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 22h comment Why does this integral not contradict Fubini's Theorem? By computing the integral of its absolute value. 22h comment Convergence of $\dot{x}(t) = -\alpha(t)x(t) + b\mathrm{e}^{-\lambda t}$ $$\int_0^t \mathrm{e}^{-A(t-\tau)} e^{-\lambda \tau} \mathrm{d}\tau=e^{-\lambda t}\int_0^t g(\tau) \mathrm{d}\tau\qquad g(s)=\mathrm{e}^{-A(s)} e^{\lambda s}$$ 23h comment Convergence of $\dot{x}(t) = -\alpha(t)x(t) + b\mathrm{e}^{-\lambda t}$ And now apply this lemma to your setting... 23h comment Evaluate the iterated integral @Omnomnomnom The function is Riemann integrable. Quickest route: for every $h>0$, $\{f\geqslant h\}$ is finite hence the upper Darboux sums can be made as small as one wishes. Since the lower Darboux sums are all zero, we are done. 1d comment Uses of step functions @IrregularUser Yes, this could even make for an exciting motto: "I am not a paragraph, I am a free man!" Oh wait, was this already used? :-) 1d comment Convergence of $\dot{x}(t) = -\alpha(t)x(t) + b\mathrm{e}^{-\lambda t}$ Hint: Can you show that if $g(t)e^{-\lambda t}\to0$ when $t\to\infty$ then $$e^{-\lambda t}\int_0^tg(s)ds\to0\ ?$$ 1d comment Generating a random variable from a uniform random variable This is a case where lack of context (in addition of going against the rules of the site) really prevents to give an appropriate answer. The "inverse CDF method" (explained by some users below) is at the same time mathematically correct and quite impractical. Other ways to simulate gaussians from uniforms exist, which are actuallly used routinely in practice, but are they what the OP is interested in, one cannot know. 1d comment Finding expectation and variance of poisson random variable. Sorry but "A poisson distributed random variable X, where X>0" is absurd since every Poisson random variable X is such that P(X=0) is nonzero. 1d comment Uses of step functions @Cathe1974 Unfortunately: aberdeenmathstutor.co.uk 1d comment Investigate convergence of $\sum_{n=1}^\infty \frac{\ln(n)}{n}$ @Dark $n>2$, actually. 1d comment What is infinity to the zeroth power? The current version is much better. One remaining problem is that you use "concept" in a bizarre way (and one which would make every philosopher cringe...). 1d comment How many students would have to take the exam to ensure with probability at least $.9$ that the class average would be within $5$ of $75$? "the distribution for the average of $n$ students is approximately normal with mean $\mu$ and variance $\frac{\sigma^2}{n}$. " Again, why is that? For which $n$, with an error of size what? And what does that mean to be approximately normal with a variance depending on $n$? 1d comment How many students would have to take the exam to ensure with probability at least $.9$ that the class average would be within $5$ of $75$? Let me follow you and let us assume this is not a maths question. Then, since it falsely masquerades as one, our urgent task as mathematicians should be to explain why this is not mathematics, instead of supporting the belief that it is. No? 1d comment How many students would have to take the exam to ensure with probability at least $.9$ that the class average would be within $5$ of $75$? "since $\frac{\bar{X} - \mu}{\sqrt{\frac{\sigma^2}{n}}}$ is approximated to be standard normal" Approximated how? This is the whole point, which makes the question a terrible exercise. 1d comment How many students would have to take the exam to ensure with probability at least $.9$ that the class average would be within $5$ of $75$? This exercise is terrible. To the attention of non probabilist readers, let us mention that it is the logical equivalent of the following: Consider some real valued sequence $(x_n)$ such that $nx_n\to1$, find some $n$ such that $x_k\leqslant\frac1{10}$ for every $k\geqslant n$. Scary, no? 1d comment How many students would have to take the exam to ensure with probability at least $.9$ that the class average would be within $5$ of $75$? "since $\frac{\bar{X} - \mu}{\sqrt{\frac{\sigma^2}{n}}}$ is standard normal" IS IT? In fact, nothing in the hypotheses implies that. 1d comment What is infinity to the zeroth power? After the (good) initial sentence, I was expecting to see the argument that $x^y$ has no limit when $x\to\infty$ and $y\to0$, unfortunately, next, we are summoned to believe that the "better" explanation of the limit is as the limit of $x^{1/x}$ (an assertion with which one can disagree). 1d comment behavior of the Linear system of an ODE model Apparently, not enough to even react. 1d comment What is the determinant of []? "I can't und(er)stand why it returns the determinant zero" vs "ans = 1" ?? 1d comment How to solve for the expectation of the Ito Integral: $\int_0^4 B_t^2 dB_t$? This is a local martingale with integrable finite variation hence a martingale hence the expectation is... Re your argument, note that if $f(t,x)=\frac13x^3$ then $\frac{\partial f}{\partial t}=0$.