(a) If <em>f(x)</em> is to be a proper density function, then its integral over the given support must evaulate to 1:
For the integral, substitute <em>u</em> = <em>x</em> ² and d<em>u</em> = 2<em>x</em> d<em>x</em>. Then as <em>x</em> → 0, <em>u</em> → 0; as <em>x</em> → ∞, <em>u</em> → ∞:
which reduces to
<em>c</em> / 2 (0 + 1) = 1 → <em>c</em> = 2
(b) Find the probability P(1 < <em>X </em>< 3) by integrating the density function over [1, 3] (I'll omit the steps because it's the same process as in (a)):
The correct answer would be 2y^5
Rewriting the equation as a quadratic equation equal to zero:
x^2 - x - 30 = 0
We need two numbers whose sum is -1 and whose product is -30. In this case, it would have to be 5 and -6. Therefore we can also write our equation in the factored form
(x + 5)(x - 6) = 0
Now we have a product of two expressions that is equal to zero, which means any x value that makes either (x + 5) or (x - 6) zero will make their product zero.
x + 5 = 0 => x = -5
x - 6 = 0 => x = 6
Therefore, our solutions are x = -5 and x = 6.
Answer: -15y
Step-by-step explanation:
6(x-2y) - 3(2x+y)
6x-12y -6x-3y
0x -15y
-15y