A point $P$ is randomly selected from the square region with vertices at $(\pm 2, \pm 2)$. What is the probability that $P$ is within one unit of the origin

If P = (X, Y) is a point in the given square, then X and Y are i.i.d random variables each with distribution

and so the joint density of X and Y is

We want to find P(X² + Y² ≤ 1). Points that satisfy this inequality lie in the set

R = {(x, y) : -1 ≤ x ≤ 1 and -√(1 - x²) ≤ y ≤ √(1 - x²)}

but we can more easily describe the region in polar coordinates by setting

x = r cos(t) and y = r sin(t)

so that the set R is identical to

R' = {(r, t) : 0 ≤ r ≤ 1 and 0 ≤ t ≤ 2π}

Integrate the joint density over R' :

687.09 dollars       happy now

step-by-step explanation: 6870

1857 times .63 equals 1169.91 subtract that from 1857 you get 687.09

step-by-step explanation:

f(x)=2x^2 -1 

translation is a function transformation that transform the function in a way of translating. by translating the function its graph is moved  up an down. moving the function f(x) 2 units to the left will be the new function g(x) with the equation: g(x)=f(x+2)=2(x+2)^2 -1