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I need serious help with these 2 questions. I already used the first attempt and have 1 left so please help me

I need serious help with these 2 questions. I already used the first attempt and have-example-1
I need serious help with these 2 questions. I already used the first attempt and have-example-1
I need serious help with these 2 questions. I already used the first attempt and have-example-2
User Tjackadams
by
2.9k points

2 Answers

10 votes

Answer:

c and full truat i nwver wrong

User Masato
by
3.9k points
4 votes

Answer:

1. 0.13

2.0.26

Explanation:

1. 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=25

m, and width, w=17 m, into the formula for the area of a rectangle, A=lw

.

A=(25)(17)=425

m2

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

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

Find the area of the trapezoid.

Substitute the known values for the bases, b1=13

m and b2=8 m, and height, h=4 m, into the formula for the area of a trapezoid, A=12(b1+b2)h

.

A=12(13+8)(4)=42

m2

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

P1=42425

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.

Substitute the known values for the base, b=4

m, and height, h=7 m, into the formula for the area of a triangle, A=12bh

.

A=12(4)(7)=14

m2

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

P2=14425

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

P=P1+P2

=42425+14425

=56425≈0.13

Therefore, the probability that a point chosen randomly inside the rectangle is either in the triangle or in the trapezoid is about 0.13

2.

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

.

User Renuz
by
3.5k points