What is the area of Shaded part?

1π

Step-by-step explanation:

suppose the radius of semicircle P is r,

then the radius of semicircle Q = (r+d)/2  ... d≤r

radius of semicircle R = (r-d)/2

area P = 1/2 (r)²π

area Q = 1/2 ((r+d)/2)² π = 1/8 (r² + 2rd + d²)π

area R = 1/2 ((r-d)/2)² π = 1/8 (r² - 2rd + d²)π

shaded area = P-Q-R = 1/2 r²π - 1/4 (r² + d²)π

                                   = ((r² - d²)/4) * π

because there is no constant r value in the question and d value changes with the r change, when the vertical segment length equal the semicircle P radius (r), r=2 and d = 0

therefore the shaded area = ((2² -0²)/4)*π = 1π

the answer   is 200

step-by-step explanation:

this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )