Convert $20\text{kg}$ to pounds, using the appropriate unit notation.

muroscamsey
2022-08-14
Answered

Convert $20\text{kg}$ to pounds, using the appropriate unit notation.

You can still ask an expert for help

Olivia Petersen

Answered 2022-08-15
Author has **16** answers

As we know that

$1\text{kg}=2.20462\text{pounds}\phantom{\rule{0ex}{0ex}}1\text{kg}=2.204621\text{lbs}$

So,

$20\text{kg}=20\times 2.20462\text{lbs}\phantom{\rule{0ex}{0ex}}=44.0924\text{lbs}$

In $20\text{kg}=44.0924\text{lbs}$

$1\text{kg}=2.20462\text{pounds}\phantom{\rule{0ex}{0ex}}1\text{kg}=2.204621\text{lbs}$

So,

$20\text{kg}=20\times 2.20462\text{lbs}\phantom{\rule{0ex}{0ex}}=44.0924\text{lbs}$

In $20\text{kg}=44.0924\text{lbs}$

asked 2022-06-14

Suppose $E[|X|]<\mathrm{\infty}$ and that ${A}_{n}$ are disjoint sets with ${\cup}_{n=1}^{\mathrm{\infty}}{A}_{n}=A$. Show that $\sum _{n=0}^{\mathrm{\infty}}E[X{\mathbf{1}}_{{A}_{n}}]=E[X{\mathbf{1}}_{A}]$

Here is what I try with Monotone Convergence Theorem:

$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty}}E[X{\mathbf{1}}_{{A}_{n}}]& =\underset{m\to \mathrm{\infty}}{lim}\sum _{n=0}^{m}E[X{\mathbf{1}}_{{A}_{n}}]\\ & =\underset{m\to \mathrm{\infty}}{lim}E[X\sum _{n=0}^{m}{\mathbf{1}}_{{A}_{n}}]\\ & =\underset{m\to \mathrm{\infty}}{lim}E[X{\mathbf{1}}_{{F}_{n}}]\\ & =E[X\underset{m\to \mathrm{\infty}}{lim}{\mathbf{1}}_{{F}_{n}}]\\ & =E[X{\mathbf{1}}_{A}]\end{array}$

where ${F}_{n}={\cup}_{i=1}^{n}{A}_{i}$, so ${\cup}_{n=1}^{m}{A}_{n}={\cup}_{n=1}^{m}{F}_{n}$, and ${\mathbf{1}}_{{F}_{n}}$ is increasing.

I am not quite sure if I get it right. My main concern is the first step: $\sum _{n=0}^{\mathrm{\infty}}E[X{\mathbf{1}}_{{A}_{n}}]=\underset{m\to \mathrm{\infty}}{lim}\sum _{n=0}^{m}E[X{\mathbf{1}}_{{A}_{n}}]$. Do we always have $\sum _{n=0}^{\mathrm{\infty}}=\underset{m\to \mathrm{\infty}}{lim}\sum _{n=0}^{m}$?

I may also have made other mistakes. Please feel free to point out. Any other hints are also welcome.

Here is what I try with Monotone Convergence Theorem:

$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty}}E[X{\mathbf{1}}_{{A}_{n}}]& =\underset{m\to \mathrm{\infty}}{lim}\sum _{n=0}^{m}E[X{\mathbf{1}}_{{A}_{n}}]\\ & =\underset{m\to \mathrm{\infty}}{lim}E[X\sum _{n=0}^{m}{\mathbf{1}}_{{A}_{n}}]\\ & =\underset{m\to \mathrm{\infty}}{lim}E[X{\mathbf{1}}_{{F}_{n}}]\\ & =E[X\underset{m\to \mathrm{\infty}}{lim}{\mathbf{1}}_{{F}_{n}}]\\ & =E[X{\mathbf{1}}_{A}]\end{array}$

where ${F}_{n}={\cup}_{i=1}^{n}{A}_{i}$, so ${\cup}_{n=1}^{m}{A}_{n}={\cup}_{n=1}^{m}{F}_{n}$, and ${\mathbf{1}}_{{F}_{n}}$ is increasing.

I am not quite sure if I get it right. My main concern is the first step: $\sum _{n=0}^{\mathrm{\infty}}E[X{\mathbf{1}}_{{A}_{n}}]=\underset{m\to \mathrm{\infty}}{lim}\sum _{n=0}^{m}E[X{\mathbf{1}}_{{A}_{n}}]$. Do we always have $\sum _{n=0}^{\mathrm{\infty}}=\underset{m\to \mathrm{\infty}}{lim}\sum _{n=0}^{m}$?

I may also have made other mistakes. Please feel free to point out. Any other hints are also welcome.

asked 2021-02-13

1.After several tries of measuring, Lydia gets the results of 2.75, 2.76, 2.30 cm. She realized that the results of measurement is closest to the actual measurement which is 3.25. What is the implication of her measurements? Is it Accurate and precise?

2.I measured the length of cabinet 3 times. The results of my measurements are 3.44 m, 3.55 m, 3.47 m. Afterwards, I compared it to the results with each other. What did I was trying to find out? Is it precision?

2.I measured the length of cabinet 3 times. The results of my measurements are 3.44 m, 3.55 m, 3.47 m. Afterwards, I compared it to the results with each other. What did I was trying to find out? Is it precision?

asked 2022-07-08

I have the following velocity measurements, where the sign of ${V}_{e}$ defines opposing directions of movement in a completely symmetric experimental setting:

$\begin{array}{|llr|}\hline {V}_{e}[cm/s]& {V}_{m}[cm/s]& \mathrm{\Delta}V[cm/s]\\ 9& 9.38& 0.38\\ 8& 8.491& 0.491\\ 7& 7.482& 0.482\\ 6& 6.502& 0.502\\ 5& 5.726& 0.726\\ 4& 4.499& 0.499\\ 3& 2.021& -0.979\\ 2& 2.34& 0.34\\ 1& 2.018& 1.018\\ 0& 0& 0\\ -1& -0.501& -0.499\\ -2& -2.328& 0.328\\ -3& -2.988& -0.012\\ -4& -3.503& -0.497\\ -5& -4.506& -0.494\\ -6& -5.762& -0.238\\ -7& -7.479& 0.479\\ -8& -7.981& -0.019\\ -9& -8.496& -0.504\\ \hline\end{array}$

In this table, $\mathrm{\Delta}V=\left|{V}_{m}\right|-\left|{V}_{e}\right|$

Running a Student t-test on $\mathrm{\Delta}V$, we find that the mean does not significantly differ from zero, under a type I error of 5%. From this result, I conclude that $\mathrm{\Delta}V$ is a random error.

The reviewer of my work (I'm an academic student) insists that my method does not account for the direction (i.e. the sign of $V$). That is indeed the case, since my test answers a precise question: Does the measurement method (${V}_{m}$) systematically over/underestimates the true velocity ($||{V}_{e}||$)?

Instead, the reviewer uses $\mathrm{\Delta}V={V}_{m}-{V}_{e}$ to show that the method significantly overestimates $Ve$, especially when ${V}_{e}<0$, using the same t-test. However, I am having trouble finding what specific question such a test answers, and the reviewer's statement is wrong in my opinion.

What is the correct way of defining $\mathrm{\Delta}V$ and discern a systematic measurement error?

$\begin{array}{|llr|}\hline {V}_{e}[cm/s]& {V}_{m}[cm/s]& \mathrm{\Delta}V[cm/s]\\ 9& 9.38& 0.38\\ 8& 8.491& 0.491\\ 7& 7.482& 0.482\\ 6& 6.502& 0.502\\ 5& 5.726& 0.726\\ 4& 4.499& 0.499\\ 3& 2.021& -0.979\\ 2& 2.34& 0.34\\ 1& 2.018& 1.018\\ 0& 0& 0\\ -1& -0.501& -0.499\\ -2& -2.328& 0.328\\ -3& -2.988& -0.012\\ -4& -3.503& -0.497\\ -5& -4.506& -0.494\\ -6& -5.762& -0.238\\ -7& -7.479& 0.479\\ -8& -7.981& -0.019\\ -9& -8.496& -0.504\\ \hline\end{array}$

In this table, $\mathrm{\Delta}V=\left|{V}_{m}\right|-\left|{V}_{e}\right|$

Running a Student t-test on $\mathrm{\Delta}V$, we find that the mean does not significantly differ from zero, under a type I error of 5%. From this result, I conclude that $\mathrm{\Delta}V$ is a random error.

The reviewer of my work (I'm an academic student) insists that my method does not account for the direction (i.e. the sign of $V$). That is indeed the case, since my test answers a precise question: Does the measurement method (${V}_{m}$) systematically over/underestimates the true velocity ($||{V}_{e}||$)?

Instead, the reviewer uses $\mathrm{\Delta}V={V}_{m}-{V}_{e}$ to show that the method significantly overestimates $Ve$, especially when ${V}_{e}<0$, using the same t-test. However, I am having trouble finding what specific question such a test answers, and the reviewer's statement is wrong in my opinion.

What is the correct way of defining $\mathrm{\Delta}V$ and discern a systematic measurement error?

asked 2022-04-14

Kindly answer both .no need to explain anything . Only name

1) there are 15 students in a college that are to be selected for the music competition. For the selection of students, find the level of measurement and which sampling method will u apply

2) George is ranked 5th in basketball team representing use team.he wears no 4 9n his t shirt .find the level of measurement

1) there are 15 students in a college that are to be selected for the music competition. For the selection of students, find the level of measurement and which sampling method will u apply

2) George is ranked 5th in basketball team representing use team.he wears no 4 9n his t shirt .find the level of measurement

asked 2022-06-22

If we have a question where we have to find the coefficient's units such as K in this case. The actual formula contains more parts but it is simply the derivatives that I am unsure about.

$K=\frac{dv}{dt}$

Where $v=m{s}^{-1}$ (metres per second) and t is time

Do I simply find the derivative of $m{s}^{-1}$ in respect to s? And whatever units of measurement I am left with are the units of K? ie.

$\frac{d(m{s}^{-1})}{ds}$

$=m\ast \frac{d({s}^{-1})}{ds}$

$=-m{s}^{-2}$

$K=\frac{dv}{dt}$

Where $v=m{s}^{-1}$ (metres per second) and t is time

Do I simply find the derivative of $m{s}^{-1}$ in respect to s? And whatever units of measurement I am left with are the units of K? ie.

$\frac{d(m{s}^{-1})}{ds}$

$=m\ast \frac{d({s}^{-1})}{ds}$

$=-m{s}^{-2}$

asked 2022-06-25

I have a matrix $A$ with null trace.

What is the minimum number of linear measurements that I have to perform in order to determine $A$?

By a linear measurement I mean that I know the quantities $A{\overrightarrow{x}}_{i}\cdot {\overrightarrow{h}}_{j}$ for given ${\overrightarrow{x}}_{i}$ and ${\overrightarrow{h}}_{j}$

What is the minimum number of linear measurements that I have to perform in order to determine $A$?

By a linear measurement I mean that I know the quantities $A{\overrightarrow{x}}_{i}\cdot {\overrightarrow{h}}_{j}$ for given ${\overrightarrow{x}}_{i}$ and ${\overrightarrow{h}}_{j}$

asked 2022-07-09

Let $X$ be a set, $\mathcal{A}$ a ring of subsets of $X$, $\mu :\mathcal{A}\to {\overline{\mathbb{R}}}_{\ge 0}$ a premeasure and ${\mu}^{\ast}$ the outer measure generated by $\mu $. (By Caratheodory)

If $E\in {\mathcal{M}}_{{\mu}^{\ast}}$ and satisfacies ${\mu}^{\ast}(E)<\mathrm{\infty}$, then for each $\epsilon $ existx $A\in \mathcal{A}$ such that ${\mu}^{\ast}(A\mathrm{\u25b3}E)<\epsilon $

I try to test this, my first idea was to give a cover of $E$, I just don't know if I can find said cover in $\mathcal{A}$, as the measure is not finite, so the extension is not unique, someone can give me a hint how to proceed?

If $E\in {\mathcal{M}}_{{\mu}^{\ast}}$ and satisfacies ${\mu}^{\ast}(E)<\mathrm{\infty}$, then for each $\epsilon $ existx $A\in \mathcal{A}$ such that ${\mu}^{\ast}(A\mathrm{\u25b3}E)<\epsilon $

I try to test this, my first idea was to give a cover of $E$, I just don't know if I can find said cover in $\mathcal{A}$, as the measure is not finite, so the extension is not unique, someone can give me a hint how to proceed?