(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)):
Triangles QST and RST are similar. Therefore, the following is true:
q s
--- = ---- This results in 10q=rs.
r 10
Also, since RST is a right triangle, 4^2 + s^2 = q^2.
Since QST is also a right triangle, s^2 + 10^2 = r^2.
4 s
Also: ---- = ----- which leads to s^2 = 40
s 10
Because of this, 4^2 + s^2 = q^2 becomes 16 + 40 = 56 = q^2
Then q = sqrt(56) = sqrt(4)*sqrt(14) = 2*sqrt(14) (answer)
The following are the general steps to draw a hexagon inscribed in a circle:
First draw a circle with center A through B:
Then the intersection points will be vertex C and vertex D of the hexagon. Doing this for each vertex (from left to right) we will get the following:
Answer:
Step-by-step explanation:
we substitute the value of "y" in the equation:
She is going to 5 classes 3×5 = 15 and 15÷3=5
