MAJORR HELP!!
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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%.
<h2>Steps</h2>
So to find the solutions, I will be factoring by grouping. Firstly, what two terms have a product of 12x² and a sum of -8x? That would be -2x and -6x. Replace -8x with -2x - 6x:
Next, factor 4x² - 2x and -6x + 3 separately. Make sure that they have the same quantity on the inside of the parentheses:
Now you can rewrite it as:
Now, apply the zero product property and solve:
<u>In short, x = 3/2 (1.5) and 1/2 (0.5).</u>