help-27

[willow helm]

Yes. and what is -b/a?

And c/a

[verbal barn]

-7/1

[willow helm]

What's a, b and c

[verbal barn]

12 / 1

[willow helm]

We have x^2 + 9x + 20?

[verbal barn]

oh ya

sry i was looking at diff question

[willow helm]

oh

no problem

[verbal barn]

alpha + beta = - 9/1

[willow helm]

Yes

$\alpha + \beta =-9$

[verbal barn]

alpha x beta = 20/1

[willow helm]

10?

Yes

$\alpha \cdot \beta = 20$

[willow helm]

Now let's look at

$\frac{1}{\alpha} + \frac{1}{\beta} - 3\alpha\beta$

[willow helm]

Can we do something instantly?

[verbal barn]

ya we can change it intoo 2/-9 - 3 x20

[willow helm]

uh

no

[verbal barn]

ok

[willow helm]

You can't add fractions like that

[verbal barn]

oh dang

[willow helm]

If they don't have the same denominator

$\frac{1}{2} + \frac{1}{3} \neq \frac{2}{5}$

[verbal barn]

but we dont know the denominator

[willow helm]

You can only add fractions which have the same denominator

Yes

But we can multiply numerator and denominator of a fraction by the same value right

And the fraction remains unchanged

$\frac{1}{2} = \frac{1 \cdot m}{2 \cdot m}$

[willow helm]

So

And we can add fractions which have the same denominator

So we must find the common denominator of these two fractions

Which is...?

[verbal barn]

umm 20 ?

[willow helm]

what

uhh

Okay

Well this is a bit of a bigger problem

😅

[verbal barn]

oh

[willow helm]

You should revisit some material on fractions

Specifically working with them, like multiplying them, adding, finding common denominators (rewriting as one fraction)

[verbal barn]

ok

[willow helm]

As that is a key part of this example, which they take for granted that you already know

[verbal barn]

kk

[willow helm]

Right

Well uh

[viral kernel]

Couldn't you solve the equation and substitute ?

[willow helm]

I currently don't have that much time to explain this, as it would take me quite a bit, sorry 🙁

[verbal barn]

ok

[willow helm]

What do you mean solve the equation

[viral kernel]

x^2+9x+20=0

[willow helm]

Yes, of course

But that is not the main idea here

[viral kernel]

oh okay

[willow helm]

The main idea is to rewrite 1/alpha + 1/beta as (alpha+beta)/(alpha*beta)

And we know what alpha+beta and alpha*beta is

And we are done in 1 step

However we stumbled onto a problem with this, as OP isn't well versed in rewriting fractions under a common denominator

I would explain everything, but I am in a hurry currently

And I don't want to rush things

So I will let someone else take over

(but finding the two solutions is absolutely one way to solve this, yes)

(just takes more time)

[jagged harbor]

$\frac1{\alpha}+\frac1{\beta}=\frac1{\alpha}\mathcolor{red}{\cdot\frac{\beta}{\beta}}+\frac1{\beta}\mathcolor{red}{\cdot\frac{\alpha}{\alpha}}$

[woven radish]

Flip

[verbal barn]

yea but we dont know the roots??

[jagged harbor]

we don't know where they are on a number line, but 1. that doesn't stop us from giving them names, and 2. they're still numbers, so we can treat them as such

now granted, we totally know the roots because that polynomial is factorable

for one reason or another we're on this ride

[verbal barn]

so wt r we suppose to multiply this with

[jagged harbor]

not sure what you mean

[verbal barn]

i know till this step but i am stuck i dont what to do after

[jagged harbor]

let me know if you agree with these two properties of real numbers

1. that $\frac xy=x\cdot\frac1y$

[woven radish]

Flip

[verbal barn]

yes

[jagged harbor]

2. that $a(b+c)=ab+ac$

[woven radish]

Flip

[verbal barn]

yea

[jagged harbor]

and secret 3. that $\frac ab\cdot\frac cd=\frac{ac}{bd}$

[woven radish]

Flip

[verbal barn]

arnt we suppose to cross multiply ?

[jagged harbor]

that's usually said when there's an equality involved, like two fractions are equal

[verbal barn]

oh ok

[late pilot]

ab^-1 * cd^-1 = ac(bd)^-1 = ac/bd

[jagged harbor]

so using 3, do you agree that the above equals $\frac{\beta}{\alpha\beta}+\frac{\alpha}{\alpha\beta}$?

[woven radish]

Flip

[verbal barn]

yes

[jagged harbor]

then you factor out the 1/(alpha * beta)

if you like

[verbal barn]

so make it 1/20 ??

[jagged harbor]

sure whenever you want to do that

[verbal barn]

ok

[jagged harbor]

I wasn't concerned with that yet but go you

[verbal barn]

ok now what

[jagged harbor]

well actually since we're doing that

what's alpha + beta? did we know that?

[verbal barn]

-9

[jagged harbor]

ok

so you have 1/(alpha * beta) * (alpha + beta)

after the substitutions, this is 1/20 * (-9)

alpha * beta = 20, alpha + beta = -9

don't forget the dropped -3alphabeta from the original question

but we can say that 1/alpha + 1/beta = -9/20 yes?

[verbal barn]

1/-180 - 60

[verbal barn]

uh yeaa suree

[jagged harbor]

who is this?

[verbal barn]

idk i multiply 20 and - 9

[jagged harbor]

hm

for fun?

I'm jesting

but that seems like an arbitrary choice

[verbal barn]

yea ig it felt right to me

[jagged harbor]

it felt wrong to me

[verbal barn]

i doono wt tht a something is

[jagged harbor]

arbitrary means it had no particular reason or significance

[verbal barn]

ohhhhhhhhhhhh

okkk

[jagged harbor]

you multiplied them for no reason

[verbal barn]

okk okk

back to -9/20

[jagged harbor]

give me a recap

pretend you're explaining what you've done so far, like how the USS enterprise and I were

[verbal barn]

uh umm i am bad at explainningg 😅

can i just send wt i wrote

[jagged harbor]

a bad explanation won't hurt you

and learning how to explain things is good

you can send what you wrote but I'm lazy and won't follow along unless you give some narration

I feel the visceral need to read an english sentence

[verbal barn]

yoo wht did ya expect i am too lazy to typee

[jagged harbor]

sloth is one of the seven deadly sins

[verbal barn]

ummm ik 🙂

[jagged harbor]

if we were also too lazy to type then we might've never gotten here

[verbal barn]

didnt ya wonder y i only types ok and k most of the time 🙂

[jagged harbor]

yes it baits helpers into overexplaining and giving you solutions

ik

ok not you specifically

[verbal barn]

omg 🥲

[jagged harbor]

but this happens so frequently

[verbal barn]

cann we come back to the questionn i have classs in 30 minss 🙂

[gloomy aurora]

Well only if you cooperate.

[jagged harbor]

don't run away from your problems

[verbal barn]

ok okk

[jagged harbor]

nobody's forcing you to stay but I can imagine you just thought of places you'd rather be lol

you should practice explaining your own work

[verbal barn]

tbh i dont wana study at all but i dont wana fail eitherr

ookk ok finee

first we found the sum and product thingyy
α+β=−b/a = -9/1
​,αβ=c/a = 20/1
​

[jagged harbor]

why does alpha + beta = -9 anyway?

[verbal barn]

-b/c its in the questionn

[jagged harbor]

who are b and c?

[verbal barn]

9 and 20

[jagged harbor]

ok

it's in the question?

[verbal barn]

yep

[jagged harbor]

this one?

[verbal barn]

yepp just chage the b into negative tht itt
​

then we used this identity 1/alpha + 1/beta = alpha+beta/apha*beta

and we got -9/20

[jagged harbor]

so the original question asserts alpha + beta = -b/a?

sorry I'm still on that lol

[verbal barn]

yess

now we have -9/20 - 3(20)

[jagged harbor]

why does this identity hold?

[verbal barn]

uhh what

cuz its 1/alpha + 1/beta

[jagged harbor]

why does 1/a + 1/b = (a+b)/ab?

[verbal barn]

alpha/alpha x beta + beta/alpha*beta = alpha+beta/alpa x beta

and we took 1 as alpha and beta

[jagged harbor]

I'm confused, where is 1/alpha + 1/beta in all this?

[verbal barn]

do i really have to explainn

not to be rude but ma class is gonna start soonn

[jagged harbor]

I'm only asking because it was something you were stuck on

[verbal barn]

oh okk

[jagged harbor]

you're not being rude, you should go to class, be a good student lol

think of an explanation for how you add two fractions

i.e. represent the sum of two fractions as a single fraction

[verbal barn]

multiplying numerater and denominator to make the denomiator the same number

ummm

[jagged harbor]

you could explain it with symbols

like how me and a ship were

[verbal barn]

a ship ?

wht ship

[jagged harbor]

the USS enterprise of course

[verbal barn]

oh ok

wdym symbolls

[jagged harbor]

a, b, c, d

[verbal barn]

i gtg class startedd

thc anyway

[jagged harbor]

glhf

[verbal barn]

i will do this after class

byee

[devout snow]

@verbal barn Has your question been resolved?

[devout snow]

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[polar nebula]

heyyyy i wanna get some help on finding the distance between SD and AC

[polar nebula]

with pyramid S.ABCD, ABCD is a square with a side length of 1

SA ⟂ (ABCD) and SA=1 too

i need to find d(SD,AC)

[young spade]

This is basically what we have, right?

[polar nebula]

yeaaaa

that's right

[polar nebula]

so confused about finding the distance between 2 skew lines

[young spade]

Well, first thing to note, we can get rid of a lot of info here

[polar nebula]

hmmm

[young spade]

Since we only care about AC and SD, we can discard basically half the figure

We know that These are basically two rectangular triangles stuck together

[polar nebula]

yea that's perfect

[young spade]

And that all other angles are 45º

[polar nebula]

yep

we have 2 ways to solve this
1 is to find a line that is perpendicular to both SD and AC
2 is to find a plane that contains one line and is parallel to another line

[young spade]

If youre versed enough about vectors any of the two should be practically trivial

[polar nebula]

oh unfortunately, i cannot use vector

im in grade 11 and vector is in grade 12

[young spade]

Well, ig this will be mildly harder then

But i was intending to make use of geometry anyways.

I want to believe you can see that these three are the main projections

The bottom one is if you look at it from the front, the top left is from above, and the top right is from the right

[polar nebula]

oh yeah i can totally understand this

[young spade]

Ill import this to desmos, gimme a sec, it will make my life easier to explain this

https://www.desmos.com/calculator/hcidxxtkfk

You can use this to also see what im about to say, lmao.

[polar nebula]

this is perfect

[young spade]

Take t0 = 0, where the points sit at D and C

The distance is, quite obviously, 1

Say you move the red point on the hypotenuse of the white triangle alone (just through imagination exercise)

[polar nebula]

yep

[young spade]

Then the point would move in all the axis away from the other point

so the distance has to increase.

This is basically a direct implication that both the points are fixed to be in the same ratio in their own lines.

Now, notice, since the two triangles are just reflections of each other stuck together, the case t0 = 0 is just a reflection of t0 = 1

You could now go ahead and test the idea, "what happens" if i start from t0 = 0 and go to t1 = 0.1

Say they end up somewhere like this

You can calculate the distance through the pythagorean theorem quite easily

[polar nebula]

oooo

[young spade]

Youll see that by varying a little t0 = 0 -> t0 = 0.1 the distance has reduced a little bit.

and since t0 = 1 is a reflection of t0 = 0

The same has to be true for t0 = 1 -> t0 = 0.9

Youll eventually find that if you repeat this again and again, the point with the least distance has to be where 1 -> 0.9 -> 0.8 ... and 0 -> 0.1 -> 0.2 ... converge into the same number

aka t0 = 0.5

Which is exactly at the middle of the two lines.

just about here.

Knowing that the points sits at the middle of AC and SD, you can, again, use pythagoras

[polar nebula]

wow

[young spade]

If you know something about functions and derivatives, theres a really neat way to show that this is true too

[polar nebula]

that's so cool, fully understand, just geometry alone soo cannot use d/dx or sth 🥹

[young spade]

https://www.desmos.com/calculator/gkk7qqvgxp
ill just leave it as part of the desmos thing

You can see that if you "flatten" out your two triangles, the absolute distance between is constant at 1

But what youre seeing there is just a flattened projection of the two legs of your rectangular triangle for pythagoras measurement

And its pretty easy to show that sqrt{ 0.5^2 + 0.5^2 } is the minimum for that relation

[devout snow]

@polar nebula Has your question been resolved?

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[daring hollow]

can i get help with this one

[daring hollow]

<@&286206848099549185>

[brave plaza]

what have you tried?

[hardy crown]

Who speaks Arabic?

[brave plaza]

if you havent already try to make a drawing

[wheat pawn]

are you expected to do it by integration, or are you allowed simple methods?

[rare kernel]

also pls ping helpers after 15 mins

[daring hollow]

drawing it out

but it didnt make sense

[versed juniper]

what doesnt make sense

[daring hollow]

idk how to get the bounds

[wheat pawn]

again, are you expected to do it by integration, or are you allowed to use simple methods?

[brittle inlet]

A hint that could be useful: if you're meant to do it integrating, which I assume you are, then dividing the region in two would be helpful

That is, obtain the area of your region of interest above the x axis and multiply by 2, trying to obtain the whole area through a single integral seems hellish

[versed juniper]

x = 1 so what is y if its a circle of radius 2

[daring hollow]

well thats just y = +- sqrt(4-x^2)

[versed juniper]

ok good you know the integrand then whats the x bounds

[daring hollow]

1 to 2

[versed juniper]

cool

keep in mind when you integrate that you're only getting the area over the x axis

so you have to add a multiplier of 2 as well

[daring hollow]

why?

i can set my bounds -sqrt(4-x^2) to sqrt(4-x^2)

[versed juniper]

uh what are you integrating then?

that doesnt really make sense

[daring hollow]

the half circle?

after x=1

[versed juniper]

are you integrating with respect to x or y

[daring hollow]

x

i mean y

[versed juniper]

oh ok

in that case what y does it start and end at

it wouldn't make sense for the bounds to contain a variable

because you'll need to sub them in later and you'll get an expression containing a variable instead of a number for the area

i hope u get why it doesnt make sense

[daring hollow]

im talking abt the inner integral

[daring hollow]

the inner integral has those bounds

[versed juniper]

inner...? oh you're doing it with two...

i mean sure

[daring hollow]

yes double integral

[versed juniper]

it works

then what is the integrand inside

[daring hollow]

f(x,y)

but my issue is with it being after the line x=1

wait can i integrate that volume and subtract it from the result

[faint gorge]

The issue is dealt by the bounds of the outer integral.

[daring hollow]

right i forgot, but theres an issue and its if i wanna convert it to polar coordinates

[faint gorge]

You can write it in polar if you want

[daring hollow]

but how am i supposed to know my bounds

[faint gorge]

You have two boundaries

The outer is just r=2

the purple boundary can be found

[daring hollow]

how

[faint gorge]

What are the equations for x and y you get after you apply polar coordinates

[daring hollow]

r^3?

[faint gorge]

How did you get that

[daring hollow]

x^2+y^2 = r^2

then theres an r dr dtheta

r * r^2 = r^3

[faint gorge]

tf

[daring hollow]

what

?

[faint gorge]

When you use polar coordinates you want to write x and y in terms of r and theta

So the equations are just: x=rcos(theta) and y=rsin(theta)

r dr d(theta) is the jacobian you add to the integral

[faint gorge]

We want the area from the region itself, so f(x,y)=1 simply

[daring hollow]

no idk why this is our subs

[faint gorge]

That's polar coordinates

[daring hollow]

why are doing that sub

[faint gorge]

That's polar coordinates

[daring hollow]

its squared

its supposed to be squared

x = rcos^2(theta) y = rsin^2(theta)

[faint gorge]

It's not

Polar coordinates

[daring hollow]

am i trippin

[daring hollow]

youre doing a simple substitution for x and y right? i mean a conversion

[faint gorge]

Yeah

[daring hollow]

x is the scalar r multiplied by the adjacent side of the triangle

[faint gorge]

and as you see, the adjacent isn't squared

[daring hollow]

oh shit idk why i said squared

[daring hollow]

ohh i said squared because i was subbing x^2 and y^2

but we want the area

[faint gorge]

We got off track just write x=1 in terms of r

[daring hollow]

is this what we were tryna do

[faint gorge]

It's equivalent

-# bruh Lex

[daring hollow]

knowing who wrote that is crazy

[faint gorge]

[ prpl frac{1}{2} int_{alpha}^{beta} r^2(theta) dd{theta} = int_{alpha}^{beta} int_0^r rho dd{rho} dd{theta} ]

[faint gorge]

We are doing the right side, since that's what you asked for

[daring hollow]

which one is easier

the only reason why i chose double integration is bc i didnt even see that they wanted the area i thought they were asking for the volume

[faint gorge]

You get either way to the same result, this is just rather a shortcut

[daring hollow]

oh okay

but how do i get my bounds

in that case

[faint gorge]

It' still the same

[faint gorge]

You have two radii, one outer (the constant one in polar coordinates) and the inner one (the constant one in cartesian coordiantes)

The lower bound can be found in terms of r and theta with x=1

Now substitute x

[daring hollow]

okay so rcos(theta) = 1?

[faint gorge]

yes

[daring hollow]

why are we looking at the right side tho

[faint gorge]

Re-read the task

[daring hollow]

no i know but

[daring hollow]

what i mean is why do the bounds work that way

[versed juniper]

further from origin is larger

[daring hollow]

yes

[versed juniper]

you're looking for right side of x so that's far away

so r should be greater

[daring hollow]

so we want the one going from the origin to the far right side of the circle

but lowest being 1

[daring hollow]

so thats out first bound?

right

whats our second

[versed juniper]

what's the one further away

from origin

[daring hollow]

i guess my problem is interpreting the bounds of the radius when it comes to the integration stipulation wise

[daring hollow]

well radius 1 and greater

[versed juniper]

ok well you have two lines that make up the area

a straight line you found and a curve

which one is further away

[daring hollow]

the curve

[versed juniper]

and what is the equation of the curve in polar

[daring hollow]

this should be the second radius

am i cookin

[versed juniper]

no

[daring hollow]

without the r

in the bottom

1/cos(theta)

[versed juniper]

R in polar IS the radius

r = 1000 draws a circle with radius 1000

so what's the polar form of the second bound

[daring hollow]

i took x=sqrt(4-y^2)

because thats the second half

converted it to polar coordinates

then solved for r

i lowkey shouldve just wrote r = 2

2 - 1/cos(theta)?

[versed juniper]

yeah

[daring hollow]

whats the point of the inequality above tho that star did

[versed juniper]

to show its the lower bound ig

[faint gorge]

It becomes more evident what's upper/lower bound algebraically rather than having only =

[daring hollow]

but in our case the lower band had to be = right

[faint gorge]

but how would you know that without for example a sketch

I am just saying how the math would lead you there

because in the end you'd end up with sec(theta)=<r=<2 which is now clear how r ranges

[faint gorge]

How did you get the upper bound of r and the value of theta

[daring hollow]

upper bound is the total radius - the one we found

[faint gorge]

The outer radius is just r=2

[daring hollow]

its not just that space

?

[daring hollow]

i know

but does that mean our outter bound is just 2

[faint gorge]

So why do you subtract sec(theta)

[daring hollow]

because i wanted the radius of the shaded region

[faint gorge]

But the lower bound handles that already

You start at r=sec(theta) which is at x=1 until r=2

which is x²+y²=4

[daring hollow]

alright

theta is wrong?

i dont think it should be wrong

[faint gorge]

Look at the green line, the angle can't be at most pi/2

pi/2 would be reaching the y-axis

[daring hollow]

oh right

do i need to solve for theta??..

[faint gorge]

Yes

Basically when x=1 and x=2cos(theta) intersect

[daring hollow]

it aint that bad actually

[faint gorge]

indeed

[daring hollow]

i thought we found our inner radus

radius

which is sec(theta) = r

[faint gorge]

We are talking about theta now

we found r

[daring hollow]

wait so sec(theta) = 1

bro why am i freezing

whats going on

im lwky lagging

just solve for theta?

arcsec(1) = theta

[faint gorge]

So you get 2cos(theta)=1

Or very rigorously 2cos(theta)>=1

[daring hollow]

we had the radius function

correct

of the inner radius

which points to x = 1

its angle is what we want

right

@faint gorge

[faint gorge]

I am here

[daring hollow]

mb

[faint gorge]

yes

[daring hollow]

so the green line is sec(theta) = 1

correct

[faint gorge]

No

[daring hollow]

yeah im trippin

its sec(theta) = r

i was thinking of plugging smth for r then solving for theta

but i dont think we even have r

[daring hollow]

wait what

[faint gorge]

and r=2

[daring hollow]

2 is the radius of the whole circle

not the green line

[faint gorge]

green line is also 2

[daring hollow]

how its shorter no??

[faint gorge]

It's literally the distance from the circle to the origin

[daring hollow]

it doesnt make sense

shouldnt it range from the top of the green line to the bottom of it rotating

wait that wont be a radius anymore

brothere

[faint gorge]

hmm?

[daring hollow]

r was supposed to be for the values connecting the origin to the line on all points right

is that right

[faint gorge]

r simply represents the distance from a point to the origin

[faint gorge]

The green line touches the circle so it has radius 2 at that particular point, nothing more

[covert drum]

o o o

[devout snow]

@daring hollow Has your question been resolved?

[devout snow]

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[ripe oxide]

I know that there are more advanced ways to solve this problem, but for my class, they want us to solve these by dividing the number of favorable solutions by the total amount of possibilities. Anyone?

[ripe oxide]

?

[ebon coyote]

@desert bluff bro <@&268886789983436800>

It's a cryptoscam dw abt it

[ripe oxide]

ty

any ideas?

[ebon coyote]

tbf it's not an advanced way - that's basically the expected way

[ripe oxide]

well you can do fancy stuff like inverting the problem

"fancy"

[ebon coyote]

It's getting late over here - but the idea would be precisely to count how many possible solutions there are, and how many possible arangements in total there are

[ripe oxide]

yea, im more struggling with the numerator

[coral zephyr]

combinatorics ?

[ebon coyote]

A solution being, you randomise the bills in a list, and you take the first 6, wherein both counterfeits find themselves

[ripe oxide]

ye

[coral zephyr]

you wanna arrange them in such a way that order doesnt matter right

[ripe oxide]

yes

[coral zephyr]

so thats gonna be a combination

[ripe oxide]

would it help if i showed u guys my work thus far

one sec

excuse the bad handwriting...

[coral zephyr]

its been a while im tryna remember how you do this

[ripe oxide]

gochu

i can just ask my teacher in class tmrw if needed

ill let it cool for 5 mins or so before i close in case anyone else knows tho

Branshi?

[vivid estuary]

you can treat this as a sampling problem without replacement where order doesnt matter, the number of elements in the sample space is given by 10 choose 6 which you have. however your calculation in the numerator for the number of events with 2 counterfeits isnt correct

think about it like this

the number of ways to choose two counterfeits out of the counterfeits available (2C2), and the number of ways to choose the remaining bills

[ripe oxide]

ok ill try that! thank you!

.close

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[worn whale]

Hey

[worn whale]

Is it just all real numbers except minus 1 cuz the question should be on a difficult part of the subject

[small geyser]

yes.

[devout snow]

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[zealous ledge]

Wouldnt this finally result in a circle that has its center at the centroid of equilateral triangle, and radius as the perpendicular distance of the centroid from a side?
My rationale is that its clear that the figure will approach a circle as steps approach infinity. Since we keep trisecting the sides, eventually the midpoints of the original triangle will become points of the polygon after infinite steps.
And the centre of the circle having these points will be the centroid of the triangle.

[small geyser]

,rccw

[zealous ledge]

I feel like the solutions has an error to it.
Particularly, i feel like it is incorrect to say that at every step, the resultant polygon will be regular with side s/3

This is because when you do the same with a hexagon, the side of corners turns out to be s/√3
-# either that or im tripping

[sand quarry]

i think you are right

but seemingly their approach still works out even under that assumption

not totally sure why rn

[zealous ledge]

The two areas dont match up tho

The circle approach gives r= a/2√3

So the area turns out to be 10π/3√3

[sand quarry]

yeah but im saying the infinite series still converges to 40/7 regardless

[zealous ledge]

🤔

Ah well

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[clever sphinx]

Hiii

[faint gorge]

Hi, do you have a math question?

[clever sphinx]

I’ve found the mean and the unbiased sample variance but I’m a bit confused on how to sue it

Use*

I found the mean as 30.7 and the unbiased sample variance as 22.4

So yh id appreciate some help🥲

[devout snow]

@clever sphinx Has your question been resolved?

[vital sedge]

which question?

[clever sphinx]

Part a

[vital sedge]

Do you know what the significance level of the test is?

[clever sphinx]

No that’s what we’re solving for

[clever sphinx]

Im not sure actually

[vital sedge]

it's the probability of rejecting the null when it actually true

so we need to find the p-value

[clever sphinx]

Ah okay

I tried doing X~N(30,22.4)
And p(x>30.7) but that was wrong

[vital sedge]

Do you know the T test?

[clever sphinx]

No

The markscheme used standardising but I don’t understand why

[vital sedge]

Have you ever calculated the p-value before?

Im not sure what you're being taught...

[clever sphinx]

Yeah

This is what we are expected to do

[vital sedge]

Alright

so we have X - N(30 , 22.4)

How would we get ... - N(0,1)

[clever sphinx]

We can use the formula to standardise

The x bar - mu/ (sigma / root(n))

[vital sedge]

Yes exactly

So do that here

We get 1.673

when do we reject the null hypothesis when the null is true?

[clever sphinx]

When the significance level is greater than the probability

[vital sedge]

Alright so what is the probability?

[clever sphinx]

0.047

[vital sedge]

So we reject when alpha > 0.047

[clever sphinx]

Yep

[vital sedge]

And so the range is from 0.047 to 1

[clever sphinx]

I don’t really understand what the point is in standardising when we have all of the values already

Surely we only need X, mu and varx

[vital sedge]

Well how did you find 0.047

[clever sphinx]

I said mu= 0 and var(x)=1

But can’t I say mu=30 var(x)=22.4 and it’s P(x>30.7)?

[vital sedge]

It's completely fine to do it that way

But in statistics usually working with distributions of the form like N(0,1) have very nice properties

There are a lot of important theorems using N(0,1) and in earlier mathematics, people calculating the values of the normal distributoins only conciderd N(0,1)

It's also usefull for comparing different results, as their significance is given by the same distribution

the N(0,1)

[clever sphinx]

Thank you

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[unreal crow]

What am I missing ?

[azure parrot]

Seems right to me

[azure parrot]

What made you think you're missing smth

Nvm the last line is wrong, r is not 2

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[faint gorge]

I am sorry, but we don't do that here usually, also try to not send files other than images

[nocturne field]

ohh sorry

[devout snow]

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[nocturne field]

.close

[faint gorge]

The channel closed now already on its own, you need a new one

At best try to formulate your doubts regarding continuity/differentiability

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[dapper fable]

Let $2021$ be a "fantastic number". Now, for any positive integer $m$, if atleast one of $m, 2m+1, 3m$ is fantastic, then all of them are. Is $2021^{2021}$ a fantastic number?

[woven radish]

Copter

[dapper fable]

i have no idea how i should start this

[polar chasm]

Well, if 2021^2021 was fantastic, you'd have to get some kind of ladder of fantastic number which climbs from 2021 towards 2021^2021

try to focus on the end of this "ladder"

[dapper fable]

it would probably be from it being 2m+1?

it cant be 3m and if it came from m it means the earlier terms would have to decrease to it

i think that can still happen tbf

[topaz axle]

sounds like collatz

[polar chasm]

oh true, i didnt really think about that option

but it can be checked pretty easily that you cant increase directly towards it, not sure whether that helps a lot tho

basically 2021^2021 isnt 3m, but its 2m+1 where m = (2021^2021 - 1) / 2, but then this m isnt divisibly by 3 and its even. So it cant be 3n nor can it be 2n+1 (this means that if 2021^2021 is fantastic, its only because some higher number is fantastic)

probably doesnt help though

[dapper fable]

there is probably a general form of the numbers

,w 2021

how do i search for the prime factors

[polar chasm]

,calc 43*47

[woven radish]

Result:

2021

[polar chasm]

the 2m+1 thing wont preserve prime factorization very well though

[dapper fable]

i was thinking smth modulo 2021

[polar chasm]

2021 is kinda arbitrary here, if you were given 1010 (or any other fantastic number) instead, it would genereate the exact same set

[dapper fable]

ooooh wait

2n is fantastic <=> n fantastic

2n => 4n+1 => 12n+3 => 6n+1 => 3n => n

[polar chasm]

oh thats very nice

[dapper fable]

and since 2n+1 is fantastic <=> n fantasic

we can reduce things until we get 1 fantastic

and i think every number is fantastic?

[polar chasm]

sounds right

[dapper fable]

cool cool

.close

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[polar chasm]

you basically have 2n+1 is fantastic iff 2n is fantastic, which is just like k is fantastic iff k+1 is fantastic

so everything is fantastic

[polar chasm]

oh wait no

[dapper fable]

oh?

[polar chasm]

how did you go from 12n+3 to 6n+1?

[dapper fable]

12n+3 = 2(6n+1) + 1

[polar chasm]

oh right im dumb

[dapper fable]

all good?

[polar chasm]

Yeah, sounds good

i cant even do arithmetic anymore

[devout snow]

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[clever sphinx]

Ok this is really stupid but what’s the null and alternative hypothesis

[clever sphinx]

Im not sure if its H0: no change in heart beat
H1: there is an increase in heart beat
Or
H0: p=0.5
H1: p>0.5

[dark tundra]

first one

but it might be best to express it more mathematically

[clever sphinx]

How would I write it?

<@&286206848099549185>

[devout snow]

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[proven anchor]

Sry for asking same question again and pinging you back @polar chasm

If we take f(x) as constant function

Will it be one-one

[polar chasm]

you dont care about f being one to one

T is a function that maps functions to functions

and you only care about T being one-to-one

anyway, looking at the channel, you need to prove that T(f) = T(g) -> f = g

you correctly noticed that if T(f) = T(g), then T(f-g) = 0

so it suffices to prove that when T(h) = 0, then h = 0 (0 as in the 0 function)

so what you need to prove is that if $\int_0^x h(t) \dd t = 0$ for all $x$, then $h(t) = 0$ for all $t$.

[woven radish]

MathIsAlwaysRight

[proven anchor]

We input a function and we get output as integration of that function

[polar chasm]

do you have any other questions btw? Or are you stuck proving it? Did you understand what I said?

[proven anchor]

I'm looking fot T(f(x)=0 implies one-one over chatgpt🫢

We take any function and we get zero function as output

[polar chasm]

we dont have to use it actually

lets just work with T(f) = T(g) -> f = g

suppose that T(f) = T(g)

[proven anchor]

Two different functions and their integration will be same

[polar chasm]

$\int_{0}^{x}f(t)\dd t=\int_{0}^{x}g(t)\dd t$

yep

[woven radish]

MathIsAlwaysRight

[polar chasm]

now we can subtract it

[proven anchor]

really?

[polar chasm]

$\int_{0}^{x}(f(t)-g(t))\dd t=0$

[woven radish]

MathIsAlwaysRight

[proven anchor]

Nice

[polar chasm]

i mean that we can subtract one integral

[polar chasm]

then it becomes this

and notice that we're where we were with the zero funciton thing

[polar chasm]

it suffices to prove that f(t) - g(t) = 0, given that its integral to all x is 0

[proven anchor]

Yeah yeah

[polar chasm]

if you want a hint on how to prove this, I'd probably use the fundamental theorem of calculus

[proven anchor]

We can take x=0?

[polar chasm]

sure but thats not very helpful

that only tells you that the integral from 0 to 0 is 0

[proven anchor]

I meant both limits are same so area would be 0

[polar chasm]

which holds for any function

[polar chasm]

yes, so we wont get any information about f-g out of it

[polar chasm]

try this

ill have to go, sorry. If you need further hints, you can just ask and someone else will answer

ill leave one more spoilered hint in here, that explains how exactly to use the FTC

[proven anchor]

I have no idea how to use FTC here?

😌😌

[polar chasm]

||By FTC, there is a function H such that H' = f - g and the integral = H(x) - H(0). But the integral is 0 everywhere...||

[proven anchor]

I'm totally confused and learning

H' ??

Derivative of it?

[polar chasm]

yes

H is the antiderivative of f - g (it exists by ftc)

so H' = f - g

[proven anchor]

H(x)=H(0)

For all x

Which meansssss

We will have input as. X=0

[polar chasm]

it holds for all x, not just x = 0

H(1/2) = H(0)
H(0.21314) = H(0)

what kind of function could H be

[proven anchor]

Yeah yeah

Constant function?

[polar chasm]

precisely

[proven anchor]

Zero function?

[polar chasm]

its always equal to H(0), which is constant

[polar chasm]

not necessarily 0 function

keep in mind that our goal is proving that f - g = 0

now you just have to connect all the dots

[polar chasm]

read this again and keep in mind that H is constant

[proven anchor]

yes it is constant

Hang on let me think

We found that H(x) is constant

we can take it out of integration

[polar chasm]

youre overthinking this a bit

[proven anchor]

🥹

Sry

I am exhausted and sad over my learning

I want to learn and skip

[polar chasm]

dw, youre doing good

[proven anchor]

Watch videos or buy a coaching course

[polar chasm]

youre just overthinking this specific thing

you need to prove that f-g is 0. What else is equal to f-g?

[proven anchor]

I can learn things but due to language barrier and maths...lots of complications

😭😭😭

[polar chasm]

im sure youll do this, but ill have to go now, sorry. If you need more hints, just ask and someone will certainly answer

[proven anchor]

I don't need solution and there is no one who will see solutions by my side I'm just learning at home for myself

[polar chasm]

bye and good luck

[proven anchor]

Thanks a lot

f-g was H' anti derivative of H

and it is constant

Of H is constant then obviously its derivative is 0

So we get H' is 0

[devout snow]

@proven anchor Has your question been resolved?

[devout snow]

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[wild eagle]

this level is a scam

[wild eagle]

i tried 51

it won't work

[untold ravine]

whats the rules

[solid perch]

||hint: consider computing and checking the differences in successive terms.||

[finite fable]

I think we need to figure the pattern out. That's the usual question in such type of problems

[dull parrot]

what app is this btw, looks similiar to the one I use

[untold ravine]

ah, ok

[finite fable]

+4, +8, +16

[untold ravine]

this completely gives it away

[finite fable]

+32 ig

Each step adds multiple of double of previous

[wild eagle]

my friend already helped me

[untold ravine]

no

[wild eagle]

it's 63

[marsh trail]

There's another method I guess
Hint: consider doubling each term and adding something to get the next term

[solid perch]

I apologize in that case, I'm not really sure how else I can phrase a hint.

[wild eagle]

also here's the game

[untold ravine]

yeah its kind of difficult too

[finite fable]

Sorry. Should be +4, +8, +16.....

Not multiple of 4

[dull parrot]

idk how you can phrase a hint for that in a way it doesn't gives it away

[crystal dawn]

considering that two helpers already directly dropped the rule Chiaki's hint is tame by comparison imo

[finite fable]

Exactly lmao

[solid perch]

that is fair. I'll step out then.

[wild eagle]

https://play.google.com/store/apps/details?id=com.BlackGames.MathRiddles

[marsh trail]

Whoops, I dropped it cause OP said someone already helped

[dull parrot]

oh yeah it's the same one :p

[crystal dawn]

!done, anyway?

[devout snow]

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[wild eagle]

yes, thanks

.close

[devout snow]

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[atomic idol]

Ive been trying to map two real number intervals (a,b) and (c,d) for a bit now
I need to show they have the same amount of elements by using a bijection

[atomic idol]

This what I have so far but it’s not working as i thought it would and I’m not sure why

The first part is supposed to map any interval to (0,1)
As a starting point so to speak

It seems to work just fine I tried it with a few examples

The second part is supposed to be the ratio of said intervals, I was hoping this would scale (0,1) to the proper size

Which also seems to work by itself

And then finally, c is supped to be the offset

[stone stump]

have you written down the bijection (0,1) -> (c,d)

f(x) = ?

[atomic idol]

That would be the same thing without part 1

Just the ratio + c I think

[stone stump]

write it out

[atomic idol]

d-c/b-a + c

Forgive the format

[stone stump]

(0,1) -> (c,d)

[atomic idol]

d-c/1-0 + c

Or just d-c + c

[stone stump]

there should be an x in there

[atomic idol]

Or just d

[atomic idol]

Huh that’s weird

I accidentally removed the first part entirely when it should’ve been x mb

[stone stump]

ok so f(x)=(d-c)x+c

what is the inverse of this

aka, what is a bijection (c,d) -> (0,1)

[atomic idol]

Y-c / d-c

That’s the same as my first factor here isn’t it

This part was supposed to map to (0,1)

[stone stump]

good

so we have a map (a,b)->(0,1) and a map (0,1)->(c,d)

how do we get (a,b)->(c,d) ?

[atomic idol]

Compose

[stone stump]

then do that

the reason you failed earlier is that you are trying to jump steps

take it slow

[atomic idol]

I was not aware of any other steps existing honestly

Thought i came up with a functional idea

[stone stump]

you did

you just went a bit too quick

[atomic idol]

So I just had to replace the second denominator?

[stone stump]

yes

[atomic idol]

Alright nice thank you

.close

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[atomic idol]

How do I prove that every positive real number is either between two natural numbers or equal to one?

[atomic idol]

x is an element of R+
n is an element of N (including 0)
And this applies:
n < x <= n+1

[uncut crow]

probably with the least upper bound principle

or archimedean principle

[atomic idol]

I have the following definition of R:

-R is a totally ordered field.
-a+c < b+c applies as long as a<b
-ab>0 is true as long as a>0, b>0
-contains Q
-fulfills the supremum axiom

[atomic idol]

Im not allowed to use it I think
Or we weren’t taught the name

[atomic idol]

That could work tho

[uncut crow]

hm ok

well it follows from the least upper bound principle anyway

[atomic idol]

For context: only what is part of the lecture is considered to be math because my prof is a dinosaur

[uncut crow]

lol well that's pretty normal for the beginning of an analysis class

[atomic idol]

It’s been like that for more than half a year tho

Welp anyway

Supremum yippe

Again

I can’t think of a way to use it here tho

[uncut crow]

let x be a real number. can you show there is an integer n such that x <= n?

[atomic idol]

Can I not assume that that is the case if N is infinite

[uncut crow]

and then would you be allowed to use the fact that the set {m \in N : x <= m} has a least element? (since it is a nonempty subset of N, if the above is true)

[atomic idol]

Not sure if I understand the notation of the set

[uncut crow]

the set of natural numbers that are greater than or equal to x

[atomic idol]

I think I am allowed to use the fact that N is infinite and must thus have a greater element

[uncut crow]

ok i'll treat this as completed then

[uncut crow]

do you see the signifiance of this?

[atomic idol]

We can always take the next best natural number after x

And subtract 1

To get another natural number
Or 0

Which is going to coat our x

In a way

[uncut crow]

yes sure that's the idea

[atomic idol]

To me it seemed done already

[uncut crow]

the "next best natural number after x" is the least element of {m \in N : x <= m}

[atomic idol]

So..?

[uncut crow]

what seemed done?

[atomic idol]

Does that make a difference?

[atomic idol]

I thought that line of reasoning was the proof

[uncut crow]

oh, sure, that's fine reasoning

let me explain the signifiance of my messages though

[atomic idol]

Alright

[uncut crow]

well it's just that we need to show this "next best natural number after x" exists. and my point is that describing it as "the least of element of {m \in N : x <= m}" does that

[atomic idol]

I thought they expressed the exact same thing

How are they different?

[uncut crow]

they pretty much do, and there is nothing wrong with your intuition here. like yes for any real number x, there is a "next best natural after x"

i am just stating it more formally as you might want to do in a proof

[atomic idol]

Ohh ok

I thought something crucial was missing in my description

This is what it looks like rn

[uncut crow]

i don't really like "N is infinite"

a better description would be "N is unbounded above" or something

[atomic idol]

|N| = Infinity?

[atomic idol]

This is literally in the script of the lecture

[uncut crow]

well my point is like... {1, 1/2, 1/3, 1/4, ...} is also infinite but it doesn't have this property where for any real number x, there is an element in it that is >= x

[atomic idol]

True

I didn’t consider that

[atomic idol]

Do I not need to prove it if I use that tho?

[uncut crow]

yes this is true, but it doesn't follow just from N being infinite

[uncut crow]

not sure, maybe

[atomic idol]

What do you suggest then

[uncut crow]

you can prove it with the least upper bound principle

for any real number x, there is an integer n with n > x

[atomic idol]

What exactly does the upper bound principle state?

There’s an entire chapter on the axiom but it doesn’t clearly give me a statement

Only definitions

(What a supremum is for example)

It doesn’t even actually have the word axiom in the chapter itself anywhere

[atomic idol]

So the axiom is just the definition of a supremum?

[uncut crow]

the axiom is this part

[atomic idol]

I see thanks

[atomic idol]

How do we apply that here?

[uncut crow]

if N is bounded above then it has a least upper bound M

[atomic idol]

But it’s not bounded

Well not yet

But I’m not sure what to do about that

[uncut crow]

yes but that's what we're trying to show

consider the number M-1

[atomic idol]

Sorry this must be painful for you

[uncut crow]

no i'm not that in that much pain

M-1 is not an upper bound of N

so there exists...?

[atomic idol]

A greater number in N

?

[uncut crow]

yes

let's call it k

there is a natural number k with k > M - 1

now recall M is the least upper bound of N (well all we will care about now is that it is an upper bound)

[atomic idol]

So k<M

[uncut crow]

well not quite. we have k <= M

[atomic idol]

Oh right

[uncut crow]

but don't worry about that. k > M - 1 is equivalent to k + 1 > M

[atomic idol]

Could we not do this over and over again claiming that k is always <M no matter how many times we do that

Showing that there are infinitely large numbers in N

[uncut crow]

no i think we should just observe that k + 1 is a natural number and is > M

[atomic idol]

Did i get the variables right

Im kinda getting confused by all the stuff we’re writing down so I wanted everything in one place

More like this I think

Got the relations wrong

[uncut crow]

these are kinda... extra

If N is bounded then it has a least upper bound M. and then we did some stuff and found a natural number k with k + 1 > M

but that can't be, because M is an upper bound of N

[atomic idol]

You mean bound?

[uncut crow]

yes oops

[atomic idol]

Indeed
That confused me too

[uncut crow]

it shows that N is bounded

[atomic idol]

How so if we found sth from beyond the bound

[uncut crow]

there can't be a bound then

for any bound of N we can find a natural number greater than that bound

N is bounded means: there exists a real number A with n <= A for all n in N
we have shown no such thing exists

[atomic idol]

Does allow us to claim that n>=x exists

[uncut crow]

yes

[atomic idol]

So from there we could just say we take said number

And subtract 1

And we should be done right

[uncut crow]

yes

[atomic idol]

Is this better then?

[uncut crow]

not enough words in the first part

[atomic idol]

What missing ?

[uncut crow]

also k := M - 1 is just not the right idea

[atomic idol]

Oh right we said M-1 < k <= M didn’t we

[uncut crow]

since M-1 is not an upper bound of N, there exists k in N with M-1 < k

something like that should probably be written in your proof

[atomic idol]

Update

[uncut crow]

sure

just btw though there is no need to write k <= M. we get that there exists k with M - 1 < k right from M-1 being not an upper bound. and yes also k <= M from M being an upper bound but we don't care about that, so no need to claim it

it's not wrong, just unnecessary

[atomic idol]

Idk I thought it made it clearer

[uncut crow]

you can leave it if you wish though

[atomic idol]

I started calculus last week fwiw

[uncut crow]

it's ok

M - 1 < k is the relevant part

[uncut crow]

oh also one thing. not just "upper bound M" but "least upper bound M"

[atomic idol]

I’ll just replace it with supremum

Fits the terminology of the script more anyway

[uncut crow]

sure