Double m, then raise the result to the 9th power
1 answer:
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Answer:
The probability that a randomly selected point within the circle falls in the white area is approximately 36.3%
Step-by-step explanation:
The given parameters of the area are;
The length of the radius of the circle, r = 4 cm
The length of the side of the inscribed quadrilateral, s = 4·√2 cm
The diagonals of the inscribed quadrilateral = 2 × The radius of the circle
∴ The diagonals of the inscribed quadrilateral are equal and given that the sides of the quadrilateral are equal, the quadrilateral is a square
The probability that a randomly selected point within the circle is white is given by the ratio of the white area to the brown square area as follows;
The white area = The total area - The area brown square
The total area, A = The area of the circle with radius, r = π·r²
∴ A = π·(4 cm)² = 16·π cm²
The area of the brown square, Asq = s² = 4·√2 cm × 4·√2 cm = 32 cm²
The white area, Aw = 16·π cm² - 32 cm² = 16·(π - 2) cm²
The probability that a randomly selected point within the circle falls in the white area is therefore;
By rounding to the nearest tenth of a percent, we have;
The probability that a randomly selected point within the circle falls in the white area, P(W) ≈ 38.3%.