Rom

4 walls  = 2 x (14 + 10) x 8 = 384 sq ft

windows = 2 x 2 x 5 = 20 sq ft

door = 4 x 7 = 28 sq ft

total = 384 - 20 - 28 = 336 sq ft.

336/350  is rounded up to 1 gallon

\((1+\sqrt{3})-(1-\sqrt{3}) = 1-1 + \sqrt{3}+\sqrt{3}=0+2\sqrt{3}\)

There's only 1 zero of f(x) in the interval [-2,4]

1 point can't enclose anything....

let's first figure out how many permutations do have the 3rd term equal to 3

The first number we have 4 choices (not 1 or 3), the second number we have 4 choices, the 3rd number is 3, the 4th there are 3 choices, 5th 2 choices.

This is 4 x 4! = 96

The total number of permutations in S is just 5 x 5! = 600 (do you see why?)

and thus

P[term 3 = 3] = 96/600 = 4/25

A = 3W2B

B = 4W3B

transfer W from A to B gives: A = 2W2B, B = 5W3B, Probability 3/5

transfer B from A to B gives : A = 3W1B, B = 4W4B, Probability 2/5

Let event X be the event that the moved marble was white

Let event Y be the event that the marble chosen from bag B was white

We want P[X | Y]

\(P[X | Y] = \dfrac{P[Y|X]P[X]}{P[Y]}\)

\(P[Y|X] = \dfrac 5 8\\ \\ P[X] = \dfrac 3 5 \\ \\ P[Y] = \dfrac 5 8 \dfrac 3 5 + \dfrac 4 8 \dfrac 2 5 = \dfrac{23}{40}\)

\(P[X|Y] = \dfrac{\frac 5 8 \cdot \frac 3 5}{\frac{23}{40}} = \dfrac{15}{23}\)

There are 4C2 = 6 ways to select 2 schools.

Given those 2 schools there are 4² = 16 ways to pair up players 1 from each school

There are 3 games played by each of those 3 pairs.

This is a total of 6 x 16 x 3 = 288 games played against other schools.

There are 4C2 = 6 ways to select a pair from the 4 people at a given school.

There are 4 schools

They play 1 game each

This is a total of 6 x 4 x 1 = 24 games played against schoolmates

A total of 288 + 24 = 312 games

Neat problem.

First off note that \(\dbinom{n}{k}=\dbinom{n}{n-k}\)

We are choosing (n-1) items from 2n possible items.  One way to do this is to split the 2n items into 2 piles of n items each, choose k items from the first pile, and (n-1-k) items from the second pile.  We do this for \(0 \leq k \leq n-1\)

This gets us

\(\dbinom{2n}{n-1} = \sum \limits_{k=0}^{n-1}~\dbinom{n}{k}\dbinom{n}{n-k-1} = \sum \limits_{k=0}^{n-1}~\dbinom{n}{k}\dbinom{n}{k+1}\)

you can recognize the right hand term as the summation shown

the above answer sort of misses the point I think.

We can notice that \(k^3 \pmod{6} = k,~k=1,2,\dots 5 \\ \\ \text{and obviously }6^3 \pmod{6} = 0\)

so we have a repeating sequence in the sum \( 1+2+3+4+5+0 + 1+2+\dots\)

there are 16 of these groups of 6 and then 4 additional elements of the sum

so we have

\(\sum \limits_{k=1}^{100}~k^3 \pmod{6} = 3\cdot 16 + 1+2+3+4 = 58 \\ \text{and} \\ \\ 58 \pmod 6 = 4 \\ \\ \text{so }\sum \limits_{k=1}^{100}~k^3 \pmod{6} = 4\)

\(f(a+2) = 3((a+2)+5) + \dfrac{4}{a+2} = 3(a+7) + \dfrac{4}{a+2} \\ \\ \text{i.e. choice B}\)