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Mar
3
comment Central Limit theorem Clarification.
We have $X_i$ is the result of the $i$th observation. In particular, each $X_i$ is a random variable, and therefore so is $(X_1+\cdots +X_n)/n$. If you do the experiment $n$ times, then do it again and again, the values of $(X_1+\cdots +X_n)/n$ will vary. We are calculating the mean and variance of this random variable.
Mar
3
answered Poisson Statistics
Mar
3
comment Differentiate the function
If I read the thing right, you are looking at $(x^{1/2}+x^{-1/3})^2$. You could expand and differentiate term by term, or use the Chain Rule.
Mar
3
comment Limit when x tends to 1
Yes. After the simplification we have $\frac{1+2y}{6(1+y)(1+y+y^2)}$.
Mar
3
answered Different arrangements - Permutation
Mar
3
comment Different arrangements - Permutation
We are looking at the $12$-letter words made up of the letters t,t,t,c,c,c,c,a,a,a,a,a.
Mar
3
comment Cards Probability and Combination
You are welcome.
Mar
3
answered Limit when x tends to 1
Mar
3
comment Cards Probability and Combination
For b) it is $(13)(12)(13)(13)$. You have a) right.
Mar
3
comment Cards Probability and Combination
a) The first can be any of $13$ cards. For each choice of first card, there are $13$ choices for second card (and so on). b) This first can be any of $13$ cards. For each choice of first card, there are $12$ choices for second card. And for each choice of first and second, there are $13$ choices for the third card. And so on.
Mar
2
comment Proving density of $\mathbb{Q}^n$ in $\mathbb{R}^n$
To get a rational point $(r_1,\dots,r_n)$ within $\epsilon$ of $(x_1,dots,x_n)$ it is enough to make $|x_i-r_i|\lt \frac{\epsilon}{\sqrt{n}}$ for all $i$.
Mar
2
comment Proving that $a^m \equiv 1 \bmod{n}$ if and only if $k\mid m$ for $k=\operatorname{ord}_n(a)$
If $a^m\equiv 1\pmod{n}$, it does imply that $k$ divides $m$. For from $a^m\equiv 1\pmod{n}$, we conclude that $1\equiv a^m=(a^k)^qa^r\pmod{n}$. But $(a^k)^q\equiv 1\pmod{n}$, and therefore $a^r\equiv 1\pmod{n}$. If $r\ne 0$, that means there is a positive integer $w\lt k$, namely $r$, such that $a^w\equiv 1\pmod{n}$. That contradicts the fact that $k$ is the order of $a$. the only way out is to have $r=0$, meaning that the remainder when you divide $m$ by $k$ is $0$, which means $k$ divides $m$.
Mar
2
comment Proving that $a^m \equiv 1 \bmod{n}$ if and only if $k\mid m$ for $k=\operatorname{ord}_n(a)$
We are told that $k$ is the order of $a$ (modulo $n$). That means, by definition, that $k$ is the smallest positive integer such that $a^k\equiv 1\pmod{n}$. That's why we can write $a^k\equiv 1\pmod{n}$, and therefore $(a^k)^q\equiv 1^q\equiv 1\pmod{n}$.
Mar
2
comment Cube root equations
It is probably pretty hopeless. Change your $8$ to a $7$. With natural manipulations we end up with a polynomial equation of degree $\gt 4$, with no symmetries to bring down the degree.
Mar
2
comment Proving that $a^m \equiv 1 \bmod{n}$ if and only if $k\mid m$ for $k=\operatorname{ord}_n(a)$
It is an if and only if question. For the easy direction, if $m=kq$, then $a^m=(a^k)^q\equiv 1^q=1$. For the harder direction, if $a^m\equiv 1$, then since $a^{kq}\equiv 1$, we get from my answer that $a^r\equiv 1$. since $k\lt r$, this contradicts the fact that $k$ is the order of $a$, unless $r=0$.
Mar
2
comment Probability question (grid)
There are $\binom{70}{20}$ ways to choose squares to fill. There are $\binom{7}{1}\binom{60}{10}-\binom{7}{2}$ ways that we can have (at least) one entire row of filled squares. So the required probability is $\dfrac{\binom{7}{1}\binom{60}{10}-\binom{7}{2}}{\binom{70}{20}}$.
Mar
2
answered Proving that $a^m \equiv 1 \bmod{n}$ if and only if $k\mid m$ for $k=\operatorname{ord}_n(a)$
Mar
2
answered Why doesn't the telescopic series $\sum_{n\geq1}\frac1{n(n+1)}$ diverge? It is a sum of harmonic series!
Mar
2
comment Number that is a certain percent between two other numbers, without subtraction
My answer said nothing about lesser or greater. We do not need to know that. All we need to know is our source point $a$, our target point $b$, and the proportion. If for example you want to know the point $c$ whose distance from $10$ is $25\%$ of its distance from $2$, you can choose $a=10$, $b=2$, $t=0.25$, and you get $(0.75)(10)+(0.25)(2)$.
Mar
2
answered Number that is a certain percent between two other numbers, without subtraction