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mina [271]
2 years ago
11

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

Mathematics
2 answers:
Westkost [7]2 years ago
6 0

Answer:

c and full truat i nwver wrong

kirill115 [55]2 years ago
4 0

Answer:

1. 0.13

2.0.26

Step-by-step 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

.

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Answer:

I answered the questions but that formatting is very confusing and discourages anyone for trying to answer.

The last question is confusing. If I got it wrong, tell me. I will try to answer in the comment section then.

Step-by-step explanation:

Kayson is looking at two buildings, building A and building B, at an angle of elevation of 73°. Building A is 30 feet away, and building B is 35 feet away. Which building is taller and by approximately how many feet?

$\text{tan}\theta=\frac{h_{A}}{30} \Rightarrow h_{A}=\text{tan}73\º \cdot 30 \Rightarrow h_{A}\approx 98.12 \text{ feet}$

$\text{tan}\theta=\frac{h_{B}}{35} \Rightarrow h_{B}=\text{tan}73\º \cdot 35 \Rightarrow h_{B}\approx 114.47 \text{ feet}$

\text{Difference} \approx 16.35

Building B is around 16.35 feet taller than building A.

A Look at the figure below: an image of a right triangle is shown with an angle labeled y If sin y° = a divided by 6 and tan y° = a divided by b, what is the value of cos y°?

$\text{sin}(y)=\frac{a}{6} $

$\text{tan}(y)=\frac{a}{b} $

a is the opposite side; 6 is the hypotenuse; b is the adjacent side.

Therefore,

$\text{cos}(y)=\frac{b}{6} $

If sin f° = eight ninths and the measure of segment YW is 24 units, what is the measure of segment YX? triangle XYW in which angle W is a right angle, angle X measure f degrees, and angle Y measures d degrees.

This seems a bit confusing. The angles don't match. We have 90\º+62\º +21\º \approx 173\º

\text{sin}(f)=0.888...

f \approx 62\º

YX is the hypotenuse of the right triangle.

$\text{cos}(21\º)=\frac{24}{YX} $

YX \approx 25.7

Considering f \approx 69\º

$\text{sin}(69\º)\approx\frac{24}{YX} $

YX \approx 25.7

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I'll make you brainliest Easyquestion 6th grade multiply and add fractions pleaze answer it ASAP
kirill115 [55]

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                         = 10.5x5.75/2

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