Rom

This problem is a bit tricky.  They are mixing mass and weight and hoping you'll notice.

The gravitational potential is given by \(U=m g h\)

where g is the acceleration due to Earth's gravity and is 9.8 m/s²

But you have to be careful in that \(weight=m g\)

So in calculating the potential here we have to sum up all the weights and multiply them by the height.

\(TotalWeight = 4 \times 1000 \times g + 6 \times 150 =\\ 4000(9.8) + 900 = 40100 n \\ \mbox{and so the potential is}\\ U=112\times 40100 = 4491200 J ~~~~~\mbox{(J is for Joules.)}\)

This isn't the greatest question in the world.

What they are trying to get at is that the number of trainees that are still employed is a binomial random variable, in this case with parameters \(n=6, p=0.2\)

You should know or can look up that the variance of a binomial random variable is given by

\(\sigma_{binomial} = n p (1-p) \\ \mbox{so here we have}\\ \sigma = 6 (0.2)(0.8) = 0.96\)

use the Gaussian elimination method

\(\begin{pmatrix}1 &0 &0 &| &1 &0 &0 \\ -1 &1 &0 &| &0 &1 &0 \\1 &1 &1 &| &0 &0 &1\end{pmatrix} \Rightarrow\)

\(\begin{pmatrix}1 &0 &0 &| &1 &0 &0 \\ 0 &1 &0 &| &1 &1 &0 \\ 0 &1 &1 &| &-1 &0 &1\end{pmatrix} \Rightarrow\)

\(\begin{pmatrix}1 &0 &0 &| &1 &0 &0 \\ 0 &1 &0 &| &1 &1 &0 \\0 &0 &1 &| &-2 &-1 &1 \end{pmatrix} \Rightarrow\)

and we read off the inverse as

\(\begin{pmatrix}1 &0 &0 \\ 1 &1 &0 \\ -2 &-1 &1\end{pmatrix} \)

as an alternative to the method heureka has used we can use Gaussian elimination and operate on an identity matrix as well.  At the end of the elimation the identity matrix will have been transformed into the inverse matrix.

\(\begin{pmatrix}0 &4 &| &1 &0\\ -6 & 5 &| &0 &1\end{pmatrix} \Rightarrow \\ \begin{pmatrix}-6 &5 &| &0 &1 \\ 0 &4 &| &1 &0\end{pmatrix} \Rightarrow \\ \begin{pmatrix}-6 &5 &| &0 &1 \\ 0 &1 &| &\frac 1 4 &0\end{pmatrix} \Rightarrow \\ \begin{pmatrix}-6 &0 &| &-\frac 5 4 &1 \\ 0 &1 &| &\frac 1 4 &0\end{pmatrix} \Rightarrow \\ \begin{pmatrix}1 &0 &| &\frac 5{24} &-\frac 1 6 \\ 0 &1 &| &\frac 1 4 &0\end{pmatrix} \Rightarrow \\\)

and we read off the inverse as

\(\begin{pmatrix}\frac 5{24} &-\frac 1 6 \\ \frac 1 4 &0\end{pmatrix}\)

\(dB = 20 \log\left(\dfrac {p_1}{p_2}\right) \\ \dfrac {dB}{20} = \log\left(\dfrac {p_1}{p_2}\right) \\ \dfrac {dB}{20} = \log(p_1)-\log(p_2) \\ \log(p_1) = \dfrac {dB}{20} + \log(p_2) \\ p_1 = 10^{\left( \frac {dB}{20} + \log(p_2)\right)}\)

\(5\dfrac 2 3 \times \dfrac 1 6 =\\ \left(5 + \dfrac 2 3\right) \times \dfrac 1 6 = \\ 5 \times \dfrac 1 6 + \dfrac 2 3 \times \dfrac 1 6 = \\ \dfrac 5 6 + \dfrac 1 9 = \dfrac {15}{18}+\dfrac {2}{18} = \dfrac {17}{18}\)

\(x(7-x) > 8 \\ 7x - x^2 -8 > 0 \\ x^2 -7x + 8 < 0 \\ \mbox{This isn't factorable so we complete the square.}\\ (x-3.5)^2 -(3.5)^2 + 8 < 0 \\ (x-3.5)^2 < 4.25 = \dfrac {17}{4} \\ -\dfrac{\sqrt{17}}{2} < x-3.5 < \dfrac {\sqrt{17}}{2} \\ \dfrac{7-\sqrt{17}}{2} < x < \dfrac {7 + \sqrt{17}}{2}\)

437 rounded to the nearest tenth is still 437.

437 rounded to the nearest 10 would be 440

\(\dfrac 2 3 = 0.666666 \dots = 0.\overline{6}\)

\(\mbox{We want to check factors of the constant term, 12, for possible roots.} \\ \mbox{1 and -1 are not roots.}\\ \mbox{2 is a root. So we can divide through by }(x-2)\\ \dfrac {2x^4-5x^3-11x^2+20x+12}{x-2}=2 x^3-x^2-13 x-6\\ \mbox{Now we look for roots of this polynomial from factors of 6.}\\ \mbox{Checking we find that -2 is a root so we can divide through by }(x+2)\\ \dfrac {2 x^3-x^2-13 x-6}{x+2}=2 x^2-5 x-3\)

\(\mbox{Again we look for roots from factors of the constant term 3.}\\ \mbox{We find 3 is a root so we can divide through by }(x-3)\\ \dfrac {2x^2-5x-3}{x-3}=2x+1 \\ \mbox{So the full factorization is }\\ 2 x^4-5 x^3-11 x^2+20 x+12 = (x+2)(x-2)(x-3)(2x+1)\)