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%.

pay attention in your classes and concentrate the best thing you can do to raise your test scores is to pay attention when you're supposed to be learning the material: in class!

take good notes.

do your homework.

use mnemonics and other tricks.

do practice tests.

note: the best i know.

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v