Note: 1 gallon = 4 quarts

A. 2 1/8 qt

B.16 1/2 qt

C.33 qt

D.66 qt

Aubrie Aguilar
2022-09-17
Answered

How many quarts are in 814 gallons?

Note: 1 gallon = 4 quarts

A. 2 1/8 qt

B.16 1/2 qt

C.33 qt

D.66 qt

Note: 1 gallon = 4 quarts

A. 2 1/8 qt

B.16 1/2 qt

C.33 qt

D.66 qt

You can still ask an expert for help

asked 2022-05-22

Let $E\subseteq {\mathbb{R}}^{d}$ be a (Lebesgue) measurable set and let $f$,$g$ be two measurable functions defined on E. I would like to show that if $\mathrm{\Phi}$ is a continuous function on ${\mathbb{R}}^{2}$, then the function $h:x\mapsto \mathrm{\Phi}(f(x),g(x))$ is measurable. The proof remains unknown to me, but I can address the problem if it is only one-dimensional. Specifically, if $\mathrm{\Phi}$ is a continuous function on $\mathbb{R}$, then I can show that $\mathrm{\Phi}\circ f$ is measurable. Indeed, since $\{\mathrm{\Phi}<a\}$ is an open set $G$, we can conclude that $\{\mathrm{\Phi}\circ f<a\}={f}^{-1}(G)$ is measurable. How about the two-dimensional problem? Does anyone have an idea? Thank you.

asked 2022-06-17

State the number of possible triangles that can be formed using the given measurements, then sketch and solve the triangles, if possible.

$m\mathrm{\angle}A={48}^{\circ},c=29,a=4$

$m\mathrm{\angle}A={48}^{\circ},c=29,a=4$

asked 2022-05-22

Suppose $(\mathrm{\Omega},\mathcal{F},\mu )$ is a measure space and $f,g$ are $(\mathcal{F},\mathcal{B})-$−measurable functions where $\mathcal{B}$ is the Borel algebra on $\mathbb{R}$. If

$\mu (\{f<g\})>0$

then are we able to find a constant $\xi \in \mathbb{R}$ such that the set $\{f\le \xi <g\}$ is also of nonzero measure? I am confident that I could find $\u03f5>0$ such that the set $\{f<g-\u03f5\}$ is of nonzero measure, but I find the existence of such a constant $\xi $ questionable.

$\mu (\{f<g\})>0$

then are we able to find a constant $\xi \in \mathbb{R}$ such that the set $\{f\le \xi <g\}$ is also of nonzero measure? I am confident that I could find $\u03f5>0$ such that the set $\{f<g-\u03f5\}$ is of nonzero measure, but I find the existence of such a constant $\xi $ questionable.

asked 2022-06-20

Let $X$ and $Y$ be random variables. The Ky Fan metric is defined as:

$d(X,Y):=min\{\u03f5>0:P(|X-Y|>\u03f5)\le \u03f5\}$

I want to show it is indeed a metric, for which I need to show that it satisfies triangle inequality.

Set $d(X,Y)={\u03f5}_{1},d(Y,Z)={\u03f5}_{2}$,and $d(Z,X)={\u03f5}_{3}$ .

I approached this by trying to show that $({\u03f5}_{1}+{\u03f5}_{2})\in \{\u03f5>0:P(|X-Z|>\u03f5)\le \u03f5\}$. But to show that, I had to show $P(|X-Y|>{\u03f5}_{1})+P(|Y-Z|>{\u03f5}_{2})\ge P(|X-Z|>{\u03f5}_{1}+{\u03f5}_{2})$, which I could not. Can you help in showing this?

$d(X,Y):=min\{\u03f5>0:P(|X-Y|>\u03f5)\le \u03f5\}$

I want to show it is indeed a metric, for which I need to show that it satisfies triangle inequality.

Set $d(X,Y)={\u03f5}_{1},d(Y,Z)={\u03f5}_{2}$,and $d(Z,X)={\u03f5}_{3}$ .

I approached this by trying to show that $({\u03f5}_{1}+{\u03f5}_{2})\in \{\u03f5>0:P(|X-Z|>\u03f5)\le \u03f5\}$. But to show that, I had to show $P(|X-Y|>{\u03f5}_{1})+P(|Y-Z|>{\u03f5}_{2})\ge P(|X-Z|>{\u03f5}_{1}+{\u03f5}_{2})$, which I could not. Can you help in showing this?

asked 2022-05-28

Is the probability model given by the following distribution function

${F}_{p}(x)=\{\begin{array}{ll}\sum _{k=1}^{\lfloor x\rfloor}p(1-p{)}^{k-1}& \text{if}x\ge 1\\ 0& \text{else}\end{array}$

dominated? (in which case exhibit a dominant measure and give the Radon-Nikodym derivative).

I know the definition of a dominated model:

A model $({P}_{\theta}{)}_{\theta}$ is dominated if there exists a measure $\mu >0$ and $\sigma -$−finite such that ${P}_{\theta}$ is dominated by $\mu $ and ${P}_{\theta}$ is absolutely continuous w.r.t $\mu $

But I don't know how to show that the model given by this distribution function is dominated.

${F}_{p}(x)=\{\begin{array}{ll}\sum _{k=1}^{\lfloor x\rfloor}p(1-p{)}^{k-1}& \text{if}x\ge 1\\ 0& \text{else}\end{array}$

dominated? (in which case exhibit a dominant measure and give the Radon-Nikodym derivative).

I know the definition of a dominated model:

A model $({P}_{\theta}{)}_{\theta}$ is dominated if there exists a measure $\mu >0$ and $\sigma -$−finite such that ${P}_{\theta}$ is dominated by $\mu $ and ${P}_{\theta}$ is absolutely continuous w.r.t $\mu $

But I don't know how to show that the model given by this distribution function is dominated.

asked 2022-05-15

Maybe anyone will help me

Let $f\ge 0$. Let

$\mu (\{x:f(x)>t\})=\frac{1}{{t}^{2}+1}$

I'm trying to compute

${\int}_{\mathbb{R}}f\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}d\mu $

To compute this, this is the approach I took. I know that the integral of a measurable function is

$\int f\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}d\mu =sup\{\int g\phantom{\rule{thinmathspace}{0ex}}d\mu :g\text{simple},0\le g\le f\}$

Further, I know that I can represent $f$ as a non-decreasing sequence of simple functions that converges to $f$, as such:

${f}_{n}=n\cdot {1}_{{B}_{n}}+\sum _{k=1}^{n\cdot {2}^{n}}\frac{(k-1)}{{2}^{n}}{1}_{{A}_{n},k}$

where

${B}_{n}=\{x:f(x)>n\}$

and

${A}_{n,k}=\{x:(k-1){2}^{-n}<f(x)\le k\phantom{\rule{thinmathspace}{0ex}}{2}^{-n}\}$

Now, since this sequence is increasing and converges to f, the sup of the set earlier would just be

$\underset{n\to \mathrm{\infty}}{lim}\int {f}_{n}$

But, I'm unsure how exactly to calculate this. I know that if g is a simple function such that

$g=\sum _{k=1}^{n}{c}_{k}$

Then

$\int g=\sum _{k=1}^{n}{c}_{k}\mu ({A}_{k})$

But, our sequence of ${f}_{n}$ has the $n\cdot {1}_{{B}_{n}}$ as an additional term outside the summation, and for the measure of ${A}_{n,k}$, I'm unsure if I'm allowed to say that

$\begin{array}{rl}\mu (\{x:(k-1){2}^{-n}<f(x)\le k\phantom{\rule{thinmathspace}{0ex}}{2}^{-n}\})& =\\ & =\mu (\{x:f(x)>(k-1){2}^{-n}\})-\mu (\{x:f(x)>k\cdot {2}^{-n}\})\end{array}$

Let $f\ge 0$. Let

$\mu (\{x:f(x)>t\})=\frac{1}{{t}^{2}+1}$

I'm trying to compute

${\int}_{\mathbb{R}}f\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}d\mu $

To compute this, this is the approach I took. I know that the integral of a measurable function is

$\int f\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}d\mu =sup\{\int g\phantom{\rule{thinmathspace}{0ex}}d\mu :g\text{simple},0\le g\le f\}$

Further, I know that I can represent $f$ as a non-decreasing sequence of simple functions that converges to $f$, as such:

${f}_{n}=n\cdot {1}_{{B}_{n}}+\sum _{k=1}^{n\cdot {2}^{n}}\frac{(k-1)}{{2}^{n}}{1}_{{A}_{n},k}$

where

${B}_{n}=\{x:f(x)>n\}$

and

${A}_{n,k}=\{x:(k-1){2}^{-n}<f(x)\le k\phantom{\rule{thinmathspace}{0ex}}{2}^{-n}\}$

Now, since this sequence is increasing and converges to f, the sup of the set earlier would just be

$\underset{n\to \mathrm{\infty}}{lim}\int {f}_{n}$

But, I'm unsure how exactly to calculate this. I know that if g is a simple function such that

$g=\sum _{k=1}^{n}{c}_{k}$

Then

$\int g=\sum _{k=1}^{n}{c}_{k}\mu ({A}_{k})$

But, our sequence of ${f}_{n}$ has the $n\cdot {1}_{{B}_{n}}$ as an additional term outside the summation, and for the measure of ${A}_{n,k}$, I'm unsure if I'm allowed to say that

$\begin{array}{rl}\mu (\{x:(k-1){2}^{-n}<f(x)\le k\phantom{\rule{thinmathspace}{0ex}}{2}^{-n}\})& =\\ & =\mu (\{x:f(x)>(k-1){2}^{-n}\})-\mu (\{x:f(x)>k\cdot {2}^{-n}\})\end{array}$

asked 2021-02-26

Juan makes a measurement in a chemistry laboratory and records the result in his lab report. The standard deviation of students lab measurements is