Rom

\(2y=\dfrac{f(2x)}{2}+2\\ y = \dfrac{f(2x)}{4}+1\\ f\left(2 \dfrac 9 2\right)=f(9)=7\\ y = \dfrac{7}{4}+1=\dfrac{11}{4}\\ (x,y) = \left(\dfrac 9 2,~\dfrac{11}{4}\right)\)

\(x+y = \dfrac{18+11}{4} = \dfrac{29}{4}\)

\(\ln(x+1) =- \sum \limits_{k=1}^\infty ~\dfrac{(-x)^k}{k}\)

\(23(0.94)^x = 15\\ (0.94)^x = \dfrac{15}{23}\\ x\log(0.94) = \log\left(\dfrac{15}{23}\right) = \log(15)-\log(23)\\ x = \dfrac{\log(15)-\log(23)}{\log(0.94)}\)

\(\text{Let }\Phi(x) \text{ be the CDF of the standard normal distribution}\\ P[n \leq N] = \Phi\left(\dfrac{N-\mu}{\sigma}\right)\\ P[N < n] = 1-P[n \leq N] \\ P[N_l \leq n \leq N_h] = \Phi\left(\dfrac{N_h-\mu}{\sigma}\right) - \Phi\left(\dfrac{N_l-\mu}{\sigma}\right)\)

now just dump your numbers in

or you can just find

 \(p\left(\dfrac 1 x\right) = 0\\ \left(\dfrac 1 x\right)^3 - \dfrac 2 x - 3 = 0\\ 1 - 2x^2 - 3x^3 = 0 \\ \text{This is the form the problem asks for, but}\\ \text{dividing both sides by -3}\\ x^3 + \dfrac 2 3 x^2 - \dfrac 1 3 = 0\\ \text{which is the form Alan wrote it in}\)

\(\text{similarly for the second part}\\ x=\dfrac{1}{2\alpha+1} \Rightarrow \alpha = \dfrac 1 2\left(\dfrac 1 x -1\right)\\ \text{so find }\\ p\left(\dfrac 1 2\left(\dfrac 1 x -1\right)\right)\\ \text{and work the same sort of algebra to get nice coefficients}\)

\(\text{The point }p \text{ is described by a 2D joint uniform distribution}\\ p(x,y) = \dfrac{1}{18},~0\leq x \leq 3,~0\leq y\leq 6\\ P[x+y \leq 4] = P[y\leq 4-x] = \\ \displaystyle \int_0^3 \int_0^{4-x}~\dfrac{1}{18}~dy~dx = \\ \displaystyle \int_0^3 \dfrac{4-x}{18}~dx = \\ \left. \dfrac{4x-\frac{x^2}{2}}{18}\right |_0^3 = \dfrac{12-\frac 9 2}{18} = \dfrac{5}{12}\)

in (b) is it really supposed to be \(\dfrac 1 \alpha + \dfrac 1 \beta\)

or does it mean that \(\dfrac 1 \alpha, ~\dfrac 1 \beta,~\dfrac 1 \gamma\) are all roots?

\(\text{The way to look at this semi formally is to consider the different cases}\\ \text{There can be 0-2 pairs of letters chosen}\\ \text{with 0 pairs there are }\dbinom{7}{4}4! =840 \text{ unique 4 letter words}\\ \text{with 1 pair there are }3\dbinom{4}{2}\dbinom{6}{2}2!=540 \text{ unique 4 letter words}\\ \text{This last bit deserves some discussion}\\ \text{The 3 comes from the fact that 3 of the letters can be chosen as a pair, I, N, V}\\ \text{The }\dbinom{4}{2} \text{ comes from the fact that we select two slots from the 4 to place this pair}\\ \text{The }\dbinom{6}{2} \text{ comes from the fact that we can select 2 letters from 6 to complete the word}\\ \text{The 6 comes from removing the numbers that are the pair, and the }\\ \text{2nd copies of the two other numbers}\\ \text{Finally the }2! \text{ comes from the fact the these last two letters are distinct and their order matters}\)

\(\text{Finally with 2 pairs there are }\dbinom{3}{2}\dbinom{4}{2} =18 \\ \text{This is choosing two pairs from the 3 available and choosing 2 slots from the 4}\\ \text{adding all these up we get}\\ 840+540+18 = 1398 \)

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