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Vlad1618 [11]
3 years ago
15

A regular hexagon is inscribed (vertices lie on the circles) in a circle.

Mathematics
1 answer:
Sladkaya [172]3 years ago
6 0
Radius of the circle: r
Area of the circle: A1=pi r^2
A1=3.141592654 r^2

Area of the regular hexagon: A2=(6)(1/2)(r)(r) sin 60°
A2=(6/2) r^2 sqrt(3)/2
A2=3 sqrt(3) r^2 /2
A2=3(1.732050808) r^2/2
A2=2.598076212 r^2

<span>a. What percent of the area of the circle is overlapped by the area of the inscribed regular hexagon?

Percentage: P=(A2/A1)*100%
P=[2.598076212 r^2 / (3.141592654 r^2)]*100%
P=(0.826993343)*100%
P=82.6993343%
P=82.7%

Answer: 82.7 </span><span>percent of the area of the circle is overlapped by the area of the inscribed regular hexagon.

</span><span>b. if the radius of the circle is tripled and a regular hexagon is inscribed in the new circle, what percent of the area of the circle is overlapped by the area of the inscribed regular hexagon?

The same percentage.

Answer: 82.7 </span><span>percent of the area of the circle is overlapped by the area of the inscribed regular hexagon.</span>
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The complete question is:

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\frac{6}{50}P(Pays bill by mail/Customer is female) = \frac{P(\text{Pays bill by Mail}\bigcap \text{ Customer is Female})}{P(\text{ Customer is Female})}

Now, Probability that customer is female =  \frac{30}{50}

Also, Probability that customer pays bill by mail and is female =   \frac{6}{50}

So, Required probability =  \frac{\frac{6}{50} }{\frac{30}{50} }

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                                         =  \frac{1}{5}   =  0.20

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