You room is 5 meters wide and 6 meter long. What is your room's area in square feet? (1 meter=3.28ft)

bergvolk0k
2022-10-14
Answered

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megagoalai

Answered 2022-10-15
Author has **22** answers

Given

Width (wide) of a room $=5m=5\times 3.28ft=16.4ft$

length of a room $=6m=6\times 3.28ft=19.68ft$

Since room has a rectangular shape

Since area of rectangular $shape=length\times width$

$=19.68ft\times 16.4ft$

$=322.752f{t}^{2}$

Hence, areas of room $=323f{t}^{2}$

Therefore, option second from the top is the answer

Width (wide) of a room $=5m=5\times 3.28ft=16.4ft$

length of a room $=6m=6\times 3.28ft=19.68ft$

Since room has a rectangular shape

Since area of rectangular $shape=length\times width$

$=19.68ft\times 16.4ft$

$=322.752f{t}^{2}$

Hence, areas of room $=323f{t}^{2}$

Therefore, option second from the top is the answer

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Who knows

If I define algebra $\mathcal{F}(A)$ generated by $A$, collection of subsets of $S$ (the universal set) as the intersection of $\mathcal{F}$, algebra superset of $A$:

$\mathcal{F}(A)=\bigcap _{algebra\text{}\mathcal{F}\supseteq A}\mathcal{F}$

What if $A$ is an infinite (either countable or uncountable) set? Algebra, unlike $\sigma $-algebra, guarantees being closed under finite Boolean operations. Here, finite(in the definition of algebra) and infinite(in the setting) confuses me. e.g. $A$ is the collection of intervals $(-\mathrm{\infty},x]$($x\in \mathbb{R}$) and $S=\mathbb{R}$, what $\mathcal{F}(A)$ be like? Any help will be appreciated!

If I define algebra $\mathcal{F}(A)$ generated by $A$, collection of subsets of $S$ (the universal set) as the intersection of $\mathcal{F}$, algebra superset of $A$:

$\mathcal{F}(A)=\bigcap _{algebra\text{}\mathcal{F}\supseteq A}\mathcal{F}$

What if $A$ is an infinite (either countable or uncountable) set? Algebra, unlike $\sigma $-algebra, guarantees being closed under finite Boolean operations. Here, finite(in the definition of algebra) and infinite(in the setting) confuses me. e.g. $A$ is the collection of intervals $(-\mathrm{\infty},x]$($x\in \mathbb{R}$) and $S=\mathbb{R}$, what $\mathcal{F}(A)$ be like? Any help will be appreciated!

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$[0,1]\subset \bigcup _{k}{\mathcal{N}}_{\mathcal{k}}\subset [-1,2]$

Where ${\mathcal{N}}_{k}=\mathcal{N}+{r}_{k}$

I get that by monotonicity, we have

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But I do not see the contradiction as to why since neither $m(\mathcal{N})=0$ nor $m(\mathcal{N})>0$ we are done by contradiction, can't the measure be strictly between 1 and 3? so 2?

$[0,1]\subset \bigcup _{k}{\mathcal{N}}_{\mathcal{k}}\subset [-1,2]$

Where ${\mathcal{N}}_{k}=\mathcal{N}+{r}_{k}$

I get that by monotonicity, we have

$1\le \sum _{k\in \mathbb{N}}m({\mathcal{N}}_{k})\le 3$

But I do not see the contradiction as to why since neither $m(\mathcal{N})=0$ nor $m(\mathcal{N})>0$ we are done by contradiction, can't the measure be strictly between 1 and 3? so 2?

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However, in this question, follows the above result, we will have ${2}^{{2}^{3}}=256$ elements in the smallest field $\mathcal{F}$. How can we represent it?

However, in this question, follows the above result, we will have ${2}^{{2}^{3}}=256$ elements in the smallest field $\mathcal{F}$. How can we represent it?