Let X be a positive continuous random variable with density fXpxq. Let Y " lnpXq. (a) Find the density fY pyq of Y in terms of the density of X. (b) If X is a uniform random variable on the interval p0, 5s, what is fY pyq?

X density = fXpxq and

Y" =InpXq

Now to find Y density FYpyq interms of the density of X we compare the density of X with Y"

fX = In

And PXq =pxq

Thus replacing x with y,

PXq = pyq

(a) Hence the density of Y is FYpyq

(b) at p0, fYpyq =fYp0q= 0

At 5s, FYpyq =5

a) lnf(x)dx =[ xlnf(x)- x] + C

b) 5(ln5pq-1)

Step-by-step explanation:

Given f(x) = pxq,

Y=ln(pxq)

(a). f(y) = ∫lnf(x)dx over range x

Using integration by parts

∫udv = uv - ∫vdu

lnf(x) = u, dv =1, > v = x

∫udv = lnf(x). x - x∫(1/pxq)pq.dx

∫udv = xlnf(x) - ∫dx

∫lnf(x)dx =[ xlnf(x)- x] + C

(b) ∫lnf(x)dx = [xln(pxq) - x], at range x= 0, and x= 5

= 5ln5pq - 5

=5(ln5pq-1)

363 square inches

step-by-step explanation:

surface area of a pyramid is b + 1/2ph

b is the area of the base: 11 x 11 = 121

p is the perimeter of the base: 11 + 11 + 11 + 11 = 44

h is the height of the pyramid: 11

121 + 1/2(44)(11)

121 + 242

363