The average of the salaries of Tim, Maida, and Aaron is $34,000 per year. Maida earns $10,000 more than Tim, and Aaron's salary is $8,000 less than twice Tim's salary. Please find the salary of each person.

Francisco Proctor
2022-08-01
Answered

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polishxcore5z

Answered 2022-08-02
Author has **14** answers

Step 1

Average salary of Tim, Maida and Aaron $=\$34000$

Average salary $=\frac{\text{Total salary}}{\text{No. of persons}}$

$Tim+Maida+Aaron=34000\times 3$

$T+M+A=102000--(1)$

Maida earn 10000 more than Tim.

$\therefore M=T+10000--(2)$

Aaron salary is 8000 less than twice Tim salary

$A=2T-8000--(3)$

Put value of M and A in (1)

$T+(T+10000)+(2T-8000)=102000\phantom{\rule{0ex}{0ex}}4T+2000=102000\phantom{\rule{0ex}{0ex}}4T=100000\phantom{\rule{0ex}{0ex}}T=25000$

Step 2

$M=25000+10000\phantom{\rule{0ex}{0ex}}M=35000$

$A=2(25000)-8000\phantom{\rule{0ex}{0ex}}A=42000$

Salary $Tim=\$25000\phantom{\rule{0ex}{0ex}}Maida=\$35000\phantom{\rule{0ex}{0ex}}Aaron=\$42000$

Average salary of Tim, Maida and Aaron $=\$34000$

Average salary $=\frac{\text{Total salary}}{\text{No. of persons}}$

$Tim+Maida+Aaron=34000\times 3$

$T+M+A=102000--(1)$

Maida earn 10000 more than Tim.

$\therefore M=T+10000--(2)$

Aaron salary is 8000 less than twice Tim salary

$A=2T-8000--(3)$

Put value of M and A in (1)

$T+(T+10000)+(2T-8000)=102000\phantom{\rule{0ex}{0ex}}4T+2000=102000\phantom{\rule{0ex}{0ex}}4T=100000\phantom{\rule{0ex}{0ex}}T=25000$

Step 2

$M=25000+10000\phantom{\rule{0ex}{0ex}}M=35000$

$A=2(25000)-8000\phantom{\rule{0ex}{0ex}}A=42000$

Salary $Tim=\$25000\phantom{\rule{0ex}{0ex}}Maida=\$35000\phantom{\rule{0ex}{0ex}}Aaron=\$42000$

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Say we fix an i.i.d. probability measure $\mu $ on ${\mathbb{R}}^{\mathbb{Z}}$. Precisely, $\mu (\prod {A}_{i})=\prod \nu ({A}_{i})$ for some probability measure $\nu $ on $\mathbb{R}$.

Is there a nice, concrete proof that $\mu $ is tight (independent of $\nu $)?

The reason I ask is tightness is somewhat counter-intuitive here. For instance, if $\nu $ has unbounded support, then, for each $M>0$, the set $\{k\in \mathbb{Z}\phantom{\rule{thinmathspace}{0ex}}\mid \phantom{\rule{thinmathspace}{0ex}}|{x}_{k}|\ge M\}$ is infinite almost surely under $\mu $ (by the 0-1 law). Thinking along those lines, it's not easy to see ${K}_{\u03f5}$ intuitively.

A classical result on measure theory in Polish spaces tells us that any probability measure $\mu $ on a Polish space $X$ is tight: given $\u03f5>0$, then there is a compact set ${K}_{\u03f5}\subseteq X$ such that $\mu ({K}_{\u03f5})>1-\u03f5$.

Hence any product probability measure on ${\mathbb{R}}^{\mathbb{Z}}$ is tight.

Say we fix an i.i.d. probability measure $\mu $ on ${\mathbb{R}}^{\mathbb{Z}}$. Precisely, $\mu (\prod {A}_{i})=\prod \nu ({A}_{i})$ for some probability measure $\nu $ on $\mathbb{R}$.

Is there a nice, concrete proof that $\mu $ is tight (independent of $\nu $)?

The reason I ask is tightness is somewhat counter-intuitive here. For instance, if $\nu $ has unbounded support, then, for each $M>0$, the set $\{k\in \mathbb{Z}\phantom{\rule{thinmathspace}{0ex}}\mid \phantom{\rule{thinmathspace}{0ex}}|{x}_{k}|\ge M\}$ is infinite almost surely under $\mu $ (by the 0-1 law). Thinking along those lines, it's not easy to see ${K}_{\u03f5}$ intuitively.

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