Find the indicated expression of 1/2m inches in feet

Martin Hart
2022-10-22
Answered

Find the indicated expression of 1/2m inches in feet

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Carly Yang

Answered 2022-10-23
Author has **19** answers

$\frac{1}{2}m\text{inches}\times \frac{12\text{inches}}{1\text{feet}}=6m\text{feet}$

Your answer is 6m feet.

Your answer is 6m feet.

asked 2022-05-13

Given a sigma algebra $\mathcal{S}$ over some set $X$ which is generated by some set $S\subset \mathcal{P}(X)$, and a probability function $P:\mathcal{S}\to \mathbb{R}$ why is it sufficient to check the conditions: $P(A)\ge 0,P(X)=1,P({\cup}_{n}{A}_{n})=\sum _{n}P({A}_{n})$ for all $A,{A}_{n}\in S$ where $({A}_{n})$ is a disjoint sequence, to determine that $P$ is a valid probability? Intuitively it is obvious that if we check the conditions to be true on a generating set then we should have the conditions true for all the elements of $\mathcal{S}$, but how do we rigorously prove this fact?

For example suppose $X=\mathbb{R},\mathcal{S}=\mathcal{B}$ (the Borel sigma algebra) and $S=\{I:I{\textstyle \text{is an interval in}}\mathbb{R}\}$. Now it is easy to check that $P(I)=\frac{1}{\sqrt{2\pi}}{\int}_{{I}_{x}}{e}^{-{x}^{2}/2}dx$ satisfies the requisite conditions. How does it follow that the conditions are now met for $P$? Is it because of the result known as the Carathéodory's extension theorem? If so, can anyone refer me to proof, preferably in the context of probability measure?

For example suppose $X=\mathbb{R},\mathcal{S}=\mathcal{B}$ (the Borel sigma algebra) and $S=\{I:I{\textstyle \text{is an interval in}}\mathbb{R}\}$. Now it is easy to check that $P(I)=\frac{1}{\sqrt{2\pi}}{\int}_{{I}_{x}}{e}^{-{x}^{2}/2}dx$ satisfies the requisite conditions. How does it follow that the conditions are now met for $P$? Is it because of the result known as the Carathéodory's extension theorem? If so, can anyone refer me to proof, preferably in the context of probability measure?

asked 2022-07-02

I have somehow confused myself with this fairly straightforward proof. We need to show that $\lambda f$ is a measurable function on $(S,\mathcal{S})$, i.e. that for any $c\in \mathbb{R}:\{s\in S:f(s)\le c\}\subset S$. If $\lambda \ne 0$, then the claim follows immediately from the measurability of $f$, namely as $c/\lambda \in \mathbb{R}$ it follows that $\{s\in S:f(s)\le c/\lambda \}=\{s\in S:\lambda f(s)\le c\}$

But then, if $\lambda =0,\lambda f=0$ and I am not sure how to convince myself that $\lambda f$ is measurable.

But then, if $\lambda =0,\lambda f=0$ and I am not sure how to convince myself that $\lambda f$ is measurable.

asked 2022-07-01

In a study looking at undergraduate students’ perceptions of sense of community at their university, a researcher hypothesizes that the farther away students live from campus (in miles), the less they feel they are part of the university community. The researcher collected data for the following two variables – miles from campus and part of community (rating from 1-10 of how much they felt part of the university community).

what is the scale of measurement? (i.e., nominal, ordinal, interval or ratio)

what is the scale of measurement? (i.e., nominal, ordinal, interval or ratio)

asked 2022-04-10

What is the effect of measurement error in Y? How is this different from the effect of measurement error in X?

asked 2022-06-24

Let $\mathbb{B}$ be a $\sigma $-algebra on a set $X$ and let $f\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{thinmathspace}{0ex}}X\to \overline{R}$ be an extended real-valued function. Define the sets $P\phantom{\rule{mediummathspace}{0ex}}={f}^{-1}(-\mathrm{\infty})$ and $N\phantom{\rule{mediummathspace}{0ex}}={f}^{-1}(+\mathrm{\infty})$.

Define a real-valued function $g\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{thinmathspace}{0ex}}X\to \mathbb{R}$ by

$g(x)=\{\begin{array}{ll}f(x),& f(x)\in \mathbb{R}\\ 0,& otherwise\end{array}.$

Prove that $f$ is $\mathbb{B}$-measurable if and only if $g$ is $\mathbb{B}$-measurable and $P,N\in \mathbb{B}$.So i tried to prove this like this:

$P,N\in \mathbb{B}$ obviously. Now assume $f$ is $\mathbb{B}$-measurable. Let $\alpha \in \mathbb{R}$. Consider

$\{x\in X|g(x)>\alpha \}$=$\{x\in X|f(x)>\alpha \}\setminus P$. In this circumstance $P\in \mathbb{B}$, $\{x\in X|f(x)>\alpha \}\in \mathbb{B}$ so $\{x\in X|f(x)>\alpha \}\setminus P\in \mathbb{B}$. So ${P}^{c}\cap \{x\in X|f(x)>\alpha \}\in \mathbb{B}$. This is because, $\mathbb{B}$ is closed under finite intersection.

$(\Leftarrow )$$g$ is $\mathbb{B}$-measurable. Then

$\{x\in X|f(x)>\alpha \}=\{\begin{array}{ll}\{x\in X|g(x)>\alpha \}\cup P,& \alpha >0\\ \{x\in X|g(x)>\alpha \}\setminus N,& \alpha \le 0\end{array}.$

Both of cases, observe that output sets belongs to $\mathbb{B}$.

But my professor said the proof was wrong. Where am I going wrong? Can you help me?

Define a real-valued function $g\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{thinmathspace}{0ex}}X\to \mathbb{R}$ by

$g(x)=\{\begin{array}{ll}f(x),& f(x)\in \mathbb{R}\\ 0,& otherwise\end{array}.$

Prove that $f$ is $\mathbb{B}$-measurable if and only if $g$ is $\mathbb{B}$-measurable and $P,N\in \mathbb{B}$.So i tried to prove this like this:

$P,N\in \mathbb{B}$ obviously. Now assume $f$ is $\mathbb{B}$-measurable. Let $\alpha \in \mathbb{R}$. Consider

$\{x\in X|g(x)>\alpha \}$=$\{x\in X|f(x)>\alpha \}\setminus P$. In this circumstance $P\in \mathbb{B}$, $\{x\in X|f(x)>\alpha \}\in \mathbb{B}$ so $\{x\in X|f(x)>\alpha \}\setminus P\in \mathbb{B}$. So ${P}^{c}\cap \{x\in X|f(x)>\alpha \}\in \mathbb{B}$. This is because, $\mathbb{B}$ is closed under finite intersection.

$(\Leftarrow )$$g$ is $\mathbb{B}$-measurable. Then

$\{x\in X|f(x)>\alpha \}=\{\begin{array}{ll}\{x\in X|g(x)>\alpha \}\cup P,& \alpha >0\\ \{x\in X|g(x)>\alpha \}\setminus N,& \alpha \le 0\end{array}.$

Both of cases, observe that output sets belongs to $\mathbb{B}$.

But my professor said the proof was wrong. Where am I going wrong? Can you help me?

asked 2022-05-25

Let $(X,\mathcal{A},\mu )$ be a measure-space and $f:X\to \mathbb{[}0,\mathrm{\infty}]$ a unsigned measurable map. Show that $\mu (\{f(x)>0\})=0\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mu (\{f(x)>t\})=0$ For every $t>0$.

Suppose that $\mu (\{f(x)>0\})=0$. Since $f=0$ a.e we have that ${\int}_{X}f=0$. Thus $\mu (\{f(x)>t\})\le \frac{1}{t}{\int}_{X}f=0$ and since the measure is positive we have that $\mu (\{f(x)>t\})=0$.

Conversely suppose that $\mu (\{f(x)>t\})=0$. Then what can be done for this case? It seems that I could write ${\int}_{X}f={\int}_{X\setminus \{f(x)>t\}}f+{\int}_{\{f(x)>t\}}f$ and the latter integral would be zero, but I don’t really get anywhere from here?

Suppose that $\mu (\{f(x)>0\})=0$. Since $f=0$ a.e we have that ${\int}_{X}f=0$. Thus $\mu (\{f(x)>t\})\le \frac{1}{t}{\int}_{X}f=0$ and since the measure is positive we have that $\mu (\{f(x)>t\})=0$.

Conversely suppose that $\mu (\{f(x)>t\})=0$. Then what can be done for this case? It seems that I could write ${\int}_{X}f={\int}_{X\setminus \{f(x)>t\}}f+{\int}_{\{f(x)>t\}}f$ and the latter integral would be zero, but I don’t really get anywhere from here?

asked 2022-05-11

Let $E\subseteq \mathbb{R}$ be a set of finite measure. Assume the sequence ${f}_{n}$ is in ${L}^{1}(E)$ and ${f}_{n}\to f$ weakly in ${L}^{1}(E)$. Show that if ${f}_{n}\ge M$ almost everywhere for a constant $M$ then $f\ge M$ almost everywhere.

I tried providing it by contradiction, supposing $f<m$ on a subset of $E$ with positive measure. Then of course for every $n\in \mathbb{N}$, there is a subset ${A}_{n}$ of $A$ such that ${f}_{n}\ge M$ on ${A}_{n}$, but I can't go any further.

I tried providing it by contradiction, supposing $f<m$ on a subset of $E$ with positive measure. Then of course for every $n\in \mathbb{N}$, there is a subset ${A}_{n}$ of $A$ such that ${f}_{n}\ge M$ on ${A}_{n}$, but I can't go any further.