Apoint in the figure is selected at random. find the probability that the point will be in the part that is not shaded.
a. about 50%
b. about 40%
c. about 25%
d. about 65%

B

Step-by-step explanation:

So the question really is what is the probability that the point is in the square? You cannot pick something surrounding the circles because you do not know anything about that region.

This is solved by comparing areas. There are 2 full circles there. 1 circle has an area of pi*r^2

Total shaded area = 2 * pi * r^2.

Let r = 3   This is a completely random choice. No matter what number you choose for r, the answer will come out the same. When you get along a little further in math, you will find that you can just use letters.

Total shaded area = 2 * 3.14 * 3^2

Total shaded area = 2 * 3.14 * 9

Total shaded area = 56.55

Total area of the square.

s = 2*r

s = 2*3

s = 6

Area of the square = 6^2 = 36

Total area of both regions = 56.55 + 36 = 92.55

Answer

% = (area of square) * 100% / Total area of both regions

% = 3600 / 92

% = 39%

Obviously the answer I get is not offered. The closest answer is B so I will choose that.

This is how this question would be done without using any value for r.

Area of shaded region = 2 pi r^2 = 6.28 r^2

Area of square  = d^2 where d = 2r

Area of square = 4 r^2

Total area = 6.28 r^2 + 4r^2 = 10.28 r^2

Area of square to total = (4r^2/10.28 r^2 ) * 100%

Area of square to total (as a %) = 4/10.28  * 100 = 39% which gives the same answer.

$30.32

step-by-step explanation:

65.00*.12=2.80

65.00-2.80=57.20

57.20*1.06=60.632

60.632/2=30.316

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