Show your work:

Express 160 pounds (lbs) in kilograms (kg). Round to the nearest hundredths.

Express 160 pounds (lbs) in kilograms (kg). Round to the nearest hundredths.

priscillianaw1
2022-09-06
Answered

Show your work:

Express 160 pounds (lbs) in kilograms (kg). Round to the nearest hundredths.

Express 160 pounds (lbs) in kilograms (kg). Round to the nearest hundredths.

You can still ask an expert for help

Bernard Scott

Answered 2022-09-07
Author has **9** answers

Answer:

$72.57kg$

Step-by-step explanation:

$1lbs=0.45359237kg$

$160lbs=160\cdot 0.45359237kg=72.5747792kg$

72.5747792 kg round to the nearest hundredths is 72.57 kg

$72.57kg$

Step-by-step explanation:

$1lbs=0.45359237kg$

$160lbs=160\cdot 0.45359237kg=72.5747792kg$

72.5747792 kg round to the nearest hundredths is 72.57 kg

asked 2022-05-29

Let $(\mathrm{\Omega},\mathcal{A},\nu )$ be a measure space. Show that if the measure over the space $\mathrm{\Omega}$ is $0<\nu (\mathrm{\Omega})<\mathrm{\infty}$, then

$Q=\frac{1}{\nu (\mathrm{\Omega})}\nu $

is a probability measure.

I don't see how this is indeed a probability measure. For it to be a probability measure indeed it has to

1. Return values in [0,1].

2. Satisfy countable additivity.

3. Implied measure of $\mathrm{\varnothing}$ should be 0.

The only thing I currently know is that $\nu $ is a measure and that it is finite (and greater than 0) over $\mathrm{\Omega}$. This one seems quite tough.

$Q=\frac{1}{\nu (\mathrm{\Omega})}\nu $

is a probability measure.

I don't see how this is indeed a probability measure. For it to be a probability measure indeed it has to

1. Return values in [0,1].

2. Satisfy countable additivity.

3. Implied measure of $\mathrm{\varnothing}$ should be 0.

The only thing I currently know is that $\nu $ is a measure and that it is finite (and greater than 0) over $\mathrm{\Omega}$. This one seems quite tough.

asked 2021-02-05

Twenty-one independent measurements were taken of the hardness (on the Rockwell C scale) of HSLA-100 steel base metal, and another 21 independent measurements were made of the hardness of a weld produced on this base metal.

The standard deviation of the measurements made on the base metal was 3.06, and the standard deviation of the measurements made on the weld was 1.41.

Assume that the measurements are independent random samples from normal populations.

Need to conclude that measurements made on the base metal are more variable than measurements made on the weld?

The standard deviation of the measurements made on the base metal was 3.06, and the standard deviation of the measurements made on the weld was 1.41.

Assume that the measurements are independent random samples from normal populations.

Need to conclude that measurements made on the base metal are more variable than measurements made on the weld?

asked 2022-05-20

I'm trying to find an answer to something.

Here $\lambda $ denotes the Lebesgue measure. Furthermore:

${f}_{n}(x)={1}_{[0,1]}({2}^{m}x-j)$

Where $n={2}^{m}+j,\text{}0\le j\le {2}^{m}-1$. Prove that $\underset{n}{lim}{\int}^{\ast}|{f}_{n}|d\lambda \to 0$ and that ${f}_{n}(x)$ doesn't converge pointwise for any $x\in [0,1]$.

The second part isn't that difficult and not of too much interest. I am rather interested in the first part. My go-to approach would be to define a new function that does the same for $x\in [0,1]$ and from there we draw a line to zero. This would then be a function which is continuous with compact support. This means that the Lebesgue-Integral would simply be the Riemann Integral.

Let us call this function ${\varphi}_{n}$. The problem I run into, however, is that

${\int}^{\ast}|{f}_{n}|d\lambda ={\int}^{\ast}\underset{n}{lim}{\varphi}_{n}d\lambda ={\int}_{0}^{1}|{2}^{m}x-j|=|{2}^{m-1}-j|$

And thus we get:

${\int}^{\ast}\underset{n}{lim}|{f}_{n}|d\lambda =\underset{n}{lim}|{2}^{m-1}-j|\nrightarrow 0$

What is going on? I would be very grateful for your help!

Here $\lambda $ denotes the Lebesgue measure. Furthermore:

${f}_{n}(x)={1}_{[0,1]}({2}^{m}x-j)$

Where $n={2}^{m}+j,\text{}0\le j\le {2}^{m}-1$. Prove that $\underset{n}{lim}{\int}^{\ast}|{f}_{n}|d\lambda \to 0$ and that ${f}_{n}(x)$ doesn't converge pointwise for any $x\in [0,1]$.

The second part isn't that difficult and not of too much interest. I am rather interested in the first part. My go-to approach would be to define a new function that does the same for $x\in [0,1]$ and from there we draw a line to zero. This would then be a function which is continuous with compact support. This means that the Lebesgue-Integral would simply be the Riemann Integral.

Let us call this function ${\varphi}_{n}$. The problem I run into, however, is that

${\int}^{\ast}|{f}_{n}|d\lambda ={\int}^{\ast}\underset{n}{lim}{\varphi}_{n}d\lambda ={\int}_{0}^{1}|{2}^{m}x-j|=|{2}^{m-1}-j|$

And thus we get:

${\int}^{\ast}\underset{n}{lim}|{f}_{n}|d\lambda =\underset{n}{lim}|{2}^{m-1}-j|\nrightarrow 0$

What is going on? I would be very grateful for your help!

asked 2022-06-26

Let $(X,\mathcal{A})$ y $(Y,\mathcal{B})$ measure spaces. Let $R:=R(\mathcal{A},\mathcal{B})$ the collection of measurable rectangles $R:=\{A\times B:A\in \mathcal{A},B\in \mathcal{B}\}.$. Prove that the algebra of subsets of $X\times Y$ generated by $R$ is the collection of finite unions of elements of $R$. I have a hint and it is in which I call $C$ to be the collection of finite unions of elements of $R$, prove that $C$ is in the algebra generated by $R$, then prove that $C$ is an algebra that contains $R$.

asked 2022-03-25

Determine which of the four levels of measurement (nominal, ordinal, interval, ratio) is most appropriate for the data below.

Class times measured in minutes

A. The nominal level of measurement is most appropriate because the data cannot be ordered.

B. The interval level of measurement is most appropriate because the data can be ordered, differences (obtained by subtraction) can be found and are meaningful, and there is no natural starting point.

C. The ratio level of measurement is most appropriate because the data can be ordered, differences can be found and are meaningful, and there is a natural starting zero point.

D. The ordinal level of measurement is most appropriate because the data can be ordered, but differences (obtained by subtraction) cannot be found or are meaningless.

Class times measured in minutes

A. The nominal level of measurement is most appropriate because the data cannot be ordered.

B. The interval level of measurement is most appropriate because the data can be ordered, differences (obtained by subtraction) can be found and are meaningful, and there is no natural starting point.

C. The ratio level of measurement is most appropriate because the data can be ordered, differences can be found and are meaningful, and there is a natural starting zero point.

D. The ordinal level of measurement is most appropriate because the data can be ordered, but differences (obtained by subtraction) cannot be found or are meaningless.

asked 2022-07-05

Suppose $(\mathbb{R},\tau )$ is the standard topological space. And $\mathbb{B}$ is the Borel $\sigma $-algebra from this space.

Define set A as:

$A=\{x\in \mathbb{R}:x={q}_{1}\sqrt{{n}_{1}}+{q}_{2}\sqrt{{n}_{2}}\text{for some}{q}_{1},{q}_{2}\in \mathbb{Q},\text{and}{n}_{1},{n}_{2}\in \mathbb{N}\}$

How can I show that the set A belongs to $\mathbb{B}$?

Define set A as:

$A=\{x\in \mathbb{R}:x={q}_{1}\sqrt{{n}_{1}}+{q}_{2}\sqrt{{n}_{2}}\text{for some}{q}_{1},{q}_{2}\in \mathbb{Q},\text{and}{n}_{1},{n}_{2}\in \mathbb{N}\}$

How can I show that the set A belongs to $\mathbb{B}$?

asked 2022-05-21

Let $X$ be a locally compact Hausdorff space and ${C}_{b}$ the set of all continuous functions with support compact and ${C}_{0}(X)$ the set of all functions with compact support. The dual of ${C}_{b}$ is the dual of ${C}_{0}(X)$?