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rewona [7]
3 years ago
8

A point in the figure is selected at random. Find the probability that the point will be in the shaded region.

Mathematics
2 answers:
lesya [120]3 years ago
6 0
Area of the figure:
Af = a² + 2 a²/4 π = 2.57 a²
Area of the shaded region:
As = 2 a²/4 π = 1.57 a²
x = (1.57 · 100) / 2.57 = 61 % ≈ 60 %
Answer: B) about 60 %
alexandr1967 [171]3 years ago
3 0

Answer:

B. <em>about 60%</em>

Step-by-step explanation:

Let us assume that the radius of the semi-circle is d units.

So the area of the semi-circle will be,

=\dfrac{\pi d^2}{2}

In total there are 4 semi circles, so the net area will be,

=4\times \dfrac{\pi d^2}{2}

=2\pi d^2

The side length of the square is twice the radius of the semi circle. So the side length of the square is 2d units.

Area of the square is,

=(2d)^2\\\\=4d^2

The total area will be,

=2\pi d^2+4d^2

So that the random point will be in the shaded region is,

=\dfrac{\text{Area of semi circles}}{\text{Area of square}}

=\dfrac{2\pi d^2}{2\pi d^2+4d^2}

=\dfrac{2\pi}{2\pi+4}

=0.6

=60\%

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Step-by-step explanation:

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Now we translate the statements into algebraic equations:

The number of hot dogs and hamburgers that were sold is 78, so d+h=78.

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Now plug this in for d in the second equation:

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Since h+d=78,

32+d=78→d=46

The concession stand therefore sold 46 hot dogs and 32 hamburgers.

3 0
3 years ago
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