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Sophia throws a dart at this square-shaped target: A square is shown with sides labeled 9. A shaded circle is shown in the center of the square. The diameter of the circle is 3. Part A: Is the probability of hitting the black circle inside the target closer to 0 or 1? Explain your answer and show your work. (5 points) Part B: Is the probability of hitting the white portion of the target closer to 0 or 1? Explain your answer and show your work. (5 points)

Answer :

Mhanifago

Step-by-step explanation:

Find the area of each part

The circle

  • C = πr² = 3.14*(3/2)² = 7.07

The square

  • S = a² = 9² = 81

The area within the square but outside of circle

  • A = S - C = 81 - 7.07 = 73.93

Part A

Probability of hitting the black circle inside the target

  • P = 7.07/81 = 0.09 (rounded)

This is closer to 0 than to 1

Part B

Probability of hitting the white portion of the target is

  • P = 73.93/81 = 0.91 (rounded)

This is closer to 1 than to 0

Semsee45go

Answer:

Formulae

Area of a square = x² (where x is the side length)

Area of a circle = πr²  (where r is the radius)

Diameter = 2r  (where r is the radius)

Given:

  • side length of square = 9 units
  • diameter of circle = 3 units

Therefore:

Area of square = 9² = 81 units²

Area of circle = π · 1.5² = 2.25π units²

Part A

[tex]\begin{aligned}\sf Probability \ of \ hitting \ black \ circle & = \sf \dfrac{area \ of \ circle}{area \ of \ square}\\ & = \sf \dfrac{2.25 \pi}{81}\\ & =0.0872664626...\end{aligned}[/tex]

Therefore, the probability of hitting the black circle is closer to 0.

Part B

[tex]\begin{aligned}\sf Probability \ of \ hitting \ white \ portion & =\sf \dfrac{area \ of \ square - area \ of \ circle}{area \ of \ square}\\ & = \sf \dfrac{81-2.25 \pi}{81}\\ & = \sf0.9127335374...\end{aligned}[/tex]

Therefore, the probability of hitting the white portion is closer to 1.

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