Rom

we start with 4x6 inches and we triple this to become 12x18 inches

we can't see what "as shown" is.  I'm going to assume they miter the corners as that is what's usually done.

So each piece of framing has to be 4 inches longer than it's associated side.

Thus we have 2 framing pieces at 16 inches, and 2 at 22 inches for a total of 76 inches.

This is 6 feet 4 inches so we much buy 7 feet of framing.

\(\dfrac 2 3 t - 1 < t+7 \leq -2t + 15 \\ \dfrac 2 3t-8

\(\dfrac{12+x}{23}=2\)

\(\displaystyle \sum \limits_{k=1}^n~\alpha^k =\dfrac{\alpha \left(\alpha ^n-1\right)}{\alpha -1}\)

letting \(\alpha=i,~n=99,~\text{and assuming }i=\sqrt{-1}\)

\(\displaystyle \sum \limits_{k=1}^{99} ~i^k = \dfrac{i(i^{100}-1)}{i-1} = 0\)

ok, this one is a bit involved.  6 Mars like planets that use 1 colony unit, 5 Earth like planets that use 2 colony units

12 total colony units.

First off we note that we only consider combinations with even numbers of Marslike planets.  Otherwise we would have a colonly unit left over.  We also note that we must colonize at least 2 Martian like planets and can only colonize up to the 6 existing Marslike planets.  Below is a table summarizing the results.

Totalling the right column we see there are 100 different combinations using all 12 of the colony units.

There are \(\dbinom{30}{3}\dbinom{20}{2} = 771400\) different combinations without regard to order

I think it's easiest to count the ways that they can consecutively sit together and subtract this from the total number of seating arrangements.

There are 6 groups of 3 consecutive seats.  Within each group there are 3!=6 ways the three fussy people can be arranged.

Then the remaining 5 people can be arranged 5!=120 ways in the remaining seats.

So there are 6*6*120 = 4320 arrangements where they DO sit consecutively

There are 8! = 40320 total arrangements.

So there are 40320 - 4320 = 36000 arrangements where the 3 do not sit consecutively.

without being able to distinguish individual frogs we only have 2 possible orderings

ggbrrr, and rrrbgg

Now if you say we are able to uniquely identify each frog then we have 2!=2 ways of arranging the green frogs and 3!=6 ways of arranging the red frogs.  So there are a total of 2*2*6 = 24 ways of arranging the frogs to their liking.

all pairs of spins have probabilty of \(\left(\dfrac 1 5\right)^2 = \dfrac{1}{25}\)

so we just have to count the number of spin pairs where Jane wins and divide it by 25.

there's no magic to this, just list them out.  If I understand what "non-negative difference" means, Jane wins with

(5,5), (5,4), (5,3), (4,5), (4,4), (4,3), (4,2), (3,5), (3,4), (3,3), (3,2), (3,1), (2,4), (2,3), (2,2), (2,1), (1,3), (1,2), (1,1)

i.e. 19 spin combos that lead to Jane winning.  Thus

\(P[\text{Jane wins}]=\dfrac{19}{25}\)

\(\sum \limits_{k=1}^\infty~x^k = \dfrac{x}{1-x} \\ \\ \text{differentiate both sides} \\ \\ \dfrac {d}{dx} \sum \limits_{k=1}^\infty~x^k = \dfrac{d}{dx} \dfrac{x}{1-x} \\ \\ \sum \limits_{k=1}^\infty~k x^{k-1} = \dfrac{1}{(1-x)^2} \\ \\ \text{mutliply both sides by }x \\ \\ \sum \limits_{k=1}^\infty~k x^k = \dfrac{x}{(1-x)^2} \\ \\ \text{let }x=\dfrac 1 2 \\ \\ \sum \limits_{k=1}^\infty~\dfrac{k}{2^k} = \dfrac{\frac 1 2}{\left(1-\frac 1 2\right)^2} = 2\\ \\\)