help-38

[wraith hinge]

yes

ah i see

but how do i use (i)?

because for (i) i just proved the amount of subsets of A = {1,2...n} where k is the maximum number

so here i am trying to prove that the sum of all subsets with maximum length i from i = 0 (empty set) to i = n-1

[edgy willow]

it doesn't start from the empty set

[wraith hinge]

ah ok it starts with 1 number

[edgy willow]

since max k = 2^(k-1), 2^i = max i+1

yea

[wraith hinge]

it starts with 1

[wraith hinge]

what do u mean by this

[edgy willow]

like to match these things you have to shift by 1

just to say the left side is max 1 to max n

[wraith hinge]

$\sum{i=0}^{n-1} = 2^{1-1}+2^{2-1}+\dots+2^{n-1} = \sum{i=1}^n 2^{i-1}$

[solid kiln]

Derivative

[wraith hinge]

welp didnt work

but yes i had done this

before

here

except i started at 0 by error

ok so now i have

$\sum_{i=1}^n 2^{i-1}$

[solid kiln]

Derivative

[wraith hinge]

which is the sum of the cardinality all subsets where "i" is the maximum number.

[edgy willow]

yea you just have to explain why this is 1 less than all subsets

[wraith hinge]

thats the hard part

haha

i will think about it

Bbc

[wraith hinge]

?

[wraith hinge]

Whats up

[wraith hinge]

good u

[wraith hinge]

it has to be the empty set

because look

let n =3. lets do the process

if i =1, we get {empty set} and {1}

if i = 2 we get {1,2} {1}, {empty set}, {2}

but notice that, in i =2, {1} was already in the previous set

so basically the subsets of a smaller i will always be in i+1

if i can explain it properly

like for i =1 we get {1} and {empty set}

if i =2 we get every subset of i=1 + other subsets

[edgy willow]

I'd put it that all nonempty sets obviously have a max

[wraith hinge]

ah yes

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[foggy belfry]

Hi! Anyone know how to find x and y? I am stuck.

30 44
----- + ------ = 10 ------------ 1
x - y x + y

40 55
----- + ------ = 13 -------------- 2
x - y x + y

[granite tree]

It's understandable, it is a nonlinear system of equations

Have you tried common denominator x²-y²?

[foggy belfry]

yeah, I've tried that, but it lead me to Hyperbola situated solutions.

[loud gale]

let 1/x-y = p and 1/x+y= q solve as linear equation in one degree

[foggy belfry]

Ok, let me try that. Also, don't mind me asking though, but how you get 1/x-y = p and 1/x+y = q?

[loud gale]

It just looked like linear equation in 2 degree in that way

Plus our teacher taught that manipulation so

[foggy belfry]

Alright i'll do it.

[vital mulch]

Is (8, 3) the solution

[glass olive]

yes i get 8,3 as well

[loud gale]

Yes 8,3 is satisfying so it must be it

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[old plank]

How do i solve these types of questions ?

And how are they saying the answer is 7,8 or 9

[steel oyster]

Well let's look at the first condition:
f(1)>0 then
f(1)=(1-a)(1-b)>0
In order to get a positive value a and b must both be greater than 1.

[old plank]

yes

[steel oyster]

Then we can look at the second condition
f(4)=(4-a)(4-b)<0

With this and the previous one we all see that a and b can't be the same numbers. I forgot to mention that before.
If a is smaller than 4 but bigger than 1 (so 2 or 3), then b has to be bigger than 4 and vice versa

[old plank]

oh ok

[steel oyster]

Third condition
f(7)=(7-a)(7-b)>0

So either number is 2 or 3. Let's say a for gernal sake.
Then one term is positive. So the other one can't be negative nor 0. So the numbers 1-6.

But the first condition said numbers have to be bigger than 1, so b must be between 2-6.

But the second condition said if a is smaller than 4 which it is, the other one has to be bigger than 4, so b must be 5 or 6

So one number is 2,3 and the other 5,6
2+5=7
2+6=8
3+5=8
3+6=9

So you really have to look at all the conditions given and try working out what that means for your numbers

[old plank]

ahhh

Bro tysm I get it now

[steel oyster]

[old plank]

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[tranquil kelp]

!help can anyone help me solving these

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To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[tranquil kelp]

I'm new to negative powers

[supple copper]

Do you know what it means to have negative powers?

[tranquil kelp]

not rlly

[supple copper]

Do you know about fractions?

[tranquil kelp]

yes

[supple copper]

Do you know about power rules?

[tranquil kelp]

Yes

[supple copper]

$(a^b)^c = a^{bc}$

[solid kiln]

Frosst

[tranquil kelp]

SO

Can you helpm

[supple copper]

You know about this one?

[tranquil kelp]

Yes

[supple copper]

Perfect

So when we write $a^{-1}$ what we mean is $\frac{1}{a}$

[solid kiln]

Frosst

[tranquil kelp]

Alright

[supple copper]

Now when we see $a^{-b}$ then we use the first rule to get $(a^b)^{-1}$

[solid kiln]

Frosst

[supple copper]

Then by the definition of this $\cdot^{-1}$ thing it is simply $\frac{1}{a^b}$

[solid kiln]

Frosst

[supple copper]

Do you think you can try these questions and see how you go?

[tranquil kelp]

um

ok

Srry I was afk

The first one is

$/frac{1}{9}$

[solid kiln]

Horrible Creator [GD]

[tranquil kelp]

what

Ohh

$\frac{1}{9}$

[solid kiln]

Horrible Creator [GD]

[tranquil kelp]

Is this the right answer to the first one?

bruh

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[twin mortar]

can someone explain why the graphs looks like this?

[brazen spruce]

the formula of the graphs are in the top left. there are two.

when you substitute numbers in for x, the results are plotted and connected in the graph on the right.

for example, when x = 0 there is no solution, so there is no dot to connect when x = 0 and thus theres no green line anywhere at zero.

keep plugging in numbers for x and you'll get graphs that look like the ones produced in the image. x = 1, or x = 2, or x=3 ect ect

[twin mortar]

when i put in different x values it only becomes a straight line

is this supposed to happen

so as long as x is defined it doesnt look like that?

[brazen spruce]

its not straight, its "almost" straight

1/x gets realllllly close to zero so it looks like a straight line when x gets larger and larger

[vagrant prism]

wdym

are you doing 6/3 or something instead of 6/x?

[twin mortar]

im trying things

i dont understand why it doesnt look like that when its defined

[twin mortar]

i know that

[brazen spruce]

but when x gets smaller and smaller the whole value of the number becomes bigger and bigger for 1/x when x is positive. thats why there is a curve portion and it goves up

[twin mortar]

ohh

[vagrant prism]

but like what are you typing

[twin mortar]

f(x)=6/3

then it looks like a straight line

[vagrant prism]

yes

because the function is constant

[twin mortar]

but when 3 is x it does that weird thing

on both sides

[vagrant prism]

for any value of x, f(x) is 6/3

so it is a straight line

[vagrant prism]

these two functions are fundamentally different

[vagrant prism]

think about it like this:
f(0) = 6/3
f(1) = 6/3
f(2) = 6/3
and so on

[twin mortar]

okay heres a task with this in it

im trying to understand it

[brazen spruce]

you are mixing up f(x) = 6/x with f(x) = 6/3 = 2

or another way to put it (the y value of the graph) = 6/x is mixed up with (the y value of the graph) = 2

when you type f(x)=6/3 into the graphing calculator it will always tell you that the y value of the graph is 2, that is why its a straight line.

when you type f(x) = 6/x the graphing calculator will show you a graph with all sorts of different x value

[twin mortar]

but the graph is confusing me

[twin mortar]

ohh

[brazen spruce]

i had to make a small change to the last sentence, oops! 😄

[twin mortar]

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[wraith hinge]

!status

[trim jolt]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[mild epoch]

3 or 2

[wraith hinge]

also why does it say "left hand side of equation converges" if its given to be equal to 1

[mild epoch]

idk weird phrasing but ig its understood

so sum to infinity is 1

[wraith hinge]

can you notice that $4^x$ is just $(2^x)^2$ and similary for 8, 16, 32...

[mild epoch]

x squared?

isnt it 2^2x , 2^3x

and so on

[solid kiln]

deltaG

[mild epoch]

yh

there

yh i get that bit

so common ratio would be 2^x/4^x

[wraith hinge]

are you aware of sum of GP till infinity

[mild epoch]

yea

a/1-r

a is the first term

and r is common ratio

[wraith hinge]

yeah

[mild epoch]

isnt common ratio 1^x/2^x

[wraith hinge]

no

[mild epoch]

oh

[wraith hinge]

common ratio will be any $\frac{n+1'th\ term}{n'th\ term}$

[solid kiln]

deltaG

[wraith hinge]

so $\frac{4^x}{2^x}$

[solid kiln]

deltaG

[mild epoch]

oh i think i got the asnwer

[wraith hinge]

which is equal to 2^x

[mild epoch]

i got x= -1

[wraith hinge]

correct

[mild epoch]

could u pls verify

oh am i rgiht

well

im sry for wasting ur time

[wraith hinge]

what

it was my choice to waste my time here(here as in, on discord), not yours, why are you sorry?

[mild epoch]

nah i thought i wouldve needed more assistance

i couldve applied a bit more of thought before coming to the conclusion of seeking help on discord

[wraith hinge]

well you should do this next time then

[mild epoch]

yh fs fs

thanks tho

appreciate it

.close

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[mild epoch]

.reopen

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✅

[worldly wing]

!status

[trim jolt]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[mild epoch]

1 or 2

i was thinking maybe solve for theta

and sub it into the equation above

but idk how to go about it

[wraith hinge]

Yea that sounds reasonable

[worldly wing]

maybe not solve for theta, but you can certainly solve for $\sin(\frac{\theta}{2})$

[solid kiln]

kheerii

[worldly wing]

you can write the given function purely in terms of sin(x/2)

[mild epoch]

yeah im doing it rn as we speak

[mild epoch]

which one theres 2

oh nvm

let me do that

4cos(2(1/2 x)) -3sin(1/2 x)

i got 4 - 8sin^2(1/2 x) - 3sin(1/2 x)

[mild epoch]

used double angle identity on cos

and converted all terms to sin

[worldly wing]

yes indeed

[mild epoch]

oh and then i reaarange for half theta

and then replace that with 1/2 x

?

[worldly wing]

sin(theta/2), yes

[mild epoch]

ok

cool

[worldly wing]

and you replace sin(x/2) with the value of sin(theta/2) that you get

from the other equation

[mild epoch]

yeah

thats what i had in mind

[worldly wing]

from there you'll have a quadratic, which you can use to solve the remaining

[mild epoch]

oke

thanks

appreciate it

b = 3

a = -8

nvm b = 10

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[modern ferry]

"Calculate the triple integral:
The area Ω is located in the first octant of space (where all coordinates are positive)
and bounded by the cylinder... and the plane ...and the coordinate plane.
Change x = 6u, y = 7v, z = 5w, and you grind tedious fraction calculations at the expense of a trivial Jacobian."

[wraith hinge]

So what it s the issue with this question?

[modern ferry]

i dont know where to start do i just switch out x y and z or what do i start with?

[wraith hinge]

So I assume you want to evaluate integral of this form. With respect to specific conditions you have in the question

[modern ferry]

yes i think so

[wraith hinge]

Wait how is that pre university math?

[modern ferry]

no it is university

[wraith hinge]

Can you give me theink to this paper or anything releated to it?

[modern ferry]

what do you mean?

[wraith hinge]

I mean is it specific topic releated question? Because if it is there is might be some important context for solving it.

https://youtu.be/dkpM9nKghY8?feature=shared

[wraith hinge]

Do you know how to do this?

[modern ferry]

hmm i think i do a little as least

[wraith hinge]

Alright we have x, y, z in terms of pther variables so we now know what they are for.

Now we need to figure out why we need this equations

So I think you can find x in terms of y or z
Which we probably need for this transformation.

But, just give me a minute

[trim jolt]

@modern ferry Has your question been resolved?

[modern ferry]

do i calculate the jacobian determinant?

[limpid dawn]

yes

[modern ferry]

yay

that becomes 6* 7* 5 right?

[limpid dawn]

so fast

[modern ferry]

haha

[limpid dawn]

if you done the partials correctly sure

and the determinent

[modern ferry]

so 210?

[limpid dawn]

achso

well 6 x 7 x 5 is surely 210

can you show what you did

[modern ferry]

the matrix is | 6 0 0, 0 7 0, 0 0 5| i think

[limpid dawn]

i see

damn i have a deja vu

[modern ferry]

becuse x = 6u and y = 7v and z=5w

why haha

[limpid dawn]

yes you are right

well then we need to sub everything

[modern ferry]

so does the integral become (6u) * 210?

[limpid dawn]

,,\iiint_\Omega 6u \cdot 210 : \dd u \dd v \dd w

[modern ferry]

haha yes

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

haha yes

[modern ferry]

but what are the bounds?

[limpid dawn]

now we need the bounds

asking the right questions

well

we know x/6 = y²/49

y = 7v

and x =6u

so in terms of u

6u/6 = (7v)²/49

and the numbers seem to fit

u = v²

so in the uv plane we have a parabola

which makes sense i think

in terms of x we would also have a parabola

[modern ferry]

sqrt(u)?

to 0?

[limpid dawn]

no it has to be u = ...

[modern ferry]

oh okay

but is that not the bounds for v?

[limpid dawn]

maybe it's from 0 to v²

hmm

[acoustic flint]

yeah 0 to v^2

[modern ferry]

oh okay why?

[limpid dawn]

the 2nd equation is the bounds for v

if you solve for y

[acoustic flint]

you are bounded in the first octant, so lower bound is 0. The upper bound is v^2 which is easy to see if you draw the projection of omega onto the uv plane

[wraith hinge]

Is this integral even have bounds?
We told it has certain area.
Do we have limits for this area?

[wraith hinge]

[acoustic flint]

imagine drawing a line from v = w =0, along the u axis until you reach an edge of omega. This edge will be u = v^2

[wraith hinge]

Is this integral need limits?

[acoustic flint]

yes because it bounded

[limpid dawn]

for the last bounds you need to find the intersection between those two equations

[wraith hinge]

So what is our original volume we apply Jacob?

Like we have x and function of x and y and y and z.
And omega somewhere than we given substitutions for x and y and z.
So sketch refers to volume we apply Jacob?

So which points of intersection goes to first integral?

[limpid dawn]

i think she got a point, if we integrate over the parabola we are leaving the first octant, so that can't be it quiet

[modern ferry]

hmm okay

dont we need the bound for v and w tho?

[limpid dawn]

it should

[acoustic flint]

what do you mean?

[limpid dawn]

v² implies that v can be also negative

[acoustic flint]

I think changing the order of integration would work better then

[limpid dawn]

yes

[acoustic flint]

to dz dy dx

[limpid dawn]

they are showing here only a part of the parabola int xy plane

which would imply a square root function

[modern ferry]

so can we not use the bounds?

[limpid dawn]

,,\int_0^{1} \int_0^{\sqrt{u}} \int_0^{1-v} 6u \cdot 210 : \dd w \dd v \dd u

So the 2nd equations implies v + w = 1 which is w = 1 - v

so 0 to 1-v

2nd one is what you suggested yummy

third one implies 0 to 6 in terms of x

so in terms of u = x/6

0 to 1

[acoustic flint]

I got a different integral

This is what I got

[limpid dawn]

what you got

[wraith hinge]

[limpid dawn]

that's approx 4.743

but i could be wrong

[modern ferry]

i think i got those too

[acoustic flint]

[ 1260\int_{0}^{1} \int_{\sqrt{u}}^{1} \int_{0}^{1-v} u \dd w \dd v \dd u ]

[solid kiln]

shsgd

[modern ferry]

but was u not from 0 to v^2?

[limpid dawn]

😂

[acoustic flint]

which yields 21

[modern ferry]

sorry no just kidding

[limpid dawn]

o shoot löast bound is wrong o

294

XD

[modern ferry]

yeah u should be from 0 to 1-v

[limpid dawn]

how did you get 1 to sqrt(u)

[acoustic flint]

let me send photo one sec

how do you rotate it

[limpid dawn]

,rccw

[acoustic flint]

right, so you project Omega onto the xy plane, and integrate that shape along z

[limpid dawn]

yes

oh I got it!

you are constraining it

you are saying do sqrt(x) but until y = 1

makes sense

[acoustic flint]

yeah

[limpid dawn]

ok then we would have gotten the same nice

good job

[acoustic flint]

21 right?

[limpid dawn]

yes!

,,\int_0^{1} \int_{\sqrt{u}}^1 \int_0^{1-v} 6u \cdot 210 : \dd w \dd v \dd u = 21

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

SO COOL

[acoustic flint]

@modern ferry Maybe this will help you

[limpid dawn]

I DIDNT LEARNT MATH FOR NOTHING

TO MAKE IT CLEARER

[modern ferry]

can i move out the 6 when integrating?

[limpid dawn]

yes

that's what they did too

[limpid dawn]

here

along with the 210

[modern ferry]

ohh right

okay do i put in the bound so that i get u(1-v)?

or wait

i am confused

[acoustic flint]

what with

[modern ferry]

how did you get 21?

no nevermind

i see it now

[acoustic flint]

I just wolfram'd it

[wraith hinge]

[modern ferry]

well only 5 questions left haha

[wraith hinge]

Till when ?

[modern ferry]

it as supposed to till this wendsday but he moved it till next week

[wraith hinge]

So next week before wendsday?

[modern ferry]

yes

[wraith hinge]

[modern ferry]

haha

"Determine the volume V and the center of gravity"

this seems crazy but maybe it is not that hard

[wraith hinge]

Are you doing physics?

[modern ferry]

no haha

this is just math

[acoustic flint]

what module is this for

[wraith hinge]

Alright it have 0 graphs should be easy

[modern ferry]

hmm i am not sure he does like mixed questions somethimes

[wraith hinge]

Oh wait it actually have graphs

[modern ferry]

do we start by finding the bounds here too?

[wraith hinge]

Yes

[modern ferry]

its an ellipoid right?

so a= 6, b= 7, c = 5?

[acoustic flint]

yeah, you probably want to use spherical coordinates

[wraith hinge]

You mean polar coordinates but in 3d?

https://en.m.wikipedia.org/wiki/Spherical_coordinate_system

https://sites.science.oregonstate.edu/math/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/jacpol/jacpol.html

[modern ferry]

wow okay

[wraith hinge]

What is u tho?

[modern ferry]

im not sure what that mean but is this not correct

u?

do you mean x?

[wraith hinge]

LHS side of the equation

I mean it is coordinates

[modern ferry]

ohh sorry yes

[wraith hinge]

But we have boundary conditions for area and we have tripli integral which we can transform, but how is LHS working here?

[modern ferry]

hmm i am not sure

is ux not 1/v triple integral x dv?

[wraith hinge]

Yeah it is

[modern ferry]

the same for y and z and then you just put in the bounds?

but what are the bounds?

[wraith hinge]

x^2+y^2+z^2
Is a sphere and we have area bounded somewhere inside the sphere

[modern ferry]

is x from 0 to 6?

y from 0 to 7 and z from 0 to 5?

[wraith hinge]

Wait are we ment to substitute x/6 and y/7 and z/5
And then use it to integrate for substituted variables?

[modern ferry]

i am not sure haha

[wraith hinge]

But it is unit sphere
So how x can be from 0 to 6?

[modern ferry]

yeah your right

maybe that is wrong then

[wraith hinge]

No I am not sure,
You definitely right in takin square root of x y z expressions
But not sure how to use them

<@&286206848099549185>

[modern ferry]

yes me neiher

@limpid dawn

hmm yeah but i dont know if that helps me haha

[limpid dawn]

Wait we integrating a vector

[modern ferry]

it seems like it haha

[limpid dawn]

spherical coordinates were the hint

yes

[wraith hinge]

https://youtu.be/89Kh4HkjoWM?feature=shared

[modern ferry]

so is one bound from 0 to 2pi?

[limpid dawn]

basically we are integrating over a quarter ellipsoid

[modern ferry]

okay

[limpid dawn]

yea because of the 2nd constraint

[modern ferry]

pi/2?

[limpid dawn]

φ is the angle between z and y axis

[modern ferry]

yes

theta

[limpid dawn]

no theta was between x and y

[modern ferry]

ohh sorry

[limpid dawn]

so i think both go from 0 to pi/2

[modern ferry]

but should that not be between 0 and p/2?

haha yes

[wraith hinge]

Is it bounded or it also 0 to 2pi?

Bruh

👍

[modern ferry]

then do we have r as well?

[limpid dawn]

we are basically doing quarter circles

[modern ferry]

from 0 to 1?

[limpid dawn]

r is our ρ

yes

[modern ferry]

yes

[limpid dawn]

0 ≤ ρ ≤ 1
0 ≤ θ ≤ π/2
0 ≤ φ ≤ π/2

[wraith hinge]

Why do we want quarter tho?

[limpid dawn]

because

we have an ellipsoid right

but the second constraint says that x y and z are all positive

which is in the first octane

almost like this

[wraith hinge]

Ah yeah quite obvious. You can't get whole sphere and even half if all of them are positive

[limpid dawn]

yes

[modern ferry]

now we can put it into ux = 1/v integral x dv?

[wraith hinge]

So this question looks more trivial then previous one

[limpid dawn]

ahahah

[modern ferry]

yes right?

haha

they just get harder from here haha

[wraith hinge]

But there is video of derivation on Internet so it is more trivial.
Ah wait there is the same one for previous question.
You just don't need to read boundary conditions from graph

[modern ferry]

oh okay

can i put in the bounds now?

or wait we can find v first right?

is v = integrate r^2 sin theta?

[limpid dawn]

dont you have a paper

like this is not spherical tho it is ellipsoid

i cant seem to find something on the internet

[modern ferry]

hmm okay

[limpid dawn]

x=rasinϕcosθ
y=rbsinϕsinθ
z=rccosϕ

[wraith hinge]

https://youtu.be/naCWsY80Ce8?feature=shared

[limpid dawn]

found something

dxdydz=r² abc sinϕ dϕdθdr

very yummy

[modern ferry]

no haha

why abc

ohh i think i get it

[limpid dawn]

,,\vec{\mu} = \frac{1}{V} \iiint_\Omega \vec{x} \cdot r^2 \cdot abc \cdot \sin\varphi : \dd \varphi \dd \theta \dd \rho

[wraith hinge]

mega?

[limpid dawn]

probably a result from the jacobian

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

so actually

[wraith hinge]

[limpid dawn]

,,\vec{\mu} = \frac{1}{V} \int_0^1 \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \begin{pmatrix} r \cdot a \sin\varphi \cos \theta \ r \cdot b \sin\varphi \sin \theta \ r \cdot c \cos\varphi \end{pmatrix} \cdot r^2 \cdot abc \cdot \sin\varphi : \dd \varphi \dd \theta \dd \rho

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

pretty yummy

[modern ferry]

No not yummy

[limpid dawn]

yes

[modern ferry]

No

How do we integrate that?

[limpid dawn]

with a little bit of hope

[modern ferry]

Okay haha

[limpid dawn]

this points to componentwise integration

my differentialgeometry course was so trashy

[modern ferry]

Oh no

[wraith hinge]

Why?

[limpid dawn]

yes it is that way we integrate componentwise

[limpid dawn]

it was all too fast and unclear

,,\vec{\mu} = \begin{pmatrix} \frac{1}{V} \int_0^1 \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} r \cdot a \sin\varphi \cos \theta \cdot r^2 \cdot abc \cdot \sin\varphi : \dd \varphi \dd \theta \dd \rho \ \ \frac{1}{V} \int_0^1 \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} r \cdot b \sin\varphi \sin \theta \cdot r^2 \cdot abc \cdot \sin\varphi : \dd \varphi \dd \theta \dd \rho \ \ \frac{1}{V} \int_0^1 \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} r \cdot c \cos\varphi \cdot r^2 \cdot abc \cdot \sin\varphi : \dd \varphi \dd \theta \dd \rho \end{pmatrix}

😂

🙏🏻 😭

[modern ferry]

[limpid dawn]

i can't hold my laugh anymore

[wraith hinge]

[solid kiln]

𝔸dωn𝓲²s

[modern ferry]

Okay

[limpid dawn]

the first two are sin^2 integral

the thirs we can use an identity

everything else are constants

[wraith hinge]

👍

[limpid dawn]

$\boxed{\int \sin^2(x) : \dd x = \frac{x}{2} - \frac{\sin(2x)}{4}}$ and $\boxed{\sin(x) \cos(x) = \frac{\sin(2x)}{2}}$

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

We can make use of that

i hope we dont start wrong

[modern ferry]

Yes for the first one we can use sin* cos

[limpid dawn]

no

phi and theta

different variables

[modern ferry]

Ohhhh

Right

[limpid dawn]

Yesss

Let's make it clearer and pull out things we don't need

[modern ferry]

3a * bc / 16V?

[limpid dawn]

,,\vec{\mu} = \frac{abc}{V} \begin{pmatrix} a\int_0^1 \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \rho^3 \sin^2\varphi \cos \theta : \dd \varphi \dd \theta \dd \rho \ \ b\int_0^1 \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \rho^3 \sin^2\varphi \sin \theta : \dd \varphi \dd \theta \dd \rho \ \ c\int_0^1 \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \rho^3 \cos\varphi \cdot \sin\varphi : \dd \varphi \dd \theta \dd \rho \end{pmatrix}

yea it kinda looks better

[limpid dawn]

Now we would apply this

oh I noticed something

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

I hope you are here

[modern ferry]

Yes I am haha

[limpid dawn]

ok let's apply

,,\vec{\mu} = \frac{abc}{V} \begin{pmatrix} a\int_0^1 \rho^3 \int_0^{\frac{\pi}{2}} \cos \theta \int_0^{\frac{\pi}{2}} \sin^2\varphi : \dd \varphi \dd \theta \dd \rho \ \ b\int_0^1 \rho^3 \int_0^{\frac{\pi}{2}} \sin \theta \int_0^{\frac{\pi}{2}} \sin^2\varphi : \dd \varphi \dd \theta \dd \rho \ \ c\int_0^1 \rho^3 \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \cos\varphi \cdot \sin\varphi : \dd \varphi \dd \theta \dd \rho \end{pmatrix}

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

I pulled out the other stuff too

[modern ferry]

Yes

[limpid dawn]

fuck XD

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

This is how we would proceed

like it looks complicated

[wraith hinge]

Yeah that should be painful to type each step

[limpid dawn]

and it's work

but it's straightforward

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

you guys still here?

[modern ferry]

Yes of course

[wraith hinge]

[limpid dawn]

can anyone tell what the last component results to

if plugged in the bounds

it's ok you can cry

[modern ferry]

Haha okay

Which one are u talking about?

[limpid dawn]

the last one with cosine

[wraith hinge]

Cos

One

Bruh I am always second

[modern ferry]

Haha the last equation?

[limpid dawn]

haha yes

just the bounds and evaluating

,, \left [ -\frac{\cos(2\varphi)}{4} \right ]_0^{\frac{\pi}{2}}

[wraith hinge]

Cospi-cos0
1/4-1/4?

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

not quiet

it's minus minus

[wraith hinge]

Oh wait 1/4+1/4

[limpid dawn]

yes

[modern ferry]

Wait I am behind

[limpid dawn]

,,\vec{\mu} = \frac{abc}{V} \begin{pmatrix} a\int_0^1 \rho^3 \int_0^{\frac{\pi}{2}} \frac{\pi}{4}\cos \theta : \dd \theta \dd \rho \ \ b\int_0^1 \rho^3 \int_0^{\frac{\pi}{2}} \frac{\pi}{4} \sin \theta : \dd \theta \dd \rho \ \ c\int_0^1 \rho^3 \int_0^{\frac{\pi}{2}} \frac{1}{2} : \dd \theta \dd \rho \end{pmatrix}

[solid kiln]

𝔸dωn𝓲²s

[wraith hinge]

So we were integrating vector and not f(x, y, z)?

[limpid dawn]

yes

that was our premise

[wraith hinge]

How is that different to doing the same to f(x, y, z)?

[limpid dawn]

it's a vectorial function

[solid kiln]

𝔸dωn𝓲²s

[wraith hinge]

[limpid dawn]

@modern ferry pls

[modern ferry]

Yes okay

[limpid dawn]

this is just $\frac{1}{4} \left ( -\cos \pi + \cos(0) \right)$

[modern ferry]

Yes i see that now

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

ok very good

Now guys

we can again pull put constants

[wraith hinge]

https://youtu.be/S1Z8_B4tdTQ?feature=shared

[modern ferry]

Pi/4?

[solid kiln]

𝔸dωn𝓲²s

[modern ferry]

Yeah!

[limpid dawn]

Now it keeps getting simpler

I will be always there in the storm

[modern ferry]

Haha

[limpid dawn]

ok so now your turn

integral of cosine

precalc

[modern ferry]

It’s sin

[limpid dawn]

sin xD

sine yes

[modern ferry]

Oh haha

[limpid dawn]

sinful function

and of sine?

it's -cosine

[modern ferry]

-cosine

[limpid dawn]

yes

[modern ferry]

Yes

[wraith hinge]

Yes

[modern ferry]

Sine(pi/2) is 1

[wraith hinge]

That's true

[modern ferry]

I am doing ux first

So we get 1 there

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

nice

[modern ferry]

Haha

[limpid dawn]

notice both get to 1

[modern ferry]

So we get r^3

[limpid dawn]

-cos too

yes

[modern ferry]

Yes

[limpid dawn]

last one we get?

[modern ferry]

Integral of r^3 is r^4/4

[limpid dawn]

😭

[modern ferry]

Hold on I am getting confused doing all of them haha

[limpid dawn]

hold on please

slow down

you are going too fast

hehe

[modern ferry]

Okay sorry haha

The last one

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

WOW THIS LOOKS SYMMETrIXAL

[modern ferry]

Haha yes

[limpid dawn]

ok you already cooked

[wraith hinge]

[limpid dawn]

you said r^4/4

(probably googled)

[modern ferry]

Noo haha

[limpid dawn]

r^4/4 from 0 to 1 is?

[wraith hinge]

1/4

[modern ferry]

1/4

[limpid dawn]

ok

[modern ferry]

You beat me haha

[wraith hinge]

My first victory

[solid kiln]

𝔸dωn𝓲²s

[wraith hinge]

So what about centre of mass?

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

idk what that is

maybe yummy knows

[modern ferry]

No

[limpid dawn]

we both for the trash

[modern ferry]

No haha

It’s just the center I guess but I don’t know how to find it

[limpid dawn]

.

[wraith hinge]

It is sphere so

[limpid dawn]

looks like

it's m_x/M

[wraith hinge]

Ah yeah it might not be centre of mass but it probably is since we don't know anything about gravity

[limpid dawn]

m_y/M

m_z/M

[modern ferry]

What is that haha

[limpid dawn]

according to google

[modern ferry]

Goodle

[limpid dawn]

yes sorry

Goodle

I will punish you

[wraith hinge]

Gooble

[limpid dawn]

m_x, m_y and m_z are the components of the vector

and M the mass

[modern ferry]

Hmm okay

[limpid dawn]

where does it say center of gravity

i dont even trust your translation skills

[modern ferry]

here haha

[limpid dawn]

tyngpunkten

[modern ferry]

or main point

but it in the center of gravity

[limpid dawn]

Oh bruh

V = M

here

[modern ferry]

oh

[wraith hinge]

https://youtu.be/JYNbdf9pvag?feature=shared

[limpid dawn]

that would make sense

i am tired and i just woke up 4 hours ago

[wraith hinge]

If you wake up then you shouldn't be tired

[modern ferry]

Haha do you know Swedish?

[wraith hinge]

Not funny sorry

[wraith hinge]

Swedish? I thought it was German

[limpid dawn]

jag vet

[modern ferry]

4 hours is pretty long time

Hahaha

No it’s Swedish

[wraith hinge]

Yeah it is

[modern ferry]

Wait so how do we find V?

[limpid dawn]

it's broken German

[modern ferry]

Ha

[limpid dawn]

that is outta my capabilities I believe

[limpid dawn]

can you post the taska gain

[modern ferry]

Yes

[wraith hinge]

Yeah we need something on LHS I guess

[wraith hinge]

.

/:

[limpid dawn]

[modern ferry]

Yeah maybe

[limpid dawn]

I mean we could solve for V maybe

[wraith hinge]

We need to substitute something for coordinates I guess.

Best@m what is it?

[modern ferry]

Determine haha

[wraith hinge]

Och

?

[modern ferry]

And

[wraith hinge]

Till

[modern ferry]

Can we not just solve out v from our answer?

To

[limpid dawn]

,, V\vec{\mu} = \frac{\pi abc}{16} \begin{pmatrix} a \ b \c \end{pmatrix}

[solid kiln]

𝔸dωn𝓲²s

[modern ferry]

Yeah and is a b and c not 6 7 and 5?

[limpid dawn]

oh yes

[wraith hinge]

Is vector u define centre of gravity?

[limpid dawn]

,, V\vec{\mu} = \frac{\pi 6 \times 7 \times 5}{16} \begin{pmatrix} 6 \ 7 \ 5 \end{pmatrix}

[solid kiln]

𝔸dωn𝓲²s

[wraith hinge]

Yeah it make sense

[modern ferry]

Wait the volume of a sphere is V = 4/3 pi R^3

But do we know the radius?

[limpid dawn]

it's an ellipsoid

[modern ferry]

Oh oops

[limpid dawn]

we don't use radius

abc

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

[modern ferry]

What haha

[limpid dawn]

,, V = \frac{4}{3}\pi abc

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

so 2 steps back

[wraith hinge]

What happened to u?

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

wdym

then you plug in abc and you have the vector

mu which is probably center of mass

and volume is just this formula

i hopefully hpe

brb

[wraith hinge]

Yes but how do we find V first?
Because we have Vu=something
And we want V=something

[limpid dawn]

we use this formula

volume of an ellipsoid

i just realized maybe we need to divide by 4 additionally

no 8

by 8

[modern ferry]

Why?

[limpid dawn]

since it's in the first octane

[modern ferry]

Oh okay

[limpid dawn]

yea

[wraith hinge]

Oh

Oh

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

yeah idk i am not that good at math

the math has been to 70 % always above what was really taught

to me

[modern ferry]

Yes you are haha

[wraith hinge]

Thats always like this /:

[modern ferry]

What happened to 64?

[limpid dawn]

divided by 8

remember

[modern ferry]

Oh haha

[limpid dawn]

yea very funy haha

XD

[wraith hinge]

So anyone good at series questions?

[modern ferry]

Movie series?

[limpid dawn]

depends

[wraith hinge]

No like summation notation

[limpid dawn]

yes

[limpid dawn]

I recommend prison break

[modern ferry]

Haha

[wraith hinge]

I have watched

[modern ferry]

Do you have a question you want to solve?

[wraith hinge]

Sort of

Small part of it

[limpid dawn]

open a channel

[modern ferry]

No so it here

[wraith hinge]

I will

[wraith hinge]

Yeah I know

[limpid dawn]

i will try to take a look at

[modern ferry]

Well I guess I will close this one then

[wraith hinge]

You have 4

Questions

To do

?

[modern ferry]

Haha I can’t finish them another day

It’s enough with two for today haha

[wraith hinge]

Is your book or whatever you have helpful in any way?

[modern ferry]

I don’t have one at the moment haha

I only have one with questions

But it’s pretty good

[wraith hinge]

What year in university are you?

[modern ferry]

First

Are u in pre university?

[wraith hinge]

Yeah

Last year

[modern ferry]

Ohh nice

Do you know what you want to study after?

Or maybe not study at all

[limpid dawn]

yes dont at all

[modern ferry]

Haha

Not math at least

[limpid dawn]

eyyyyyy

you cant run away from math

[modern ferry]

I know

[wraith hinge]

Math and physics and whatever I don't care what I will do after university

[modern ferry]

Oh okay haha

[limpid dawn]

How old are you?

[wraith hinge]

17

[limpid dawn]

[wraith hinge]

?

[limpid dawn]

Indian 100 %

[wraith hinge]

Nuh what

I am not good at math to be indian

And not good at chess either

[limpid dawn]

AJR uirhnfgwessgt er

opztkuomkuzo0ktuea+wAQ D*ÖPGHB;MM

[modern ferry]

You seem to know a lot of math

[limpid dawn]

we europeans are doomed

you still kept up here

and now your series questions

[wraith hinge]

No I don't, I am trying to prepare for step and I don't think I will get anything higher than 3 or 2, that's sucks

[modern ferry]

Yeah you know more than me haha

[limpid dawn]

you are at least calc 2 level

[wraith hinge]

Ah no I just relaxing from revision by asking questions about problems here

[limpid dawn]

ahhh

relaxing

sure

[modern ferry]

Haha

[wraith hinge]

Well it is I just watch random videos and that's it.

Math at university escalet too quickly

[limpid dawn]

did you have to rpove the riemann hypothesis

[wraith hinge]

No

[modern ferry]

I don’t even know what that is haha

.close

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Closed by @modern ferry

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fickle ledge]

I drew a radius from each of the centers to perpendicularly bisect PQ and RS

Also extended the radii to intersect each other but I don't think they're necessarily equal

,rotate

[fickle ledge]

,rotate 180

[trim jolt]

@fickle ledge Has your question been resolved?

[fickle ledge]

<@&286206848099549185>

[trim jolt]

@fickle ledge Has your question been resolved?

[trim jolt]

@fickle ledge Has your question been resolved?

[main drift]

use power of a point

this is one of the most well known applications of it

[fickle ledge]

ok I'll try it rn

if I need to prove they are all on a circle I'd assume RX*XS = (PX)(XQ) so I'll see what else I can do to prove that

Hmm I still don’t know what to do next

Wait I might have smth

[main drift]

radical axis

[fickle ledge]

What’s that

@main drift

[main drift]

Look it up

[fickle ledge]

Yep I am

But aops website is kinda hard to understand

[main drift]

and read this

https://maa.org/sites/default/files/pdf/ebooks/pdf/EGMO_chapter2.pdf

theorem 2.9 is your problem

[fickle ledge]

what is this?

hopefully its this

[main drift]

power of P with respect to omega

omega is just a circle

[fickle ledge]

why are these equal?

[main drift]

read the chapter

everything is explained

[fickle ledge]

yep my bad

[fickle ledge]

ok I kind of get it now

since the distance to the radial axis is the same, the power of point for each of the circles is the same so (PX)(XQ) = (RX)(XS)

where did you get the pdf from

@main drift

[main drift]

Look up EGMO chapter 2

It’s one of the few publicly available chapters

[fickle ledge]

thank you

.close

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Channel closed

Closed by @fickle ledge

Use .reopen if this was a mistake.

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[untold void]

what rule did he use here im confused

[untold void]

how it went outside the bracket

[acoustic flint]

multiply them out then factor dy/dx

[untold void]

oh ok

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@untold void Has your question been resolved?

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Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[slim igloo]

what is 1 + 1

[slim igloo]

i don't know 😦

.close

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[lusty delta]

<@&268886789983436800> |

[north forge]

@slim igloo please dont abuse help channels

[wintry swallow]

He's bored

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[fresh gazelle]

Can anyone help me with the test statistic pls?

I've gotten 5.49447, but they got 5.491

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@fresh gazelle Has your question been resolved?

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Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

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[scenic fractal]

there has to be more information, is B a midpoint of AC? they want you to assume it ig because its clearly visible

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[gusty valley]

LOGIC/PROBABILITY PROOF
there are 15 premises (you probably only need 4 of them tho)

1. P(A | B & C) ≈ 1
2. P(A | B) ≈ 1
3. P(~A | ~B & C) ≈ 1
4. P(~A | ~B) ≈ 1
5. P( ~A & C | ~B) ≈ 1
6. P(C & A | B) ≈ 0
7. P(~C & ~A | ~B) ≈ 0
8. P(C | A & ~B) ≈ 0
9. P(C | ~A & B) ≈ 1
10. P(C | ~A & ~B) ≈ 1
11. P( ~C | A) ≈ 1

we want to conclude that ¬(P(B | ¬A & C) ≈ 1) it's very important that these are approximate premises. This for example, means that you cannot use P(C | ~A) ≈ 1 to mean if ~A then C, that makes an assumption that you would need to support, i.e. P(C | ~A) ≈ 1 ≠ P(C | ~A) ≈ 1. Curious to hear what you come up with. Ping me if you want some preliminary results

[gusty valley]

1. P(ABC) ≈ P(BC)
2. P(AB) ≈ P(B)
3. P(~A~BC) ≈ P(~BC)
4. P(~A~B) ≈ P(~B)
5. P(~A~BC) ≈ P(~B)
6. P(ABC) << P(B)
7. P(~A~B~C) << P(~B)
8. P(A~BC) << P(A~B)
9. P(~ABC) ≈ P(~AB)
10. P(~A~BC) ≈ P(~A~B)
11. P(A~C) ≈ P(A)
    this is in a different format
    Along with the proposition to prove:
    P(~ABC) << P(~AC)
    (note that I just removed the & signs for brevity, ~AB means ~A & B, not ~(A & B))

QUESTION
Is the following a countermodel?? I don't know how to check countermodels, so I need help from someone with that skill

Ω=[0,1] with the standard measure
A=[0, 1-ε²-ε³]
B=[ε⁴, 1-ε³]
C=[1-ε, 1]

Then we find:

P(A|B&C) = (1-ε²-ε³)/(1-ε²) ≈ 1
P(~A&~C|~B) = 0
P(C|A&~B) = 0
P(C|~A&B) = 1

(These are the four hardest ones to check I think, you can check through the rest yourself pretty easily)

On the other hand, we find that
P(B|~A&C) = 1/(1+ε) ≈ 1

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[gusty valley]

<@&286206848099549185> anyone around with a solid basis in probability theory/logic? Who can help me check this countermodel? I don’t know how to do it, but really wanne know whether it works

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[jagged phoenix]

hmm, you can simply put in a concrete small value like 1/1000 for epsilon and just calculate

but honestly looks like 1/10 is enough

[gusty valley]

got it, thanks I will try

[gusty valley]

anyone here interested in the recent formalisations of the boltzmann brain paradox? I am taking a stab at it with some other guys (DM me if your interested in this project), but im stuck, so wondering if there is some genius here who can find a proof:

There are 11 premises (you probably only need 4-6 of them tho) [P(~AB) is short for P(~A&B)]

1. P(ABC) ≈ P(BC)
2. P(AB) ≈ P(B)
3. P(~A~BC) ≈ P(~BC)
4. P(~A~B) ≈ P(~B)
5. P(~A~BC) ≈ P(~B)
6. P(ABC) << P(B)
7. P(~A~B~C) << P(~B)
8. P(A~BC) << P(A~B)
9. P(~ABC) ≈ P(~AB)
10. P(~A~BC) ≈ P(~A~B)
11. P(A~C) ≈ P(A)

we want to prove ¬(P(B | ¬A & C) ≈ 1). It's very important that these are approximate premises. This for example, means that you cannot use P(C | ~A) ≈ 1 to mean if ~A then C: that makes an assumption that you would need to support, i.e. P(C | ~A) ≈ 1 ≠ P(C | ~A) = 1.

Countermodel
Ω=[0,1] with the standard measure
A=[0, 1-ε²-ε³]
B=[ε⁴, 1-ε³]
C=[1-ε, 1]

These are the hardest to satisfy:
P(A|B&C) = (1-ε²-ε³)/(1-ε²) ≈ 1
P(~A&~C|~B) = 0
P(C|A&~B) = 0
P(C|~A&B) = 1
The full check:

1. P(ABC) ≈ P(BC): ABC=[1-ε, 1-ε²-ε³], BC=[1-ε, 1-ε³] ✓
2. P(AB) ≈ P(B): AB=[ε⁴, 1-ε²-ε³], B=[ε⁴, 1-ε³] ✓
3. P(~A~BC) ≈ P(~BC): ~A~BC=[1-ε²-ε³, 1], ~BC=[1-ε³, 1] ✓
4. P(~A~B) ≈ P(~B): ~A~B=[1-ε²-ε³, 1], ~B=[0, ε⁴]∪[1-ε³, 1] ✓
5. P(~A~BC) ≈ P(~B): - same as above -
6. P(ABC) << P(B): ABC=[1-ε, 1-ε²-ε³], B=[ε⁴, 1-ε³] ✓
7. P(~A~B~C) << P(~B): ~A~B~C=∅, ~B≠∅ ✓
8. P(A~BC) << P(A~B): A~BC=∅, A~B≠∅ ✓
9. P(~ABC) ≈ P(~AB): ~ABC=~AB=[1-ε²-ε³, 1-ε³] ✓
10. P(~A~BC) ≈ P(~A~B): ~A~BC=~A~B=[1-ε³, 1] ✓
11. P(A~C) ≈ P(A): A~C=[0, 1-ε], A=[0, 1-ε²-ε³] ✓

anticonclusion:
~ABC=[1-ε²-ε³, 1-ε³], ~AC=[1-ε²-ε³, 1]
=> P(~ABC) ≈ P(~AC)
=> P(B | ~AC) ≈ 1

additional premise
however, we found an additional premise, namely: P( ~A | ~B~C) >> P( A | ~B~C), which contradicts the countermodel:
P( ~A | ~B~C) >> P( A | ~B~C) means (by def. conditional probability) that P(~A~B~C) >> P(A~B~C): ~A~B~C=∅, A~B~C=[0,ε⁴] ❌

I'm super stuck... So wondering if anyone here can find a proof with this additional premise: a proof would solve the Boltzmann brain paradox or is there are countermodel? Curious to hear your thoughts <3

[winter egret]

Hey guys I'm Dutch, what's this channel for

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[gusty valley]

<@&286206848099549185> is there anyone familiar with probability theory?

[latent gyro]

Nope

[vapid pollen]

Only basics

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[spice ferry]

hey I need help calculating the p-value for my dataset (in picture). I can do everything else like r value mean, standard deviation, but hypothesis testing just doesn’t make sense to me. Thanks.

[spice ferry]

I asked earlier today but couldn’t wait around, so I’m hoping I can lock in bc my papers due in a couple hours 😶

<@&286206848099549185>

Help pls..

[haughty tundra]

what is p value

definition

[spice ferry]

It’s a value used to describe the likelihood your results are statistically significant

Basically lower p-value = less chance the results were random and the null hypothesis is true

And vice versa

most tests use (p<0.05)

meaning 95% of your data is confidently statistically significant

Or 95% chance your results are fine

I’ve looked up a bunch of tutorials and stuff but can’t grasp onto the concept of stuff like tailed tests z scores and such

[haughty tundra]

you can ask chatgpt idk lol

[spice ferry]

damn

I feel like no one here likes stats just like me

[haughty tundra]

honestly its hard to get help here

[spice ferry]

yea well idk I joined just for this and saw everyone else’s stuff getting answered within 10 mins

I asked the same question earlier today and waited like 30 minutes while everyone else got answers

[haughty tundra]

lol

cause everyone else asking easy questions

[spice ferry]

yea ig lol

Ok tbh I’m at a dead end this paper is due in 30 and I don’t have enough time 😔😔

I’ll just say I used an online calculator and eat some points off

[haughty tundra]

let me ask chat gpt

[spice ferry]

No it’s ok I already did and it’s kinda too complicated for me

I don’t think I have time for it anyways lol it’s alright

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[fair meteor]

uhm where do i begin with this