$\frac{2,000\text{ton}}{1\text{lb}}$

$\frac{1\text{ton}}{2,000\text{lb}}$

$\frac{2,000\text{lb}}{1\text{ton}}$

$\frac{1\text{lb}}{2,000\text{ton}}$

Convert between the two measurement.

$7,000\text{lb}=\text{\_\_\_tons}$

Dana Russo
2022-10-05
Answered

What conversion rate if used to convert pounds to tons?

$\frac{2,000\text{ton}}{1\text{lb}}$

$\frac{1\text{ton}}{2,000\text{lb}}$

$\frac{2,000\text{lb}}{1\text{ton}}$

$\frac{1\text{lb}}{2,000\text{ton}}$

Convert between the two measurement.

$7,000\text{lb}=\text{\_\_\_tons}$

$\frac{2,000\text{ton}}{1\text{lb}}$

$\frac{1\text{ton}}{2,000\text{lb}}$

$\frac{2,000\text{lb}}{1\text{ton}}$

$\frac{1\text{lb}}{2,000\text{ton}}$

Convert between the two measurement.

$7,000\text{lb}=\text{\_\_\_tons}$

You can still ask an expert for help

procjenomuj

Answered 2022-10-06
Author has **8** answers

We know that 2000 pounds = 1 ton.

Hence to convert pounds into ton , we simply divide by 2000 and get the answer in ton.

$x\text{pounds}=x\text{lb}\frac{1\text{ton}}{2000\text{lb}}$

Hence conversim rate used is

$\frac{1\text{ton}}{2000\text{lb}}$

Second oprim is correct

Given $7000\text{lb}$

$7000\text{lb}=7000\text{lb}\times \frac{1\text{ton}}{2000\text{lb}}$

Hence to convert pounds into ton , we simply divide by 2000 and get the answer in ton.

$x\text{pounds}=x\text{lb}\frac{1\text{ton}}{2000\text{lb}}$

Hence conversim rate used is

$\frac{1\text{ton}}{2000\text{lb}}$

Second oprim is correct

Given $7000\text{lb}$

$7000\text{lb}=7000\text{lb}\times \frac{1\text{ton}}{2000\text{lb}}$

asked 2022-07-14

Draw, in standard position, the angle whose measure is given:

$-{315}^{\circ}$

$-{315}^{\circ}$

asked 2022-06-16

Given that $\sum _{k=1}^{\mathrm{\infty}}{\mu}_{0}({S}_{n}^{k})<{\mu}^{\ast}(A)+\frac{\u03f5}{{2}^{n}}$, how do we justify the non-strict inequality $\sum _{k=1}^{\mathrm{\infty}}{\mu}_{0}({S}_{n}^{k})\le \sum _{n=1}^{\mathrm{\infty}}{\mu}^{\ast}({A}_{n})+\u03f5$? Is there anything other to this than if $a<b$, then necessarily $a\le b$?

asked 2022-08-19

The formula to convert Fahrenheit to Celsius is $C=\frac{5}{9}(F-32)$. Convert 18c to Fahrenheit. Round to the nearest degree

asked 2022-07-01

Let $$ be some measurable domain and $f:E\to \mathbb{R}$ a measurable map. Let $B\in \mathrm{Bor}(\mathbb{R})$ be a Borel set. Show that ${f}^{-1}(B)$ is measurable.

I'm advised to define $\mathcal{A}=\{A\mid {f}^{-1}(A)\text{measurable}\}$. Now $\mathcal{A}$ consists of sets whose preimage is measurable, and since $f$ is continuous these sets are open. This collection seems to form a $\sigma $-algebra on $\mathbb{R}$, but I'm confused about the construction here as it seems that $\mathcal{A}$ is the smallest $\sigma $-algebra containing open sets, but that would mean that it's equal to $\mathrm{Bor}(\mathbb{R})$?

I'm advised to define $\mathcal{A}=\{A\mid {f}^{-1}(A)\text{measurable}\}$. Now $\mathcal{A}$ consists of sets whose preimage is measurable, and since $f$ is continuous these sets are open. This collection seems to form a $\sigma $-algebra on $\mathbb{R}$, but I'm confused about the construction here as it seems that $\mathcal{A}$ is the smallest $\sigma $-algebra containing open sets, but that would mean that it's equal to $\mathrm{Bor}(\mathbb{R})$?

asked 2022-08-22

A tree is 57 inches tall. How tall is it in feet and inches?

asked 2022-05-23

MathJax(?): Can't find handler for document
MathJax(?): Can't find handler for document
I participate the stochastic course and we now speak about summable families. There we have the following definition:

Let $\mathrm{\Omega}$ be countable and $a:\mathrm{\Omega}\to {\mathbb{R}}_{+}\cup \{\mathrm{\infty}\}$ be a map. Then we define

$\sum _{\mathrm{\Omega}}a(\omega ):=\underset{F\subset \mathrm{\Omega},|F|<\mathrm{\infty}}{sup}\sum _{F}a(\omega )$

Now our Prof said that we can consider $\sum _{\mathrm{\Omega}}a(\omega )$ as the integral of math xmlns="http://www.w3.org/1998/Math/MathML">a over $\mathrm{\Omega}$ to get a better connetion to measure theory afterwards when we speak about Fatou's lemma, Beppo Levi theorem ect. Because all this theorems we have seen with integrals last semester.

But somehow I don't see why this sum is equal to the integral. So I know from measure theory that if we have a simple function $f:\mathrm{\Omega}\to {\mathbb{R}}_{+}\cup \{\mathrm{\infty}\}$ where $f(\mathrm{\Omega})=\{{b}_{1},...,{b}_{n}\}$ finately many then

${\int}_{\mathrm{\Omega}}f\text{}\mathsf{d}\mu =\sum _{i=1}^{n}f({b}_{i})\cdot \mu ({\mathrm{\Omega}}_{i})$

where ${\mathrm{\Omega}}_{i}={f}^{-1}(\{{b}_{i}\})$. So but here I don't think that this has to do something with simple functions right?

Therefore I wanted to ask you if someone could explain me why we can see this sum as the integral of $a$ over $\mathrm{\Omega}$.

Let $\mathrm{\Omega}$ be countable and $a:\mathrm{\Omega}\to {\mathbb{R}}_{+}\cup \{\mathrm{\infty}\}$ be a map. Then we define

$\sum _{\mathrm{\Omega}}a(\omega ):=\underset{F\subset \mathrm{\Omega},|F|<\mathrm{\infty}}{sup}\sum _{F}a(\omega )$

Now our Prof said that we can consider $\sum _{\mathrm{\Omega}}a(\omega )$ as the integral of math xmlns="http://www.w3.org/1998/Math/MathML">

But somehow I don't see why this sum is equal to the integral. So I know from measure theory that if we have a simple function $f:\mathrm{\Omega}\to {\mathbb{R}}_{+}\cup \{\mathrm{\infty}\}$ where $f(\mathrm{\Omega})=\{{b}_{1},...,{b}_{n}\}$ finately many then

${\int}_{\mathrm{\Omega}}f\text{}\mathsf{d}\mu =\sum _{i=1}^{n}f({b}_{i})\cdot \mu ({\mathrm{\Omega}}_{i})$

where ${\mathrm{\Omega}}_{i}={f}^{-1}(\{{b}_{i}\})$. So but here I don't think that this has to do something with simple functions right?

Therefore I wanted to ask you if someone could explain me why we can see this sum as the integral of $a$ over $\mathrm{\Omega}$.

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