(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)):

I believe the answer is B, but I'm not 100% sure. :)
Difference = 48-19
= 29
production decrease = 29 × 100/ 48
= 60.41%
Answer:
y-6=-1/3(x+6)
Step-by-step explanation:
y-y1=m(x-x1)
m=(y2-y1)/(x2-x1)
m=(3-6)/(3-(-6))
m=-3/(3+6)
m=-3/9
simplify
m=-1/3
y-6=-1/3(x-(-6))
y-6=-1/3(x+6)