Clifland
2021-01-27
Answered

Draw a diagram to represent the situation. Start by drawing a segment to represent the distance between Sam and the base of the building. Place a point on the segment and label the portion from Sam to this point as 1.22 m and the other portion as 7.32 m. Then draw two right triangles that have right angles at the endpoints of the first segment. Label the height for Sams

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i1ziZ

Answered 2021-01-28
Author has **92** answers

The angle of reflection is the same from the mirror to Sam and from the mirror to the window so the two triangles are similar. The corresponding side lengths must then be proportional. Therefore:

The height of the window is then 10.92 m.

asked 2022-06-05

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Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous bounded function, $({X}_{k}{)}_{k}$ a sequence of i.i.d random variables such that

$$

Let $x\in \mathbb{R},{Y}_{k}=f(x+\sum _{q=1}^{k}{X}_{q}),{\mathcal{F}}_{k}=\sigma ({X}_{1},...,{X}_{k}).$

Verify that:

$$

and deduce that for every $y\in \text{supp}{P}_{{X}_{1}}:=\{u,\mathrm{\forall}r>0,{P}_{{X}_{1}}([y-r,y+r])>0\},f(x+y)=f(x).$

One way to prove the first question is to note that

$$

For part 2 we can deduce that for ${P}_{{X}_{1}}$-almost every $y\in \mathbb{R},f(x+y)=f(x),$, how to conclude using continuity that $f(x+y)=f(x)$ for every $y\in \text{supp}{P}_{{X}_{1}}$ ?

How to deduce part 2 ?

$$

Let $x\in \mathbb{R},{Y}_{k}=f(x+\sum _{q=1}^{k}{X}_{q}),{\mathcal{F}}_{k}=\sigma ({X}_{1},...,{X}_{k}).$

Verify that:

$$

and deduce that for every $y\in \text{supp}{P}_{{X}_{1}}:=\{u,\mathrm{\forall}r>0,{P}_{{X}_{1}}([y-r,y+r])>0\},f(x+y)=f(x).$

One way to prove the first question is to note that

$$

For part 2 we can deduce that for ${P}_{{X}_{1}}$-almost every $y\in \mathbb{R},f(x+y)=f(x),$, how to conclude using continuity that $f(x+y)=f(x)$ for every $y\in \text{supp}{P}_{{X}_{1}}$ ?

How to deduce part 2 ?

asked 2022-04-10

What is the equation of line that passes through the point $(-7,-1)$ and has a slope of $1$?

asked 2022-05-21

Meaning of $\frac{P(X\cap Y)}{P(X)P(Y)}$

Imagine that we have a set $\mathrm{\Omega}$ and X and Y are events that can happen, I mean, $P(X),P(Y)>0$. Then, what does it mean the ratio $\frac{P(X\cap Y)}{P(X)P(Y)}$?

I know that $\frac{P(X\cap Y)}{P(X)P(Y)}=\frac{P(X|Y)}{P(X)}=\frac{P(Y|X)}{P(Y)}$ and if that ratio is equal to 1 then X,Y are independent events, but I can't figure out what exactly it means... please give simple examples.

I found this when reading about lift-data mining.

Imagine that we have a set $\mathrm{\Omega}$ and X and Y are events that can happen, I mean, $P(X),P(Y)>0$. Then, what does it mean the ratio $\frac{P(X\cap Y)}{P(X)P(Y)}$?

I know that $\frac{P(X\cap Y)}{P(X)P(Y)}=\frac{P(X|Y)}{P(X)}=\frac{P(Y|X)}{P(Y)}$ and if that ratio is equal to 1 then X,Y are independent events, but I can't figure out what exactly it means... please give simple examples.

I found this when reading about lift-data mining.

asked 2022-07-09

Let's say I have a family of Borel measures $({\mu}_{t}{)}_{t\in [0,1]}$ over ${R}^{d}$ such that the map $t\to {\mu}_{t}(B)$ is borel measurable for each Borel set $B$, and let's say I want to associate to this family a finite Borel measure $\mu $ over ${R}^{d}\times [0,1]$ clearly such that if $f\in {C}_{0}^{\mathrm{\infty}}({R}^{d}\times [0,1])$ the following holds:

${\int}_{{R}^{d}\times [0,1]}f(x,t)d\mu ={\int}_{0}^{1}{\int}_{{R}^{d}}f(x,t)d{\mu}_{t}dt.$

Is this always possible?

edit: the family ${\mu}_{t}$ is a family of finite borel measures

${\int}_{{R}^{d}\times [0,1]}f(x,t)d\mu ={\int}_{0}^{1}{\int}_{{R}^{d}}f(x,t)d{\mu}_{t}dt.$

Is this always possible?

edit: the family ${\mu}_{t}$ is a family of finite borel measures

asked 2020-12-17

Write a real-world problem that could be represented by the bar diagram below. Then solve your problem.

asked 2022-02-04

How do you evaluate the expression $8+6r$ given $r=7$ ?

asked 2022-06-27

Finding x with other variables involved

How do you find $x$ when there are also other variables like $P,A,t$

$P=\frac{A}{1+xt}$

Here's what I did, I'm not sure if I'm correct

LCD: 1 + xt

$1+xt(P=\frac{A}{1+xt})$

$(1+xt)P=1+xt(\frac{A}{1+xt})$

$P+Ptx=A$

Is the above correct? I couldn't continue it because I don't know what to do next.

How do you find $x$ when there are also other variables like $P,A,t$

$P=\frac{A}{1+xt}$

Here's what I did, I'm not sure if I'm correct

LCD: 1 + xt

$1+xt(P=\frac{A}{1+xt})$

$(1+xt)P=1+xt(\frac{A}{1+xt})$

$P+Ptx=A$

Is the above correct? I couldn't continue it because I don't know what to do next.