Henry eats 1/3 of a loaf of bread in 2 days. How many loaves of bread will he eat in 12 days?

Raegan Bray
2022-07-21
Answered

Henry eats 1/3 of a loaf of bread in 2 days. How many loaves of bread will he eat in 12 days?

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eri1ti0m

Answered 2022-07-22
Author has **11** answers

Answer:

2 loaves of bread.

Step-by-step explanation:

We have been given that Henry eats 1/3 of a loaf of bread in 2 days.

First of all, we will find amount of loaves eaten by Henry in one day by dividing 1/3 by 2 as:

$\frac{1}{2}\xf72\phantom{\rule{0ex}{0ex}}\frac{1}{2}\xf7\frac{2}{1}$

Convert in multiplication problem:

$\frac{1}{3}\times \frac{1}{2}$

$\frac{1}{6}$

To find the loaves of bread eaten by Henry in 12 days, we will multiply $\frac{1}{6}$ by 12.

$\frac{1}{6}\times 12=2$

2 loaves of bread.

Step-by-step explanation:

We have been given that Henry eats 1/3 of a loaf of bread in 2 days.

First of all, we will find amount of loaves eaten by Henry in one day by dividing 1/3 by 2 as:

$\frac{1}{2}\xf72\phantom{\rule{0ex}{0ex}}\frac{1}{2}\xf7\frac{2}{1}$

Convert in multiplication problem:

$\frac{1}{3}\times \frac{1}{2}$

$\frac{1}{6}$

To find the loaves of bread eaten by Henry in 12 days, we will multiply $\frac{1}{6}$ by 12.

$\frac{1}{6}\times 12=2$

asked 2022-07-03

All examples of a dense and co-dense set I have seen are either of full Lebesgue measure or of measure zero. For instance, in restriction to the unit interval $\mathbb{I}=[0\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{1px}{0ex}},\phantom{\rule{thinmathspace}{0ex}}1]$, we could have respectively $\mathbb{I}\cap \mathbb{Q}$ or $\mathbb{I}\setminus \mathbb{Q}$. What I am looking for is a dense and co-dense subset $A\subset \mathbb{I}$ such that

$\mathrm{m}(A)=\mathrm{m}(\mathbb{I}\setminus A)={\textstyle \frac{1}{2}}.$

I have attempted this task sequentially by, ever more finely, nibbling holes out of subintervals of $\mathbb{I}$ and partially back-filling the previously created holes. It's easy to approach half measure at each step, but I can't see how to to get convergence.

$\mathrm{m}(A)=\mathrm{m}(\mathbb{I}\setminus A)={\textstyle \frac{1}{2}}.$

I have attempted this task sequentially by, ever more finely, nibbling holes out of subintervals of $\mathbb{I}$ and partially back-filling the previously created holes. It's easy to approach half measure at each step, but I can't see how to to get convergence.

asked 2022-07-08

I have seen this line in a computation and I wonder if that makes sense : $\mathbb{E}(X|Y)=\int \mathbb{E}(X|Y=y)dP(Y=y)$ where P is a probability measure, $X,Y$ are random variables, so is $\mathbb{E}(X|Y)$. But I'm not familiar with this notation. If this is right, how can I understand it ?

asked 2022-04-12

At work, there is a statistical process going on, which I feel is probably mathematically incorrect but can't quite put my finger on what is wrong:

They are totalling up the number of hours people work per week (in minimum units of 15 minutes), and then producing averages for the whole department per week. Obviously, the results come out to be non-integer numbers of hours with a long number of decimals.

Then, they are judging the result of certain productivity-boosting techniques and displaying the findings in "minutes gained/lost"...in some cases producing productivity gains of as little as a minute or two minutes per week.

So to summarise, they are calculating units of quarters of an hour, but then presenting the average productivity gains in minutes...is this presuming an accuracy which is not present in the initial measurement? I think it is, but don't know how to argue it to my boss.

They are totalling up the number of hours people work per week (in minimum units of 15 minutes), and then producing averages for the whole department per week. Obviously, the results come out to be non-integer numbers of hours with a long number of decimals.

Then, they are judging the result of certain productivity-boosting techniques and displaying the findings in "minutes gained/lost"...in some cases producing productivity gains of as little as a minute or two minutes per week.

So to summarise, they are calculating units of quarters of an hour, but then presenting the average productivity gains in minutes...is this presuming an accuracy which is not present in the initial measurement? I think it is, but don't know how to argue it to my boss.

asked 2022-06-21

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Let $\u0427$ and $Y$ be absolutely continuous and discrete random variables, respectively. Does random vector $(X,Y)$ have to be absolutely continuous with respect to the product measure on the respective supports of $\u0427$ and $Y$?

I think this is true because the following equality holds

$$

so ${f}_{X,Y}(x,y)$ is the probability density function with respect to the product measure. However, I am not able to prove it formally. I would appreciate some help on this. Thanks.

I think this is true because the following equality holds

$$

so ${f}_{X,Y}(x,y)$ is the probability density function with respect to the product measure. However, I am not able to prove it formally. I would appreciate some help on this. Thanks.

asked 2022-05-23

I was thinking of the example in Folland on page 61 i.e. $\mu (\mathbb{R})=\mathrm{\infty}$. Let ${f}_{n}=n{\chi}_{[0,1/n]}\to 0$ a.e. Then ${f}_{n}\to 0$ in measure. My inclination is that this is true so it requires proof. Also the same question is asked but for when $\mu (X)=1$. My inclination is that this is not true and a counterexample can be provided.

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 2022-06-24

In the context of the Lévy-Khintchine formula, I have certain integral

$\begin{array}{}\text{(1)}& {\int}_{{\mathbb{R}}^{p}}{\textstyle (}{e}^{i{t}^{\prime}x}-1-i\frac{{t}^{\prime}x}{1-{x}^{\prime}x}{\textstyle )}d\nu (x),\phantom{\rule{1em}{0ex}}t\in {\mathbb{R}}^{p}\end{array}$

where $\nu $ is a measure defined on ${\mathbb{R}}^{p}$.

Suppose that

${\int}_{{\mathbb{R}}^{p}}|x{|}^{2}d\nu (x)<\mathrm{\infty}$

Define:

$\kappa (E)={\int}_{E}|x{|}^{2}d\nu (x),\phantom{\rule{1em}{0ex}}E\subset {\mathbb{R}}^{p}.$

I want to get $d\kappa (x)$ as a function of $d\nu (x)$ in order to do a substitution in (1), but I don't know how to differentiate $\kappa $:

$d\kappa (x)=?d\nu (x)$

$\begin{array}{}\text{(1)}& {\int}_{{\mathbb{R}}^{p}}{\textstyle (}{e}^{i{t}^{\prime}x}-1-i\frac{{t}^{\prime}x}{1-{x}^{\prime}x}{\textstyle )}d\nu (x),\phantom{\rule{1em}{0ex}}t\in {\mathbb{R}}^{p}\end{array}$

where $\nu $ is a measure defined on ${\mathbb{R}}^{p}$.

Suppose that

${\int}_{{\mathbb{R}}^{p}}|x{|}^{2}d\nu (x)<\mathrm{\infty}$

Define:

$\kappa (E)={\int}_{E}|x{|}^{2}d\nu (x),\phantom{\rule{1em}{0ex}}E\subset {\mathbb{R}}^{p}.$

I want to get $d\kappa (x)$ as a function of $d\nu (x)$ in order to do a substitution in (1), but I don't know how to differentiate $\kappa $:

$d\kappa (x)=?d\nu (x)$