Milk weighs 8.605 pounds to the gallon. How much does 12.2 gallons weigh?

Janessa Benson
2022-10-08
Answered

Milk weighs 8.605 pounds to the gallon. How much does 12.2 gallons weigh?

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Mario Monroe

Answered 2022-10-09
Author has **12** answers

12.2 gallons of milk weigh 104.981 pounds.

You get this by simply multiplying the weight of 1 gallon of milk to the 12.2 gallons of milk. The answer is the weight of 12.2 gallons of milk.

You get this by simply multiplying the weight of 1 gallon of milk to the 12.2 gallons of milk. The answer is the weight of 12.2 gallons of milk.

asked 2022-05-24

Suppose ${Y}_{1},{Y}_{2},\dots $ is any sequence of iid real valued random variables with $E({Y}_{1})=\mathrm{\infty}$ . Show that, almost surely, $\underset{n}{lim\u2006sup}(|{Y}_{n}|/n)=\mathrm{\infty}$ and $\underset{n}{lim\u2006sup}(|{Y}_{1}+...+{Y}_{n}|/n)=\mathrm{\infty}$.

I have solved the first part by considering non-negative integer iid r.vs ${X}_{n}=\text{floor}(|{Y}_{n}|)$ and using $E(X)=\sum _{0}^{\mathrm{\infty}}P(X\ge n)$ then doing some clever tricks so I can apply the (2nd) Borel-Cantelli lemma, but I'm not really sure how I can use the same approach to solve the second part seeing as it is tempting to set ${S}_{n}=\text{floor}(|{Y}_{1}+...+{Y}_{n}|)$ but then the ${S}_{i}$ are not iid. I'm pretty sure its gonna be Borel-Cantelli again (since limsup) so I need to come up with the right events. Please can someone nudge me in the right direction.

Hints only please

EDIT: Suppose $\underset{n}{lim\u2006sup}|{a}_{1}+...+{a}_{n}|/n<\mathrm{\infty}$. Then set ${S}_{n}=\sum _{1}^{n}{a}_{k}$

$\frac{|{a}_{n}|}{n}=\frac{|{S}_{n}-{S}_{n-1}|}{n}\phantom{\rule{0ex}{0ex}}\le \frac{|{S}_{n}|}{n}+\frac{|{S}_{n-1}|}{n-1}$

bounded

I have solved the first part by considering non-negative integer iid r.vs ${X}_{n}=\text{floor}(|{Y}_{n}|)$ and using $E(X)=\sum _{0}^{\mathrm{\infty}}P(X\ge n)$ then doing some clever tricks so I can apply the (2nd) Borel-Cantelli lemma, but I'm not really sure how I can use the same approach to solve the second part seeing as it is tempting to set ${S}_{n}=\text{floor}(|{Y}_{1}+...+{Y}_{n}|)$ but then the ${S}_{i}$ are not iid. I'm pretty sure its gonna be Borel-Cantelli again (since limsup) so I need to come up with the right events. Please can someone nudge me in the right direction.

Hints only please

EDIT: Suppose $\underset{n}{lim\u2006sup}|{a}_{1}+...+{a}_{n}|/n<\mathrm{\infty}$. Then set ${S}_{n}=\sum _{1}^{n}{a}_{k}$

$\frac{|{a}_{n}|}{n}=\frac{|{S}_{n}-{S}_{n-1}|}{n}\phantom{\rule{0ex}{0ex}}\le \frac{|{S}_{n}|}{n}+\frac{|{S}_{n-1}|}{n-1}$

bounded

asked 2022-07-03

A ring is closed under intersection. And delta-ring is closed under countable intersection. According to my understanding, if a family of sets is closed under intersection, than it should also be closed under countable intersection.

Let's say $R$ is a ring. To $R$ be a delta-ring, ${A}_{n}\cap {A}_{n+1}\cap {A}_{n+2}\cap ...$ should also be in $R$. And, as a ring is closed under intersection, ${A}_{n}\cap {A}_{n+1}$ is in $R$, then ${A}_{n}\cap {A}_{n+1}\cap {A}_{n+2}$ is in $R$, and so on. So, any ring is a delta-ring.

What am I missing? I guess I don't properly understand what countable intersection is. Thank you for reading!

Let's say $R$ is a ring. To $R$ be a delta-ring, ${A}_{n}\cap {A}_{n+1}\cap {A}_{n+2}\cap ...$ should also be in $R$. And, as a ring is closed under intersection, ${A}_{n}\cap {A}_{n+1}$ is in $R$, then ${A}_{n}\cap {A}_{n+1}\cap {A}_{n+2}$ is in $R$, and so on. So, any ring is a delta-ring.

What am I missing? I guess I don't properly understand what countable intersection is. Thank you for reading!

asked 2022-06-28

Let $u\in {\mathcal{L}}^{\mathcal{1}}\mathcal{(}\mu \mathcal{)}$, $T$ A measure preserving map s.t.: $\int (u)d\mu =\int u\circ Td\mu $. ${A}_{n,\u03f5}=\{\frac{|u({T}^{n}(x))|}{{n}^{2}}>\u03f5\}$. ${T}^{n}=T\circ T\circ T\dots $ n times.

Prove $\sum _{n\ge 1}\mu ({A}_{n,\u03f5})$ is finite.

My attempt:

By Markov inequality we get: $\mu ({A}_{n,\u03f5})\le \frac{1}{\u03f5}\int \frac{|u({T}^{n}(x))|}{{n}^{2}}d\mu $, since $T$ is measure preserving we get: $|u({T}^{n}(x))|\in {\mathcal{L}}^{\mathcal{1}}\mathcal{(}\mu \mathcal{)}$. Summing over n gives:

$\sum _{n\ge 1}\mu ({A}_{n,\u03f5)}\le \sum _{n\ge 1}{\u03f5}^{-1}{n}^{-2}\int |u({T}^{n}(x))|d\mu .$

This is the part where I am stuck. I know that all the integrals are finite but how do I ensure that summing over all the integrals stays finite? Can I make the integral independent of n? Or can I just state by assumption: $\int |u({T}^{n}(x))|d\mu =\int |u|d\mu $?

Prove $\sum _{n\ge 1}\mu ({A}_{n,\u03f5})$ is finite.

My attempt:

By Markov inequality we get: $\mu ({A}_{n,\u03f5})\le \frac{1}{\u03f5}\int \frac{|u({T}^{n}(x))|}{{n}^{2}}d\mu $, since $T$ is measure preserving we get: $|u({T}^{n}(x))|\in {\mathcal{L}}^{\mathcal{1}}\mathcal{(}\mu \mathcal{)}$. Summing over n gives:

$\sum _{n\ge 1}\mu ({A}_{n,\u03f5)}\le \sum _{n\ge 1}{\u03f5}^{-1}{n}^{-2}\int |u({T}^{n}(x))|d\mu .$

This is the part where I am stuck. I know that all the integrals are finite but how do I ensure that summing over all the integrals stays finite? Can I make the integral independent of n? Or can I just state by assumption: $\int |u({T}^{n}(x))|d\mu =\int |u|d\mu $?

asked 2022-05-19

Is there a logical way to distinguish what is assigned versus measured?

Or is this a scientific rather than a mathematical issue? If there is a logical distinction, it would be nice if it covered both frequentist and Bayesian concepts of probability.

(Maybe a starting point could be the distinction between the modes $X\to A$ versus $A\to X$ where, loosely, $A$ is better characterized than $X$)

Or is this a scientific rather than a mathematical issue? If there is a logical distinction, it would be nice if it covered both frequentist and Bayesian concepts of probability.

(Maybe a starting point could be the distinction between the modes $X\to A$ versus $A\to X$ where, loosely, $A$ is better characterized than $X$)

asked 2022-03-27

Define Unit of Measurement.

asked 2022-11-23

To fit between two windows the width of a bookshelf must be no greater than 6 1/2 feet Mrs Aguilar purchases a bookshelf that is 77 inches wide which statement doescribes the relationship between the width of the bookshelf and and the distance between the windows

asked 2022-05-29

Suppose we have a sequence of positive random variables ${X}_{1},{X}_{2},...,X$. I am trying to prove a characterization of almost sure convergence.

It states that ${X}_{n}\to X$ almost surely iff for every $\u03f5>0$, $\underset{n\to \mathrm{\infty}}{lim}P[\underset{k\ge n}{sup}\frac{{X}_{k}}{X}>1+\u03f5]=0$ and $\underset{n\to \mathrm{\infty}}{lim}P[\underset{k\ge n}{sup}\frac{X}{{X}_{k}}>1+\u03f5]=0$.

If I assume almost sure convergence, then the implication is easy but I am not being able to prove the other way round.

It states that ${X}_{n}\to X$ almost surely iff for every $\u03f5>0$, $\underset{n\to \mathrm{\infty}}{lim}P[\underset{k\ge n}{sup}\frac{{X}_{k}}{X}>1+\u03f5]=0$ and $\underset{n\to \mathrm{\infty}}{lim}P[\underset{k\ge n}{sup}\frac{X}{{X}_{k}}>1+\u03f5]=0$.

If I assume almost sure convergence, then the implication is easy but I am not being able to prove the other way round.