Rom

\(\left(\dfrac{1}{13}\right)\left(\dfrac{4}{51}\right)\left(\dfrac{4}{50}\right)= \dfrac{16}{33150} \approx 4.83\times10^{-4}\)

first off we regroup things so that we have 5 Republicans and 8 non-Republicans

this gets us

\(P[R]=\dfrac{5}{13},~P[!R] = \dfrac{8}{13}\)

now we have a simple Binomial distribution with parameters  \(n=3,~p=\dfrac{5}{13}\)

\(P[3] = \left(\dfrac{5}{13}\right)^3 = \dfrac{125}{2197}\)

This was wrong.  Melody's approach is correct.

The 10 tulips, 5 of one color and 5 of another is a bit of a red herring.

You are basically choosing 2 colors out of n.

\(\text{so you need to find }n \ni \dbinom{n}{2}=15 \\ \\ \text{a bit of testing quickly reveals that }n=6\)

\((4n+1)\cdot 90^\circ = n360^\circ + 90^\circ \\ \\ \text{trig functions are all periodic with period }360^\circ \\ \text{so}\csc((4n+1)\cdot 90^\circ) = \csc(90^\circ) \\ \csc(90^\circ) = \dfrac{1}{\sin(90^\circ)} = \dfrac{1}{1} = 1\)

12 inches = 72 cents

1 inch = 6 cents

37 inches = 37 x 6 cents = 222 cents

a + (-b) = a - b

(-a) + b = b - a

-23423423423 + 1 + 1 < 0

-14 is an integer

1 3/4 = 7/4 is a rational number

\(\sqrt{2}\) is an algebraic number

\(-14 < \sqrt{2} < \dfrac 7 4\)

In general suppose you have an odd function o(x), and an even function e(x).

o(x)o(x) will be even

o(x)e(x) will be odd

e(x)e(x) will be even

here f(x) is even, g(x) is odd, and 1/x is odd

so we have the product of 2 odd and 1 even functions which by above will be even