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Re: Statistics The parameters are as follows: • N is the total population • K is the number of "tagged" objects (defective objects in your example) • n is the size of our sample. The hypergeometric distribution pmf Drongoski wrote (in terms of the parameters N, K, n) then gives the...

$\noindent Try with $a = - \frac{1}{2} - \frac{1}{2}i$ and $R = \frac{1}{\sqrt{2}}$.$

What progress have you made? Part (a) will fall out pretty much as a consequence of the definition of sup (note that a is an upper bound of E (and since R has the least-upper bound property, sup E exists in R).) For part (b), it's essentially because sup E may turn out to actually equal a...

$\noindent It is clear from a sketch of the graph of $y = |x|$ that the area under the curve between $-2$ and $2$ is $4$ (it's the area of a square with side length $2$).$ $\noindent Thus $\int_{-2}^{2}|x|\, \mathrm{d}x = 4$.$

Re: MATH2601 Linear Algebra/Group Theory Questions That can only happen if a = e (the identity). Take a to be any other element in G (there must be at least one other element since G has prime order, which implies |G| is at least 2), and the result will follow (since |H| won't be able to be 1).

Re: Several Variable Calculus You can try first computing an expression for ∂f/∂y (using first principles to find the value of this at the origin). Then try computing ∂2f/∂x∂y (i.e. ∂/∂x (∂f/∂y)) at the origin from first principles and show that the limit does not exist.

Re: MATH2601 Linear Algebra/Group Theory Questions Well H has the same number of elements as G and is a subset of G, so H = G. (If S is a set that has only a finite number of elements, then the only subset of S with the same number of elements as S is S itself.)

Re: HSC 2017 MX2 Marathon $\noindent It's of the form $x = A\sec \theta, y = B\tan \theta$, which as we know is a parametric representation of the hyperbola $\frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1$.$

Re: MATH2601 Linear Algebra/Group Theory Questions $\noindent When you went from $2\mathbf{v} = \mathbf{0}$ to concluding that $\mathbf{v} = \mathbf{0}$, you were essentially assuming what had to be proved (i.e. that if $\alpha \mathbf{v} = \mathbf{0}$ and $\alpha$ is a non-zero scalar, then...

Re: MATH2601 Linear Algebra/Group Theory Questions $\noindent All you need to do to prove the claim is assume $\alpha \mathbf{v} = \mathbf{0}$ and show that if $\alpha \neq 0$, then $\mathbf{v} = \mathbf{0}$. To do this, simply multiply both sides of the assumption ($\alpha \mathbf{v} =...

Re: MATH2601 Linear Algebra/Group Theory Questions Since V is finite dimensional, by definition V has a finite spanning set S, and this set S also spans W, so W is also finite dimensional.

It's from within the HSC maths syllabus (3U Projectile motion probably).

Re: Several Variable Calculus $\noindent Let $\mathbf{a} \in \mathbb{R}^{n}$ be fixed and arbitrary. By definition of continuity, RTP:$ $for all $\varepsilon > 0$ there exists an open ball $B\left(\mathbf{a},\delta\right)$ (i.e. some radius $\delta > 0$) such that...

$\noindent Let the initial speed be $V$ and angle of launch $\theta$ $\Big{(}$which is an angle between $0^{\circ}$ and $90^{\circ}$, i.e. between $0$ and $\frac{\pi}{2}$ in radians$\Big{)}$. Assume the launch occurs from the origin (where $x = 0$ and $y = 0$) as usual. By symmetry of parabolic...

Re: HSC 2017 MX2 Marathon $\noindent Note that $x$ must be non-real, since $x+\frac{1}{x}$ is always at least $2$ in absolute value for real $x$. Try substituting $x = \mathrm{cis}\left(\theta\right)$. (If you initially substitute the more general $x = r\,\mathrm{cis}\left(\theta\right)$...

Re: ATAR Notes vs Bored of Studies Why are they doing that?

$\noindent Note that it is equivalent to show that $\frac{B}{G}$ cannot exceed $\frac{4}{3}$, where $B$ is the total number of boys in the class and $G$ is that for girls. (It is clear from the hypotheses that $G > 0$). Suppose $g$ girls go on trip $1$ and $h$ go on trip $2$, and wlog $g \leq...

Even if there's just 1 boy and 9 girls say, the hypotheses can be satisfied but the conclusion not. In general if there's an upper bound on the proportion of boys in each trip, we could just have arbitrarily small proportion of boys in each trip and get an overall arbitrarily small proportion of...

A class with for example no boys and only girls would satisfy the hypotheses of the question but not the conclusion.

Yeah, you can. But better to just factor k+1 from the LHS and go from there (saves you from having to expand so much).