How can I convert square root of 32 into a decimal without using a calculator?

Zack Chase
2022-09-25
Answered

How can I convert square root of 32 into a decimal without using a calculator?

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hampiova76

Answered 2022-09-26
Author has **5** answers

$\sqrt{32}=\sqrt{16\cdot 2}=4\sqrt{2}=4\cdot 1.414=5.656$

$\sqrt{2}=1.414\phantom{\rule{0ex}{0ex}}\sqrt{3}=1.732\phantom{\rule{0ex}{0ex}}\sqrt{5}=2.236$

$\sqrt{2}=1.414\phantom{\rule{0ex}{0ex}}\sqrt{3}=1.732\phantom{\rule{0ex}{0ex}}\sqrt{5}=2.236$

asked 2022-07-04

Let $({E}_{\alpha}{)}_{\alpha \in I}$, where $I$ is an index set (for example, [0,1]). Can we define the corresponding liminf, for example,

$\underset{\alpha \searrow 0}{lim\u2006inf}{E}_{\alpha}=\bigcup _{r>0}\bigcap _{0<\alpha <r}{E}_{\alpha}.$

My desired result is the following continuous parameter type Fatou's lemma: Let $\mu $ be a finite meaure and each ${E}_{\alpha}$ is measurable, then

$\mu (\underset{\alpha \searrow 0}{lim\u2006inf}{E}_{\alpha})\le \underset{\alpha \searrow 0}{lim\u2006inf}\mu ({E}_{\alpha}).$

I am not sure whether these make sense. Any comments are welcome! Many thanks!

$\underset{\alpha \searrow 0}{lim\u2006inf}{E}_{\alpha}=\bigcup _{r>0}\bigcap _{0<\alpha <r}{E}_{\alpha}.$

My desired result is the following continuous parameter type Fatou's lemma: Let $\mu $ be a finite meaure and each ${E}_{\alpha}$ is measurable, then

$\mu (\underset{\alpha \searrow 0}{lim\u2006inf}{E}_{\alpha})\le \underset{\alpha \searrow 0}{lim\u2006inf}\mu ({E}_{\alpha}).$

I am not sure whether these make sense. Any comments are welcome! Many thanks!

asked 2022-06-06

Measurements of some variable X were made at an interval of 1 minute from 10 A.M. to 10:20 A.M.. The data, thus, obtained is as follows:

X:60,62,65,64,63,61,66,65,70,68,63,62,64,69,65,64,66,67,66,64,50.

The value of $X$ which is exceeded 50% of the time in the duration of measurement, is

(i) 69

(ii) 68

(iii) 67

(iv) 66

Answer:

It is a multiple choice question and it is general knowledge question.

The answer is 68.

But I am unable to understand the question. What exactly says the question?

X:60,62,65,64,63,61,66,65,70,68,63,62,64,69,65,64,66,67,66,64,50.

The value of $X$ which is exceeded 50% of the time in the duration of measurement, is

(i) 69

(ii) 68

(iii) 67

(iv) 66

Answer:

It is a multiple choice question and it is general knowledge question.

The answer is 68.

But I am unable to understand the question. What exactly says the question?

asked 2022-03-25

Determine whether the data described below are qualitative or quantitative and explain why.

The heights of subjects in a clinical trial of a new drug

Choose the correct answer below.

A. The data are quantitative because they consist of counts or measurements.

B. The data are qualitative because they consist of counts or measurements.

C. The data are qualitative because they don't measure or count anything.

D. The data are quantitative because they don't measure or count anything.

The heights of subjects in a clinical trial of a new drug

Choose the correct answer below.

A. The data are quantitative because they consist of counts or measurements.

B. The data are qualitative because they consist of counts or measurements.

C. The data are qualitative because they don't measure or count anything.

D. The data are quantitative because they don't measure or count anything.

asked 2022-04-05

Find the temperature at which the Celsius measurement and the Fahrenheit measurement are the same number.

asked 2022-06-24

Suppose I have two positive Borel measures on $[0,\mathrm{\infty}]$, say $\mu $ and $\nu $ and I know that $\nu ([0,x])=0$ implies $\mu ([0,x])=0$ for any $x$ and that $\nu ((a,b])=0$ implies $\mu ((a,b])=0$ for any pair $a,b$.

Does this imply that $\mu <<\nu $? I would think so but I have trouble proving it, I thought that the monotone class theorem was the answer but I don't see how the sets with the property $\nu (E)=0\u27f9\mu (E)=0$ are closed under monotone intersections.

We may assume that the measures are finite, if that helps anything.

Does this imply that $\mu <<\nu $? I would think so but I have trouble proving it, I thought that the monotone class theorem was the answer but I don't see how the sets with the property $\nu (E)=0\u27f9\mu (E)=0$ are closed under monotone intersections.

We may assume that the measures are finite, if that helps anything.

asked 2022-05-23

A really strange question here in my opinion:

Let $A,B$ be two measurable subsets of ${\mathbb{R}}^{1}$. Define $f(x)=|(A-x)\cap B|$. Evaluate ${\int}_{{\mathbb{R}}^{1}}fdx$. Here $|\cdot |$ refers to the measure.

Here $f$ is clearly non-negative, so I tried using the definition of Lebesgue integral:

${\int}_{{\mathbb{R}}^{1}}fdx=sup\{{\int}_{{\mathbb{R}}^{1}}sdx:0\le s\le f\},$

where s are simple functions. But I realised that I couldn't come up with any simple functions.

I tried breaking up ${\mathbb{R}}^{1}$ into two parts:

${E}_{1}=\{y\in {\mathbb{R}}^{1}:y\in A\cap B\},{E}_{2}=\{y\in {\mathbb{R}}^{1}:y\notin A\cap B\}$

and then considering the sum

${\int}_{{E}_{1}}fdx+{\int}_{{E}_{2}}fdx$

and then I'm clueless as to how to proceed. I'm guessing the answer should be something intuitive like $|A\cap B|$. Any help would be appreciated.

Let $A,B$ be two measurable subsets of ${\mathbb{R}}^{1}$. Define $f(x)=|(A-x)\cap B|$. Evaluate ${\int}_{{\mathbb{R}}^{1}}fdx$. Here $|\cdot |$ refers to the measure.

Here $f$ is clearly non-negative, so I tried using the definition of Lebesgue integral:

${\int}_{{\mathbb{R}}^{1}}fdx=sup\{{\int}_{{\mathbb{R}}^{1}}sdx:0\le s\le f\},$

where s are simple functions. But I realised that I couldn't come up with any simple functions.

I tried breaking up ${\mathbb{R}}^{1}$ into two parts:

${E}_{1}=\{y\in {\mathbb{R}}^{1}:y\in A\cap B\},{E}_{2}=\{y\in {\mathbb{R}}^{1}:y\notin A\cap B\}$

and then considering the sum

${\int}_{{E}_{1}}fdx+{\int}_{{E}_{2}}fdx$

and then I'm clueless as to how to proceed. I'm guessing the answer should be something intuitive like $|A\cap B|$. Any help would be appreciated.

asked 2022-07-03

Setting We work on a filtered probability space with finite time horizon $T$. The filtration is assumed to be complete. Let $X$ be a stochastic process that satisfies a property (A) a.s. For example, if property (A) is being nonnegative at the time $T$, then $X$ satisfies $P[{X}_{T}\ge 0]=1$.

Question I would like to obtain that $X$ (up to indistinguishability) satisfies property (A) for each $\omega $.

My attempt Define a process $Y$ to be equal to 0 on the event $\{{X}_{T}<0\}$ and define it to be equal to $X$ on $\{{X}_{T}\ge 0\}$. Then ${Y}_{T}\ge 0$. Is it then correct that $X,Y$ are indistinguishable, i.e. $P[{X}_{t}={Y}_{t},\mathrm{\forall}t\in [0,T]]=1$? Is this approach possible whenever the event where the stochastic process does not satisfy the desired propert has probability 0?

Question I would like to obtain that $X$ (up to indistinguishability) satisfies property (A) for each $\omega $.

My attempt Define a process $Y$ to be equal to 0 on the event $\{{X}_{T}<0\}$ and define it to be equal to $X$ on $\{{X}_{T}\ge 0\}$. Then ${Y}_{T}\ge 0$. Is it then correct that $X,Y$ are indistinguishable, i.e. $P[{X}_{t}={Y}_{t},\mathrm{\forall}t\in [0,T]]=1$? Is this approach possible whenever the event where the stochastic process does not satisfy the desired propert has probability 0?