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How much progress have you made on any of those questions yourself?

How do you know? And is the answer's source a reliable one?

Are you sure the answer isn't 90 seconds? davidgoes4wce said he had a source where the answer was 90 seconds. Anyway, why would they tell us about the tap closing off at the end if we weren't meant to use that (if we do use that, we get 90 seconds)?

$\noindent a) We have$ $$\begin{align*}F_{Y} \left(y\right) &= \mathbb{P}\left(Y \leq y\right) \\ &= \mathbb{P}\left(\phi\left(X\right) \leq y\right) \\ &= \mathbb{P}\left(X \leq \phi ^{-1} \left(y\right) \right) \quad (\text{as }\phi \text{ is increasing}) \\ &= F_{X}\left( \phi ^{-1}...

$\noindent Note that the CDF of an $\mathrm{Exp}\left(\lambda\right)$ random variable is $F_{T} (t) = 1-e^{-\lambda t}$, for $t>0$. So the complementary CDF is $\overline{F}_{T}\left(t\right)=e^{-\lambda t}$ (probability of $T$ being greater than a given $t$).$ I can't see the initial Q., but...

The time between Poisson events follows an exponential distribution. $\noindent If $X \sim \mathrm{Poisson}\left(\lambda\right)$, and $T$ is the time between two (consecutive) of these Poisson events, then $T\sim \mathrm{Exponential}\left(\lambda\right)$, with pdf $f_{T} \left(t\right) =...

Re: MATH1131 help thread Subtracting row 4 from row 5, the fifth row in the right-hand column becomes $$\begin{align*}r_5 - r_1 - \left(r_4 + r_2 - 2r_1 \right) &= r_5 - r_1 - r_4 -r_2 + 2r_1 \\ &= r_5 - r_4 - r_2 + r_1,\end{align*}$$ as they have.

Re: Multivariable Calculus I think you lost a factor of 2 in your second line in the part where you're finding a second derivative (the 2 in front of the ∂z/∂y), but the method is right. (Or maybe you typo'ed it and left it out but did the remaining parts right, haven't checked it too closely.)

Re: Multivariable Calculus I used subscript numerals for the notation because otherwise it'd be more confusing due to so many variables being present.

Here's what I get. $\noindent We are given diameter $d=2\text{ mm}$, which in metres is $d=2\times 10^{-3} \text{ m}$. Hence the area of the cross-section is $A = \frac{\pi d^2}{4}$. We are given the length to be $\ell = 0.3 \text{ m}$. Also, the resistivity is $\rho = 1.7 \times 10^{-10} \...

Re: Multivariable Calculus $\noindent So basically, let $z=f(u,v)$, where $u=f(x,y)$, and $v=f(x,y)$. Then we want to find $\frac{\partial z}{\partial x}$. Note that $\frac{\partial u}{\partial x}=f_{1}(x,y)=\frac{\partial v}{\partial x}$. So we have$ $$\begin{align*}\frac{\partial...

Re: Multivariable Calculus $\noindent Need to do it with two variables $u$ and $v$, and make them both equal $f(x,y)$. This is because $f$ may depend differently on its first and second variables, so its partial derivatives won't be the same necessarily.$

Re: Multivariable Calculus Yeah that's correct.

Re: Multivariable Calculus $\noindent The difference is that$f_{2}$ is the partial derivative of the function wrt its \textsl{second} designated variable ($v$ here), whereas $f_{1}$ is that wrt the first variable. The reason that term is 0 though is because $\frac{\partial v}{\partial x}=0$.$

Re: Multivariable Calculus $\noindent Consider $z=f(u,v)$, where $u=3x$, $v=2y$. Then from the chain rule, $\frac{\partial z}{\partial x}= f_{1}\left(u,v\right) \frac{\partial u}{\partial x}+ f_{2} \left(u,v\right) \frac{\partial v}{\partial x} = f_{1}\left(u,v\right) \times 3 + 0 \Rightarrow...

Re: Multivariable Calculus $\noindent The total differential approximation tells us that an estimate for the max. error will be $\left |\Delta z \right |_{\text{max.}} \approx \left |\frac{\partial f}{\partial x} \right |\left |\Delta x\right | + \left |\frac{\partial f}{\partial y} \right |...

$\noindent At the point $x=-7.4$, you can solve to find the values of $y$ from the circle's equation (note it is a cirlce). Then note that at any point $(x,y)$ on the circle (with a defined slope of normal), the normal is just the line through the origin to the point $(x,y)$, which has slope...

Why does that mean they are disadvantaged?

A justification like that for why there's no solution would require us to put the augmented matrix into row-echelon form first, and then inspect it. Another way to see there's no solution is to note that if there is a solution, the coefficient t needs to be -1/2, in order for the first entries...

$\noindent There's an easier way to find the area of a triangle in $\mathbb{R}^{3}$: take the length of the cross product of two adjacent vectors making up the triangle, and then halve it (without halving, it's the area of the \textsl{parallelogram} those vectors span rather than the triangle).$...