Round $604.2978$ to the hundredths place.

dannyboi2006tk
2022-10-02
Answered

Round $604.2978$ to the hundredths place.

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Piper Pruitt

Answered 2022-10-03
Author has **9** answers

$604.2978$

The 9 is in the hundredths place.

So if we look at the numbers to the right of it, they are 78. So we have to round up.

So $604.2978$ rounded to the nearest hundredths is $604.30$.

$604.2978$ rounded to the nearest hundredths is NOT $604.3$.

The 9 is in the hundredths place.

So if we look at the numbers to the right of it, they are 78. So we have to round up.

So $604.2978$ rounded to the nearest hundredths is $604.30$.

$604.2978$ rounded to the nearest hundredths is NOT $604.3$.

asked 2022-07-01

Let ${I}_{1},{I}_{2}...$ be an arrangement of the intervals $[\frac{i-1}{{2}^{n}},\frac{i}{{2}^{n}})$ in a sequence. If ${X}_{n}=n$ on ${I}_{n}$ and 0 elsewhere then $su{p}_{n}E{X}_{n}<\mathrm{\infty}$ but $P(\underset{n}{sup}{X}_{n}<\mathrm{\infty})=0$. My basic space is [0,1] with Lebesgue measure.

$A:=\{\underset{n}{sup}{X}_{n}<\mathrm{\infty}\}=\{\omega \in \mathrm{\Omega}:\mathrm{\exists}M>0,\mathrm{\forall}n\in \mathbb{N},{X}_{n}(\omega )\le M\}$

hence the complement is

$\begin{array}{rl}\overline{A}& =\{\omega \in \mathrm{\Omega}:\mathrm{\forall}M\in \mathbb{N},\mathrm{\exists}n\in \mathbb{N},{X}_{n}(\omega )>M\}\\ & =\bigcap _{M\in \mathbb{N}}\bigcup _{n\in \mathbb{N}}\{{X}_{n}>M\}\\ & =\bigcap _{M\in \mathbb{N}}\bigcup _{n\ge M}\{{X}_{n}>M\}\end{array}$

How do i go on from here?

$A:=\{\underset{n}{sup}{X}_{n}<\mathrm{\infty}\}=\{\omega \in \mathrm{\Omega}:\mathrm{\exists}M>0,\mathrm{\forall}n\in \mathbb{N},{X}_{n}(\omega )\le M\}$

hence the complement is

$\begin{array}{rl}\overline{A}& =\{\omega \in \mathrm{\Omega}:\mathrm{\forall}M\in \mathbb{N},\mathrm{\exists}n\in \mathbb{N},{X}_{n}(\omega )>M\}\\ & =\bigcap _{M\in \mathbb{N}}\bigcup _{n\in \mathbb{N}}\{{X}_{n}>M\}\\ & =\bigcap _{M\in \mathbb{N}}\bigcup _{n\ge M}\{{X}_{n}>M\}\end{array}$

How do i go on from here?

asked 2022-06-24

Pratt's Lemma is : $\xi ,\eta ,\zeta $ and ${\xi}_{n},{\eta}_{n},{\zeta}_{n}$ such that:

${\xi}_{n}\to \xi ,{\eta}_{n}\to \eta ,{\zeta}_{n}\to \zeta ,\text{convergence in probability}$

and ${\eta}_{n}\le {\xi}_{n}\le {\zeta}_{n}$, $E{\zeta}_{n}\to E\zeta ,E{\eta}_{n}\to E\eta $, and $E\zeta ,E\eta ,E\xi $ are finite, prove :

If ${\eta}_{n}\le 0\le {\zeta}_{n}$, then $E|{\xi}_{n}-\xi |\to 0$.

I know how to prove $E{\xi}_{n}\to E\xi $, but $E|{\xi}_{n}-\xi |\to 0$ seems not easy proved from it.

My first question is how to prove $E|{\xi}_{n}-\xi |\to 0$.

And I don't know why should emphasize the condition "${\eta}_{n}\le 0\le {\zeta}_{n}$", so my second question is: Is there any example that $E|{\xi}_{n}-\xi |\to 0$ is wrong if the condition is not fullfilled.

${\xi}_{n}\to \xi ,{\eta}_{n}\to \eta ,{\zeta}_{n}\to \zeta ,\text{convergence in probability}$

and ${\eta}_{n}\le {\xi}_{n}\le {\zeta}_{n}$, $E{\zeta}_{n}\to E\zeta ,E{\eta}_{n}\to E\eta $, and $E\zeta ,E\eta ,E\xi $ are finite, prove :

If ${\eta}_{n}\le 0\le {\zeta}_{n}$, then $E|{\xi}_{n}-\xi |\to 0$.

I know how to prove $E{\xi}_{n}\to E\xi $, but $E|{\xi}_{n}-\xi |\to 0$ seems not easy proved from it.

My first question is how to prove $E|{\xi}_{n}-\xi |\to 0$.

And I don't know why should emphasize the condition "${\eta}_{n}\le 0\le {\zeta}_{n}$", so my second question is: Is there any example that $E|{\xi}_{n}-\xi |\to 0$ is wrong if the condition is not fullfilled.

asked 2022-06-16

If $F$ is a continuous distribution function on $(\mathbb{R},\mathcal{B},{\mu}_{\mathcal{L}})$ with distribution ${\mu}_{F}$, use Fubini's theorem to show that

1. ${\int}_{\mathbb{R}}F(x)\phantom{\rule{thinmathspace}{0ex}}d{\mu}_{F}(x)=\frac{1}{2}$

2. If ${X}_{1},{X}_{2}$ are i.i.d random variables with common distribution $F$, then $P(\{{X}_{1}\le {X}_{2}\})=1/2$ and $\text{E}(F({X}_{1}))=1/2$.

My Attempt:

I don't really understand bs_math's answer so I have been trying to write my own. I just deleted an attempt here that was (I think) completely nonsensical. I am working on another attempt.

For example, I don't understand what's going on in line 4 of bs_math's answer.

1. ${\int}_{\mathbb{R}}F(x)\phantom{\rule{thinmathspace}{0ex}}d{\mu}_{F}(x)=\frac{1}{2}$

2. If ${X}_{1},{X}_{2}$ are i.i.d random variables with common distribution $F$, then $P(\{{X}_{1}\le {X}_{2}\})=1/2$ and $\text{E}(F({X}_{1}))=1/2$.

My Attempt:

I don't really understand bs_math's answer so I have been trying to write my own. I just deleted an attempt here that was (I think) completely nonsensical. I am working on another attempt.

For example, I don't understand what's going on in line 4 of bs_math's answer.

asked 2022-05-24

Let $\mu ,{v}_{n}$ be measures on $(X,\mathcal{B})$ with $\mu $ being $\sigma $-finite and $sup\{{v}_{n}(X)\}<\mathrm{\infty}$. Note that $v=\sum _{n\ge 1}{2}^{-n}{v}_{n}<<\mu $ if and only if ${v}_{n}<<\mu $. Let the Lebesgue decomposition of ${v}_{n}$ with respect to $\mu $ be ${v}_{n}={v}_{n,a}+{v}_{n,s}$, where $\frac{d{v}_{n,a}}{d\mu}={f}_{n}$.

Note that $v$ is a finite (hence $\sigma $-finite) measure, so it has a Lebesgue decomposition. I want to find the Lebesgue decomposition of $v$ with respect to $\mu $ and calculate $\frac{d{v}_{a}}{d\mu}$. I believe the Lebesgue decomposition will just be $v={v}_{a}+{v}_{s}=\sum _{n\ge 1}{2}^{-n}{v}_{n,a}+\sum _{n\ge 1}{2}^{-n}{v}_{n,s}$. Then my question is this:

Are Radon Nikodym derivatives countably additive? Can I say that $\frac{d{v}_{a}}{d\mu}=\sum _{n\ge 1}{2}^{-n}{f}_{n}$? Thanks.

Note that $v$ is a finite (hence $\sigma $-finite) measure, so it has a Lebesgue decomposition. I want to find the Lebesgue decomposition of $v$ with respect to $\mu $ and calculate $\frac{d{v}_{a}}{d\mu}$. I believe the Lebesgue decomposition will just be $v={v}_{a}+{v}_{s}=\sum _{n\ge 1}{2}^{-n}{v}_{n,a}+\sum _{n\ge 1}{2}^{-n}{v}_{n,s}$. Then my question is this:

Are Radon Nikodym derivatives countably additive? Can I say that $\frac{d{v}_{a}}{d\mu}=\sum _{n\ge 1}{2}^{-n}{f}_{n}$? Thanks.

asked 2022-08-09

A tree is 62 inches tall. How tall is it in feet and inches?

asked 2022-04-09

The variable age has the

a. interval scale of measurement

b. ratio scale of measurement

c. ordinal scale of measurement

d. nominal scale of measurement

a. interval scale of measurement

b. ratio scale of measurement

c. ordinal scale of measurement

d. nominal scale of measurement

asked 2022-06-23

I have 2 position estimates (along with their measurement error) and a difference in time between estimates. I estimate velocity using

Velocity = (PosA - PosB)/DeltaT

I am trying to estimate the error in my velocity estimate, but I can't seem to find any ways to calculate this. I assume it has to use Sigma_PosA and Sigma_PosB. I would also assume it's relative to DeltaT and/or abs(PosA - PosB). What is the velocity measurement variance/standard deviation?

Velocity = (PosA - PosB)/DeltaT

I am trying to estimate the error in my velocity estimate, but I can't seem to find any ways to calculate this. I assume it has to use Sigma_PosA and Sigma_PosB. I would also assume it's relative to DeltaT and/or abs(PosA - PosB). What is the velocity measurement variance/standard deviation?