we want 1 {4}, so the possibility of one round is $\frac{1}{6}$, we want 2 ${2}$, so the possibility is $\frac{1}{6} \cdot \frac{1}{6}$ and at the remaining rounds the probability is $\frac{4}{6}$ since we allow any number except than 2,4... am I right? @Semiclassical
Is the probability to get two {2} and one {4} when throwing a dice equal to $\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{4}{6} \cdot \frac{4}{6} \cdot \frac{4}{6}$ ?
If a rent a car company has 700 cars and the average per rental is 9 days, then we muliply $2.3$ by $700$ in order to find the expected number of bookings per month.
So the expected number of bookings is equal to $700 \cdot \frac{700}{9 \cdot 30} $, right ?
Why does this happen? Could you explain to me the formula?
Let $f: \mathbb{R} to \mathbb{R}$ with the properties: $f(x)>0, \forall x \geq 0$, $f$ is decreasing and $f'(0)=0$. I want to prove that $f''(x)=0$ for some $x>0$.
So we choose $\epsilon'=\epsilon |a_2-a_1|$ and and we have that $\forall \epsilon'>0$ $\exists N:=m_0$ so that when $n, m \geq N$ then $|a_n-a_m| \leq \epsilon'$.
And thuis holds since $\epsilon$ is arbitrary and so we can get any number , right? @Axoren
@Axoren Using the definition, we have from $\lim_{m \to +\infty} \frac{\theta^{m-1}}{1-\theta}=0$ that $\forall \epsilon>0$, $\exists m_0 \in \mathbb{N}$ such that $\forall m \geq m_0$, $\left| \frac{\theta^{m-1}}{1-\theta}\right|< \epsilon$.
How do we show that it's Cauchy? $(x_n)$ is Cauchy if $\forall \epsilon>0$, $\exists N$ so that when $n,m \geq N$ we have that $|a_n-a_m|\leq \epsilon$.
In our case, we have a specific number, say b, such that $|a_n-a_m| \leq b$... @AkivaWeinberger
