Draw, in standard position, the angle whose measure is given:

$-{315}^{\circ}$

$-{315}^{\circ}$

kolutastmr
2022-07-14
Answered

Draw, in standard position, the angle whose measure is given:

$-{315}^{\circ}$

$-{315}^{\circ}$

You can still ask an expert for help

Tamia Padilla

Answered 2022-07-15
Author has **16** answers

$-{315}^{\circ}$

Answer:

Answer:

Jeffrey Jordon

Answered 2022-07-26
Author has **2581** answers

asked 2022-05-21

Consider a linear measurement process with some noise:

$y=Hx+v$

with

$v\sim \mathcal{N}(0,\mathrm{\Sigma})$

the covariance matrix $\mathrm{\Sigma}$ is not a diagonal matrix. As we know, using LS, the $\hat{x}$ is

$\hat{x}=({H}^{T}H{)}^{-1}{H}^{T}y$

Does it change when $\mathrm{\Sigma}$ is not a diagonal matrix?

$y=Hx+v$

with

$v\sim \mathcal{N}(0,\mathrm{\Sigma})$

the covariance matrix $\mathrm{\Sigma}$ is not a diagonal matrix. As we know, using LS, the $\hat{x}$ is

$\hat{x}=({H}^{T}H{)}^{-1}{H}^{T}y$

Does it change when $\mathrm{\Sigma}$ is not a diagonal matrix?

asked 2022-06-14

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!

asked 2022-03-13

What is the relationship between the model (right measurement) and the actual object (left measurement) in each of these examples?

10 ft : 24 in$\text{Undefined control sequence boxempty}$

1 mi : 5,280 ft$\text{Undefined control sequence boxempty}$

20 cm : 1,000 mm$\text{Undefined control sequence boxempty}$

10 ft : 24 in

1 mi : 5,280 ft

20 cm : 1,000 mm

asked 2022-07-10

Let $(X,\mathcal{M},\mu )$ be a measure space with positive measure $\mu $ such that $\mu (X)=1$. Let $f$ be an integrable function. I have been working on a proof for $\underset{p\to \mathrm{\infty}}{lim}||f|{|}_{{L}^{p}}=||f|{|}_{{L}^{\mathrm{\infty}}}$, and I have it all worked out except one piece:

If $||f|{|}_{{L}^{p}}<\mathrm{\infty}$ for all $p\ge 1$, then $||f|{|}_{{L}^{\mathrm{\infty}}}<\mathrm{\infty}$

Here I am using the definition $||f|{|}_{{L}^{\mathrm{\infty}}}=inf{\textstyle \{}\lambda :|f|\le \lambda \phantom{\rule{thickmathspace}{0ex}}a.e.{\textstyle \}}$. Any help would be greatly appreciated!

If $||f|{|}_{{L}^{p}}<\mathrm{\infty}$ for all $p\ge 1$, then $||f|{|}_{{L}^{\mathrm{\infty}}}<\mathrm{\infty}$

Here I am using the definition $||f|{|}_{{L}^{\mathrm{\infty}}}=inf{\textstyle \{}\lambda :|f|\le \lambda \phantom{\rule{thickmathspace}{0ex}}a.e.{\textstyle \}}$. Any help would be greatly appreciated!

asked 2022-06-26

Let ${\mu}_{t}$ be the law of a random process ${X}_{t}$. Let $\nu $ and ${\nu}_{t}$ be arbitrary measures in $\mathrm{\Omega}$ with the sole restriction that ${\mu}_{t}$ be absolutely continuous w.r.t. ${\nu}_{t}$, and ${\nu}_{t}$ be absolutely continuous w.r.t. $\nu $.

Then (according to a paper I'm reading)

$\int {\mathrm{\partial}}_{t}\mathrm{log}\left(\frac{{\nu}_{t}(x)}{\nu (x)}\right)d{\nu}_{t}=\int \frac{{\mu}_{t}(x)}{{\nu}_{t}(x)}{\mathrm{\partial}}_{t}\left(\frac{{\nu}_{t}(x)}{\nu (x)}\right)d\nu =-\int {\mathrm{\partial}}_{t}\left(\frac{{\mu}_{t}(x)}{{\nu}_{t}(x)}\right)d{\nu}_{t}.$

The second inequality looks like an integration by parts, but I don't know what it means to take "$dv$" = ${\mathrm{\partial}}_{t}\left(\frac{{\nu}_{t}(x)}{\nu (x)}\right)d\nu $ where the "$dv$" derivative is with respect to a different variable (t) than the integral is taken over (\nu) -- usually "$dv$" is something like $\frac{d}{dx}v(x)dx$. And I don't understand the first inequality at all.

I'm also not sure what the notation $\frac{{\nu}_{t}(x)}{\nu (x)}$ is when we're talking about measures - I know it's not simply a fraction but something like a Radon-Nikodym derivative.

Then (according to a paper I'm reading)

$\int {\mathrm{\partial}}_{t}\mathrm{log}\left(\frac{{\nu}_{t}(x)}{\nu (x)}\right)d{\nu}_{t}=\int \frac{{\mu}_{t}(x)}{{\nu}_{t}(x)}{\mathrm{\partial}}_{t}\left(\frac{{\nu}_{t}(x)}{\nu (x)}\right)d\nu =-\int {\mathrm{\partial}}_{t}\left(\frac{{\mu}_{t}(x)}{{\nu}_{t}(x)}\right)d{\nu}_{t}.$

The second inequality looks like an integration by parts, but I don't know what it means to take "$dv$" = ${\mathrm{\partial}}_{t}\left(\frac{{\nu}_{t}(x)}{\nu (x)}\right)d\nu $ where the "$dv$" derivative is with respect to a different variable (t) than the integral is taken over (\nu) -- usually "$dv$" is something like $\frac{d}{dx}v(x)dx$. And I don't understand the first inequality at all.

I'm also not sure what the notation $\frac{{\nu}_{t}(x)}{\nu (x)}$ is when we're talking about measures - I know it's not simply a fraction but something like a Radon-Nikodym derivative.

asked 2022-07-03

I'm wondering whether the following statement holds:

Let ${f}_{n},f:\mathbb{R}\to {\mathbb{R}}_{0}^{+}$ be functions with $\int {f}_{n}(x)dx=\int f(x)dx=1$ and for every bounded and countinuous function $g:\mathbb{R}\to \mathbb{R}$ the following integral-convergence

$\int g(x)\cdot {f}_{n}(x)dx{\to}_{n}\int g(x)\cdot f(x)dx$

holds. Then it follows that ${f}_{n}\to f$ almost everywhere.

Intuitively the statement looks false, but I can't find a counterexample. If it doesn't hold: changes the further assumption that the ${f}_{n},f$ have to be continuous anything?

Kind regards

Let ${f}_{n},f:\mathbb{R}\to {\mathbb{R}}_{0}^{+}$ be functions with $\int {f}_{n}(x)dx=\int f(x)dx=1$ and for every bounded and countinuous function $g:\mathbb{R}\to \mathbb{R}$ the following integral-convergence

$\int g(x)\cdot {f}_{n}(x)dx{\to}_{n}\int g(x)\cdot f(x)dx$

holds. Then it follows that ${f}_{n}\to f$ almost everywhere.

Intuitively the statement looks false, but I can't find a counterexample. If it doesn't hold: changes the further assumption that the ${f}_{n},f$ have to be continuous anything?

Kind regards

asked 2021-03-04

Determine the level of measurement of the variable.

Favorite color

Choose the correct level of measurement.

A) Interval

B) Nominal

C) Ratio

D) Ordinal

Favorite color

Choose the correct level of measurement.

A) Interval

B) Nominal

C) Ratio

D) Ordinal