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ANTONII [103]
2 years ago
5

From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probabilit

y that at least one of the sides of the triangle is also a side of the octagon
Mathematics
1 answer:
nataly862011 [7]2 years ago
4 0

The probability that at least one of the sides of the triangle is also a side of the octagon is 5/7

<h3>How to determine the probability?</h3>

A regular octagon has 8 sides, and a triangle has 3 sides

Assume that there is a point on the triangle.

Then we select (3 - 1) points from the (8 - 1) sides.

8 - 1 = 7

3 - 1 = 2

So, the total number of selection is:

Selection = 7C2

Evaluate the combination expression

Selection = 21

Adjacent triangles have 5 lines (one line in common).

So, we have:'

n = 5 - 1

n = 4

The number of ways this points can be selected is:

Selection = 4C2

Evaluate

Selection = 6

The probability that the triangle and the octagon do noth have the same side is:

P= 6/21

The probability that at least one of the sides of the triangle is also a side of the octagon is calculated using the following complement rule

P = 1 - 6/21

Evaluate the difference

P = (21 -6)/21

P = 15/21

Reduce fraction

P = 5/7

Hence, the probability that at least one of the sides of the triangle is also a side of the octagon is 5/7

Read more about probability at:

brainly.com/question/251701

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You have to set up 2 proportions.

If the 6.9 goes with the 5 then the proportion will look like this.
6.9/5 = x/6 Cross multiply
6 * 6.9 = 5x
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If on the other hand the 6.9 goes with the 6 then the second proportion is
6.9/6 = x/5
6.9 * 5 = 6X
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Answer: D <<<<<<<<======

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alukav5142 [94]

Answer:

C

Step-by-step explanation:

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What is the area of the figure below​
jek_recluse [69]

Answer:

Option A

Step-by-step explanation:

Area of the composite figure = Area of ΔABH + Area of BCGH + Area of ΔDEF

Area of ΔABH = \frac{1}{2}(\text{Base})(\text{Height})

                        = \frac{1}{2}(6)(8-5)

                        = 9 in²

Area of the trapezoid = \frac{1}{2}(CG+BH)(\text{Distance between these lines})

                                    = \frac{1}{2}(6+4)(5)

                                    = 25 in²

Area of ΔDEF = \frac{1}{2}(8)(6)

                       = 24 in²

Therefore, area of the given figure = 9 + 25 + 24

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Option A will be the correct option.

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yanalaym [24]

\huge {\mathfrak{Answer :  }}

we know,

\boxed{\cos( \theta)  =  \dfrac{base}{hypotenuse}}

so,

  • \cos(30)  =  \dfrac{24}{x}

  • \dfrac{ \sqrt{3} }{2}  =  \dfrac{24}{x}

  • x =  \dfrac{24 \times 2}{ \sqrt{3} }

  • x = 8 \sqrt{3}  \times 2

  • x = 16 \sqrt{3}

now,

\boxed{ \tan( \theta) =  \frac{perpendcular}{base}  }

  • \tan(30)  =  \dfrac{y}{24}

  • \dfrac{1}{ \sqrt{3} }  =  \dfrac{y}{24}

  • y =   \dfrac{24}{ \sqrt{3} }

  • y = 8 \sqrt{3}
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