9.61π

m2

Find the probability that a point chosen randomly inside the rectangle is in the circle.

P2=9.61π188.8

Sum the individual probabilities to find the probability that a point chosen randomly inside the rectangle is either in the triangle or in the circle.

P=P1+P2

=18188.8+9.61π188.8

=18+9.61π188.8≈0.26

Therefore, the probability that a point chosen randomly inside the rectangle is either in the circle or in the triangle is about 0.26

.

The probability that a point chosen randomly inside the rectangle is either in the triangle or in the circle is 0.26. Option fourth is correct.

What is probability?

It is defined as the ratio of the number of favourable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.

It is the problem of geometric probability:

Area of rectangle = 16×11.8 = 188.8 square meters

Area of circle = π(6.2/2)² = 30.190 square meters

Area of triangle = (1/2)(8×[11.8-7.3]) = 18

P(triangle) = 18/188.8 = 0.095

P(circle) = 30.190/188.8 = 0.159

P = P(triangle) + P(circle) = 0.095 + 0.159  = 0.254 ≈ 0.26

Thus, the probability that a point chosen randomly inside the rectangle is either in the triangle or in the circle is 0.26. Option fourth is correct.

Learn more about the probability here:

brainly.com/question/11234923

#SPJ2

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