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Note that P_2 has dimension 3 and P_3 has dimension 4, so the matrix should be a 4x3 one.

Re: Discrete Maths Sem 2 2016 ('Subject to' will often be seen in areas like constrained optimisation, for example.)

Re: Discrete Maths Sem 2 2016 Only problem is sometimes s.t. could also mean "subject to". Best to just write the full thing if you want/need to avoid ambiguity (from the context it'll often be clear though which was meant).

n = 2 is generally considered very small.

Ah right. Were they for extremely large values, as Drsoccerball mentioned? In these cases, the continuity correction would make little difference.

Reason is due to which part of the graph you miss out on by using the continuous approximation. If X* is the continuous approximtion r.v. to X, then clearly, to approximate P(X > a), we need to use the continuity correction P(X* > a – 0.5), whereas for P(X < a), it'd be P(X* < a + 0.5). Don't...

There's nothing really universal, these are mainly rules of thumb. If in doubt, I guess just use the continuity correction (with "1/2" as the epsilon). The main thing is to make sure you do the correction the right way. The thing you originally wrote (P(X < 12) being approximated with the...

$\noindent This is called a \textbf{continuity correction}. If we're using a normal distribution to approximately calculate a discrete distribution probability, if we wanted $\mathbb{P}\left( X < 12\right)$ and $X$ is discrete, treating $X$ as continuous and doing it for $X < 12$ makes us miss...

Re: MATH1231/1241/1251 SOS Thread Not really, because to conclude just from T(something) = 0 that something = 0 (which is what you seemed to want to do), we'd have to know T was one-to-one, but T can be any linear map (and the result holds regardless). You'd end up having to do it by...

Re: MATH1231/1241/1251 SOS Thread He was more asking about the other direction, i.e. whether he could conclude from T(something) = 0 that something = 0. This is guaranteed if and only if the linear map T is a one-to-one map. If T is not one-to-one, it'll have a non-trivial kernel.

Re: MATH1231/1241/1251 SOS Thread $\noindent We'll show the contrapositive, i.e. that if $S$ is dependent, then $R$ is dependent (we'll do it for arbitrary $n$ elements, since it works the same).$ $\noindent Suppose $S$ is dependent, then there exist scalars $\alpha_{1},\ldots...

Probably for HSC Science purposes, it's safest to try doing a smooth curve of best fit.

The Marking Guidelines are dodgy. Plus, "line of best fit" is really supposed to refer to a straight line too (not sure whether HSC Science adheres to this though). What the Sample Answer did was some sort of curve of best fit. I guess just memorise what the sample answers do then (try searching...

Re: MATH1231/1241/1251 SOS Thread $\noindent As I said before, those answers are equivalent. Multiplying your answer through by $3$ makes it $3x^{2}y + y^{3} = 3C$. But since $C$ was an arbitrary constant, $3C$ is just as arbitrary, so we can just replace it with $A$ to signify an arbitrary...

Here's the link to the 2015 HSC Chemistry Exam Marking Guidelines to save anyone the trouble of having to search it up, for those who want to see it: https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/guides/2015-hsc-mg-chemistry.pdf (Question 22(a)'s guidelines are on page 3 of that...

$\noindent What do you mean? By \emph{definition}, $T(\vec{v}) = \lambda \vec{v}$, where $\lambda$ is the eigenvalue for eigenvector $\vec{v}$ (this is precisely what it means for a non-zero vector $\vec{v}$ to be an eigenvector corresponding to eigenvalue $\lambda$). If you meant something...

Re: MATH1231/1241/1251 SOS Thread $\noindent I realised you also mis-integrated the $f_{x}(x,y) = 2xy$. Partially integrating this wrt $x$ yields $f(x,y) = x^{2}y + h(y)$ (you did $xy^{2}$ instead of $x^2 y$).$

Re: MATH1231/1241/1251 SOS Thread $\noindent For your second attempt at the question, you equated the $f_{y} (x,y) = 2xy + h'(y)$ to $f_{x}(x,y)$ instead of to $f_{y}(x,y)$, which is why it yielded the wrong answer.$

Re: MATH1231/1241/1251 SOS Thread Your answer of "x^2 y + y^3/3 = C" is the same as the given answer (just multiply yours through by 3 and you get theirs, noting that 3C can just be written as an arbitrary constant itself, like A).

$\noindent If $T\sim \mathrm{Exp}(\lambda)$, the PDF is $f(t) = \lambda e^{-\lambda t}, t \geq 0$. So $\mathbb{E}\left[T^2\right] = \int _0 ^\infty t^2 \cdot \lambda e^{-\lambda t}\,\mathrm{d}t$.$