Will give brainliest Find the probability that a randomly
selected point within the circle falls
in the white area.
W
r = 4 cm
4√2 cm
[? ]%
Round to the nearest tenth of a percent.

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%.

a

step-by-step explanation:

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !