Rom

I second CPhil's answer

It's correct.

The function isn't very well specified.  It's just fortunate that each expression produces the same value for 3.

sim shows 8/15 is correct.

Ok, it's not that big a deal.

You choose the first sock with probability 1 and the second unmatched sock with probability 4/5

For clarity let's say you chose (r,w) as the first pair.  We are left with (r,w,b,b)

For the third sock you can choose, either r or w, and be forced to choose b as the 4th, or

You can choose b, and then be forced to choose r or w as the 4th.  This is

2/4 x 2/3 + 2/4 x 2/3 = 2/3

I.e. there is a 2/3 probability you will choose a non-matching pair for the 3rd and 4th socks.

P[non-matching pair] = 4/5 x 2/3 = 8/15

\(c = \lambda \nu\\ c = 3\times 10^8 m/s\\ \nu = 530 \times 10^3 ~s^{-1}\\ \text{I trust you can multiply}\)

\(\text{There are }M=(n-2)(n+8) \text{ total marchers}\\ \text{let }D\text{ be the number of drummers},~D\geq 4\\ M-D = n(2n-3)\)

\(n^2+6n-16 = 2n^2 - 3n+D\\ n^2 -9n+(16+D)=0\\ n = \dfrac{9\pm \sqrt{81-4(16+D)}}{2} \in \mathbb{N}\)

\(81 - 4(16+D) = k^2,~k \in \mathbb{N}\\ \text{The only possible value for }D \text{ is 4}\\ n = \dfrac{9 \pm 1}{2} = 5,4\\ 4+5=9\)

\(F= \dfrac{260N}{4} = 65N\\ d = 4(0.5m)= 2m\\ \text{Note that the work done }F\cdot d \text{ is constant}\)

\(\text{What are we supposed to vary here in order to get the max and min values?}\\ \text{The order of the numbers? The parens, or lack of them?}\\ \text{This is a very unclear question}\)

\(\text{I have a simpler answer.}\\ \dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}+\dfrac{1}{(2k)^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}}=1\)

\(\sum \limits_{k=1}^\infty \dfrac{1}{(2k)^3} = \sum \limits_{k=1}^\infty \dfrac{1}{8k^3} = \dfrac 1 8 \sum \limits_{k=1}^\infty \dfrac{1}{k^3}\)

\(\dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}+\dfrac 1 8 \sum \limits_{k=1}^\infty\dfrac{1}{k^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}}=1\\ \dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}} + \dfrac 1 8 = 1\\ \dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}} = \dfrac 7 8\)

Plot the point (2,1)

Plot the point 3 left and 2 up from it, i.e. (-1,3)

Draw a line through those two points.

The point 3 right and 2 down works as well, i.e (5,-1)

incorrect answer deleted