Improve Your Algebra Skills with Practice Problems

How to express the distribution of measurements ($X_m\sim <!--\; \sim \; -->{\mathcal{D}}_{\mathrm{u}\mathrm{n}\mathrm{k}\mathrm{n}\mathrm{o}\mathrm{w}\mathrm{n}}$) given that the underlying physical process being measured follows a Gaussian distribution ($X_p\sim <!--\; \sim \; -->\mathcal{N}({\mathrm{\mu <!--\; \mu \; -->}}_p,{\mathrm{\sigma <!--\; \sigma \; -->}}_p)$) and the measurement uncertainty follows a Gaussian distribution ($\mathcal{N}({\mathrm{\mu <!--\; \mu \; -->}}_u,{\mathrm{\sigma <!--\; \sigma \; -->}}_u)$) too?
Note: ${\mathrm{\mu <!--\; \mu \; -->}}_u$ here is variable and takes the value of the specific instance/sample $X_p$.
Update:
Can one start in terms of PDFs as follows?
Given the PDF of the physical process:
$f_{X_p}(x)=\frac{1}{{\mathrm{\sigma <!--\; \sigma \; -->}}_p\sqrt{2\mathrm{\pi <!--\; \pi \; -->}}}{e}^{-<!--\; -\; -->\frac{1}{2}{\left(\frac{x-<!--\; -\; -->{\mathrm{\mu <!--\; \mu \; -->}}_p}{{\mathrm{\sigma <!--\; \sigma \; -->}}_p}\right)}^2}$
In order to express the PDF of the measurements affected by measurement error ($\mathcal{N}(X_p,{\mathrm{\sigma <!--\; \sigma \; -->}}_u)$), the idea is to use an specific sample t of the physical process as the mean of the measurement uncertainty distribution. Since the specific sample t follows a PDF $f_{X_p}(t)$, therefore integration is used over the whole range of t, and weighted by the PDF:
$f_{X_m}(x)={\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{\mathrm{\infty <!--\; \infty \; -->}}\frac{1}{{\mathrm{\sigma <!--\; \sigma \; -->}}_u\sqrt{2\mathrm{\pi <!--\; \pi \; -->}}}{e}^{-<!--\; -\; -->\frac{1}{2}{\left(\frac{x-<!--\; -\; -->t}{{\mathrm{\sigma <!--\; \sigma \; -->}}_u}\right)}^2}{f}_{X_p}(t)dt$
Is the above formulation reasonable?
Update after David K's answer: This question can also serve to get a more intuitive understanding of the convolution of two normal distributions.

I have 2 position estimates (along with their measurement error) and a difference in time between estimates. I estimate velocity using

Velocity = (PosA - PosB)/DeltaT

I am trying to estimate the error in my velocity estimate, but I can't seem to find any ways to calculate this. I assume it has to use Sigma_PosA and Sigma_PosB. I would also assume it's relative to DeltaT and/or abs(PosA - PosB). What is the velocity measurement variance/standard deviation?

I know that we can find the first weak derivative for $f(x)=|<!--\; |\; -->x|<!--\; |\; -->$, which is
$w(x)=\{\begin{array}{ll}-<!--\; -\; -->1& x<0\\ 0& x=0\\ 1& x>0\end{array}.$
But why is there no second weak derivative? If I try to find the first weak derivative of $w(x)$, I get that
${\int <!--\; \int \; -->}_{\mathbb{R}}v(x)\mathrm{\psi <!--\; \psi \; -->}(x)dx=-<!--\; -\; -->{\int <!--\; \int \; -->}_{\mathbb{R}}w(x){\mathrm{\psi <!--\; \psi \; -->}}^{\prime }(x)dx=2\mathrm{\psi <!--\; \psi \; -->}(0).$
What does this mean? Why is there no v such that $Dw=v$? I feel like I'm missing something and I'm not sure how to interpret this result.

How do you give one value that makes the inequality ${\displaystyle \frac{p}{8}-2>6}$

If $-<!--\; -\; -->\frac{1}{5}$ of a number n is at most 21, what are the possible values of n? How do you write and solve an inequality to find n?

Interpolation and approximation from Numerical Methods
I was reading about interpolation and approximation in Numerical Methods and came across this statement in my course material, "for n data points, there is one and only one polynomial of order (n − 1) that passes through all the points" for example, we have 3 data points on a straight line then how can a second order polynomial satisfy it?

How do you divide 0.12 by 8?

Consider sequences of random variables $(X_n)$,$(Y_n)$ with $|X_n|<Y_n$. Suppose $(Y_n)$ converges in probability to $Y$, and $E{Y}_n\to <!--\; \to \; -->EY$. I have to show that, if $(X_n)$ converges to $X$ in probability, $E{X}_n\to <!--\; \to \; -->EX$. I have no idea of where to start. I have seen a question here in stackexchange that answers this for the case where $(Y_n)$ is constant, and would like some tips on how to do this. Can anyone help me?

Struggling to reach answer with algebraic fractions
The question is
$\frac{2x^3-<!--\; -\; -->x^{-<!--\; -\; -->3}}{x^3}$
and the answer is
$2-<!--\; -\; -->x^{-<!--\; -\; -->6}$
I have been trying but cannot get to the correct answer using the methods I know. If anyone can explain to me how to do it I would be extremely grateful.

Fraction Transforms
Here's a number theory problem I'm having some difficulty with:
Say we transform a fraction by the following rule: we start with some fraction $\frac{m}{n}$ with $m>n$ and then convert it to $\frac{fr}{n}$, where $f=\lfloor <!--\; \lfloor \; -->m/n\rfloor <!--\; \rfloor \; -->$ and $r=mmodn$. We repeat the process until our fraction is less than 1 or an integer. Given an n is there an m so that we end up with a fraction less than 1?
For $n=2$ it isn't that hard, 3, 7, 15, 31, etc. work, but what about other values of $n$?

How do you solve ${\displaystyle w–3<-9}$

How can the GCF be used to write a fraction?

Inequality $\frac{3}{\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}}\ge <!--\; \ge \; -->1+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ for positive $a$
MyTry::We can write $A$ and $B$ as
$=\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=\frac{3}{A}$
and
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{B}$
Using cauchyschwarz Inequality
$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge <!--\; \ge \; -->\frac{3^2}{1+a+1+b+1+c}=\frac{9}{3+a+b+c}$
Now How can i solve after that , Help Required, Thanks

Independent and Identically distributed, conditional independent and Naive bayes
I'm reading about Naive Bayes classification concept, noting that we make the conditionally independence assumption. But isn't this the general assumption that is always made dealing with machine learning algorithms?
Suppose we have a supervised binary classification problem setup, with a dataset $\mathcal{D}=\{(x_1,t_1),\dots <!--\; \dots \; -->,(x_n,t_n)\}$(xn,tn)} where $x_i\in <!--\; \in \; -->{\mathbb{R}}^D$ and $t_i\in <!--\; \in \; -->\{0,1\}\mathrm{\forall <!--\; \forall \; -->}i=1,\dots <!--\; \dots \; -->n$
I've read everywhere that we always make the assumption that data are iid (independent and with the same probability dstribution, this would mean that $p((x_i,t_i),(x_j,t_j))=p((x_i,t_i))p((x_j,t_j))$..right?). At this point it is reasonable to think of a Bernoulli distribution to model the data. Let $p(\mathcal{D}|\mathrm{\theta <!--\; \theta \; -->})$ the likelihood function: then we want to find
$\stackrel{^<!--\; ^\; -->}{\mathrm{\theta <!--\; \theta \; -->}}=arg\underset{\mathrm{\theta <!--\; \theta \; -->}}{max}p(\mathcal{D}|\mathrm{\theta <!--\; \theta \; -->})$
where
$p(\mathcal{D}|\mathrm{\theta <!--\; \theta \; -->})=p((x_1,t_1),\dots <!--\; \dots \; -->,(x_n,t_n)|\mathrm{\theta <!--\; \theta \; -->})=p((x_1,t_1)|\mathrm{\theta <!--\; \theta \; -->}),\dots <!--\; \dots \; -->,p((x_n,t_n)|\mathrm{\theta <!--\; \theta \; -->})$
Here we should use a conditional independence hypothesis in order to go on. So in every situation we use the naive bayes hypothesis? I'm having troubles trying to distinguish..

Solving for x [not homework]
How do I bring the remaining x to the LHS?
$\pm <!--\; \pm \; -->x=\frac{(2-<!--\; -\; -->x)\sqrt{|q_2|}}{\sqrt{|q_1|}}$
to get
$x=\frac{2\sqrt{|q_2|}}{\sqrt{|q_2|}\pm <!--\; \pm \; -->\sqrt{|q_1|}}$
I'm just not sure about the background operations to isolate the x.

My book says that ratio levels of measurement is the highest form of measurement and adheres to the same rules as as interval level measurement (distances between intervals of the scale are numerically equal), but it can have an absent property.
The examples are weight, height, blood pressure, pulse, etc.
How can weight or height be a ratio measurement? What thing being measured for weight has not weight? Or has no height? I understand no blood pressure or pulse.
Thanks :)

Here is an excercise of measure theory and real analysis. can you help me with it?
Let (X,M,$\mathrm{\mu <!--\; \mu \; -->}$) be a measure space and $f:X\to <!--\; \to \; -->[0,\mathrm{\infty <!--\; \infty \; -->}]$ a measurable function. Prove that
${\int <!--\; \int \; -->}_{X}fd\mathrm{\mu <!--\; \mu \; -->}>0\iff <!--\; \iff \; -->\mathrm{\mu <!--\; \mu \; -->}(\{x\in <!--\; \in \; -->X\mid <!--\; \mid \; -->f(x)>0\})>0.$
I tried to use the definition of the integral by the approximation theorem of measurable functions by simple functions, but I haven't been able to come to the conclusion.

How do you solve ${\displaystyle x-7<1}$ ?

Is there a positive integer n such that the fraction $(9n+5)/(10n+3)$is not in the lowest term?
I found $n=2$ be a solution. Is it correct?

Derivative of $f(x)=\frac{\sqrt{x^2-<!--\; -\; -->1}}{x}$
I have the function following:
$f(x)=\frac{\sqrt{x^2-<!--\; -\; -->1}}{x}$
And here is what I did:
$f(x)=\frac{\sqrt{x^2-<!--\; -\; -->1}}{x}$
$=\frac{(x^2-<!--\; -\; -->1{)}^{\frac{1}{2}}}{x}$
$f^{\prime }(x)=\frac{x\cdot <!--\; \cdot \; -->\frac{1}{2}(x^2-<!--\; -\; -->1{)}^{-<!--\; -\; -->\frac{1}{2}}(2x)-<!--\; -\; -->(x^2-<!--\; -\; -->1{)}^{\frac{1}{2}}}{x^2}$
$=\frac{x^2(x^2-<!--\; -\; -->1{)}^{-<!--\; -\; -->\frac{1}{2}}-<!--\; -\; -->(x^2-<!--\; -\; -->1{)}^{\frac{1}{2}}}{x^2}$
And I'm pretty sure that this is wrong, and the answer book says it isn't either.
I think I messed up somewhere, or didn't do it properly.
I tried using the quotient rule. Do I need to make it into an exponent, and solve it as a chainrule?
Please help me find the steps and answer to this question. Thank you.
Or is there any other steps so it can match this?:
$\frac{1}{x^2\sqrt{x^2-<!--\; -\; -->1}}$