Where are you stuck? Show us what you have so far... Most of these problems are applying basic stats concepts on a given example so I'm not sure exactly what you need help with. I'd like to see some effort in your part or know where you are because I don't like to just complete people's homework, instead I prefer to lead them in the correct direction.

I guess I'm just not very confident in how to do them. Here are the answers I got so far..

a. I found that c=1/15
b. E(x) = 11.917 and E[X²] = 85 so σ² = 85 - (11.917²) = -57?
c. not really sure what to do from here on

(a) You want f(x) to be non negative for every x, and integral of f over R equals 1.
Observe that integral of f is made of a trapezoid, over the interval [6;10], and of a triangle, over the interval [0;6].
Area of the trapezoid is 4/5, so 1/5 remains for the triangle.
Dimensions for the triangle are :
6 for the base
1/15 for the heigth (the value you found apparently).
To conclude, observe that the heigth is the Y-coordinate of the point with X-coordinate 6.
That leads to the equation : 6c = 1/15, hence c = 1/90.

(b) Use the definition of E(X) = ∫ x.f(x) dx
Cut the integral at 6 :
∫ x.f(x) dx over [0;6] is ∫ x^2 /90 dx = 6^3 /270 = 4/5
∫ x.f(x) dx over [6;10] is ∫ x^2 /40 dx = (10^3 - 6^3)/120 = 25/3 - 9/5
Add up : E(X) = 22/3

V(X) = E(X^2) - E(X)^2 = ∫ x^2.f(x) dx - (22/3)^2
∫ x^2.f(x) dx over [0;6] is ∫ x^3 /90 dx = 6^4 /360 = 18/5
∫ x^2.f(x) dx over [6;10] is ∫ x^3 /40 dx = (10^4 - 6^4)/160 = 272/5
Add up : 290/5 = 58

Hence, V(X) = 58 - (22/7)^2 = 2358/49
... if I do no mistake.
pm if you still need help with the rest.

Just an aside to what is said above, you don't need to do this to find c=1/90 (which is correct and used to compute rest of problem) if you wish you can use the fact that the integral of your pdf function which Is your cdf is one. You have a function defined over two seperation intervals so you can just solve for c this way. I've included a pic below.

http://postimg.org/image/wp736ccut/

This post was edited by Xx Shin3d0wn xX on Feb 19 2015 11:08pm