help-10

[pearl rune]

which the closest answer is .718

and then i did .718/.92

[spiral maple]

The probability it's marked A isn't .92

that's P(A|Orange)

You want to look at all the A marbles

[pearl rune]

so it would b .92*.13?

[spiral maple]

No

P(A) is P(Green and A)+P(Orange and A)

P(Green and A)=P(A|Green)P(Green)

[pearl rune]

P(Green A)=.13*.22

[spiral maple]

Yes.

[pearl rune]

P(A)=(.92*.78)+(.13*.22)

?

[spiral maple]

yes.

[pearl rune]

so it will be .9622 at the end?>

oh nvm It's still wrong because it's not on the option 😦

[spiral maple]

What are the options?

Cause you likely used intermediate rounding

[pearl rune]

omg... im so sorry

one of the option was .96

it was a scrolling answer so I didnt see 😦

sorry about that!

thank you so much for helping me!

.close

[obtuse pebble]

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[subtle quiver]

my brain is not working right at 1am I have no idea where to even begin, I think its neutons first law but with vectors but I am unsure

[obtuse pebble]

@subtle quiver Has your question been resolved?

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[tawdry pewter]

I'm a little lost as to why you cant just times a11 by B11

[misty terrace]

wdym times a11 by B11

[tawdry pewter]

so 1 times 2

ab11=2

Times the corresponding numbers with the corresponding locations

[misty terrace]

that's not how we define matrix multiplication

[tawdry pewter]

I know but, why because i'm really having hard time wrapping my head around it

[jagged dagger]

if the matrices aren't the same dimensions, that logic fails

imagine a=3x2 matrix

b=2x3 matrix

there's no b3,2

[tawdry pewter]

Yes fair, but even if it was the same dimensions we cant do it

[misty terrace]

i think fighterman0 is asking about why we define matrix multiplication the way it is

[jagged dagger]

exactly so it won't work

[tawdry pewter]

Yea

[jagged dagger]

eh thats a bit deeper. You might just want to look up a proof about matrix multi

i think axler has a few

[tawdry pewter]

thats fine, then I will do that later but now i need help understanding this

[jagged dagger]

Matrix vector multiplication

[tawdry pewter]

How do you this? I got the answers but i still cant wrapp my head around it

isnt it 2*1?

[jagged dagger]

?? A = 2x3 but B = 2x1

[tawdry pewter]

yet some how my tutur got 3*1

[jagged dagger]

I don't you can multiply these 2

lol

[tawdry pewter]

this is what he did

but i dont get it

[jagged dagger]

I mean he says it DNE in your first pic

[tawdry pewter]

yea but he still works it out??

[jagged dagger]

Look, the first picture has A as a 2x3 matrix

your second picture has A as a 3x2 matrix

The matrices aren't even the same

[tawdry pewter]

ooo yea, I think he just rearranged it?

[jagged dagger]

no, they are completely different questions

[tawdry pewter]

because this rule?

[jagged dagger]

look at the numbers

[jagged dagger]

This rule defines matrix multiplication

[tawdry pewter]

oooo yes yes, you right

Just to clarify we always times rows by columns?

like there is no exception to that rule?

[junior inlet]

yep. if you want the reason matrix multiplication is defined like this, maybe watch https://www.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab. (it's also much easier to remember LA formulas with this intuition)

[tawdry pewter]

Awesome will do, thank you!

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[obtuse pebble]

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[dreamy ember]

how can i get the period

[nocturne minnow]

Period is the peak to peak or trough to trough measurement

[obtuse pebble]

@dreamy ember Has your question been resolved?

[dreamy ember]

so what is the period here

[nocturne minnow]

You can determine that. I told you what you needed to know

[dreamy ember]

ohhh

[main lodge]

Can I get a source for that pls

[nocturne minnow]

Why?

You're not the OP

[main lodge]

Because what you've stated sounds like wavelength

And im wondering why its the same as period

[nocturne minnow]

You can look up period of sine/cosine and it'll say what I stated

[nocturne minnow]

And FYI, period is the time it takes to complete a cycle, and wavelength is the distance during one period

[main lodge]

I was confusing physics terminology where the wavelength is the distance from peak to peak or trough to trough and the period is wavelength/velocity

[nocturne minnow]

Wavelength is distance, period is time

[obtuse pebble]

@dreamy ember Has your question been resolved?

[nocturne minnow]

Were you not able to determine period with the info I gave?

[obtuse pebble]

@dreamy ember Has your question been resolved?

[obtuse pebble]

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[neat dust]

I'm having trouble with solving this AOPS question: If a^3 - b^3 = 24 and a - b = 2, then find all possible values of a + b.

[neat dust]

so far I've tried factoring a^3 + b^3 into (a-b)(a^2+ab+b^2

[jagged dagger]

well its a^3-b^3

[neat dust]

and I get ab as 8

yeah

and then I get (a-b)^2 + ab = 12

[pine sail]

Oh wait

[neat dust]

yes, but the book tells me a different answer

[pine sail]

So listen.

You have a-b

[pine sail]

True.

[neat dust]

I get (a+b)^2 -ab = 12

[pine sail]

From here, you can get the value of ab.

[neat dust]

ye

I get ab as 8

[pine sail]

ab is 8, yes.

Now you have

(a-b)^2=a^2-2ab+b^2

Get the value of a^2+b^2

Then use it to calculate (a+b)^2

[neat dust]

can't I just use this tho?

[pine sail]

How'd you get that?

[neat dust]

(a+b)^2 - ab = (a-b)^2 + ab

yes I get a+b as +-sqrt(20)

[pine sail]

What no.

[neat dust]

but the answer in the book doesn't match my answer

[pine sail]

$$(a+b)^2 -ab = a^2+ab+b^2$$
$$(a-b)^2+ab=a^2-ab+b^2$$

[neat dust]

ah

[warm shale]

Sakata Yaksha

[neat dust]

ahhh

forget the signs were different 😄

my bad

[pine sail]

Uh huh.

[neat dust]

thank you

[pine sail]

You're welcome.

[neat dust]

.close

[obtuse pebble]

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[lone blade]

how to do the induction? or is this correct/enough?

@compact shadow lmao last one last one

[obtuse pebble]

@lone blade Has your question been resolved?

[compact shadow]

It is already written in your picture

[lone blade]

so it is correct?

[compact shadow]

Yes

[lone blade]

thanks

.close

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[lusty palm]

circles have radius 1

how do i find the length of the square

the line is straight, top 2 points are tangent

[compact shadow]

(a,2a) is on the cycle (x-1)^2+(y-1)^2=1

(a-1)^2+(2a-1)^2=1, quadratic, solve a, the answer is 2a

Two roots then rule out one root

[pine sail]

Don't give the answer right away atleast.

...

[compact shadow]

Edited

[pine sail]

You legit gave the answer even now though.

Right, now it works.

[compact shadow]

👍

[pine sail]

[obtuse pebble]

@lusty palm Has your question been resolved?

[lusty palm]

all good

[lusty palm]

wait wdym

[compact shadow]

Use coordinates

[lusty palm]

ah got it

ty

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[timid silo]

Yo I have a physics question I dont get

[timid silo]

‘An object (with mass 40 kg) is lifted through a rope. Its acceleration is 2 m/s^2 (to the bottom). What is the force that the rope applies to the object?’

And the answer is 232 N. Im not getting it tbh

What I did is
Fy = (40*2) - (40*9.81)

But its completely wrong

yo what does it mean by the acceleration is 2ms^-2 downwards

isnt the downwards acceleration the weight force

as the mass is being lifted shouldnt it mean that the net acceleration of the body is 2 ms^-2 upwards

I have no idea

I thought that just meant that the lifting is getting slower but idk

hmm

ah ok i got it

so you need to use two formulae

when it says the acceleration is downwards that is your first formula

You know that the net vertical force is equal to the tension in the rope minus the downwards force.

Meaning you have two unkowns

$F_y=T-ma$

you need to derive another formula to find the tension force in the string

[timid silo]

note: the downwards acceleration took into account weight, that is why you can safely say the net force equals the tension take away the downwards force (F = ma) of the object, where a is given

[timid silo]

does that make any sense 🤔, there is just one more formula you need to find T

[warm shale]

OGC

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Im not sure why we need to remove ma from the found Tension

So what I did is
Fy = ma
Which means
F + 392,4 = -80
Where -80 is ma and 392,4 is 40*9.81

In which case I get F = 312
And 312 - 80 is indeed the solution

Tho I dont understand this last step

Ye, so this problem was set such that it directly gave you the net upwards force (the unkown) and the net downwards force... You know this, because when it says string attached to a mass, you clearly know tension is the only force up, however, when it gives you the acceleration it refers to the net downawards acceleration as the problem specifiically referred to. Given Newton's second law, you know that this net force can be expressed as F=ma

So you only have two forces to work with in the vertical direction, and taking the vector sum you end up with Fy = T - ma.

[timid silo]

In this scenario you didnt take into account the tension force acting in the vertical direction.

you considered the net force as the net force of the entire system, whereas the question refers to the downward acceleration as downwards, hence, opposing the tension.

The difference between these forces is what generates the upwards force, so if the forces were equal you would have an equilibrium

Brb 5 min ill try to read everything very carefully lol

Thanks meanwhile btw very helpful 🙂

ye np take ur time 👍

just try to consider that there are only two forces: the tension and the downwards force, and the question made it easier for you in giving you the net acceleration, which allows you to find the net downwards force

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Oh wait I think I can see it now

Because the question wants us to find the force that is applied, which is the summ of all forces on the y axis

So since the acceleration is negative F == 40*-2 which is -80

But the weight force downwards is still the same as I calculated, which is 40*9.81

So the sum between -80 and (40*9.81) and then remove 80

$F_y=T-ma$

[warm shale]

OGC

[timid silo]

this is the net vertical force ye

Yeah that

so u have everything except T

find one more equation

it also would have ma in it

its a more general equation which u may be used to in tension/pulley problems

Got it now, thanks again 👍

[obtuse pebble]

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[timid silo]

ye all good

[obtuse pebble]

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[tawdry pewter]

Who do you do b) it makes little sense. What I did was I found 2A, 3C and 4B

when By the look of it if i add 2A+3C it doesnt = 4B

[fallen mantle]

Subtract the multiplication of 4 with B and multiplication of 2 with A

The divide the whole thing with 3

[tawdry pewter]

but why?

[fallen mantle]

(4B-2A)/3

[tawdry pewter]

I cant do it with 2A+3C?

seems a little weird

I see what you did

[fallen mantle]

Firstly, the C that you calculated in question a or c is not applicable for b

[tawdry pewter]

Am i finding the value of C?

O

[fallen mantle]

the Matrix C in all the 3 questions a, b and c have different values

[tawdry pewter]

Okay I thought C would be the same for whatever reason. Thanks for clarifying

[fallen mantle]

Since 2A+3C = 4B,
By solving algebraically,
3C = 4B-2A
C = (4B-2A)/3

[tawdry pewter]

Yea

Yep awesome got the same answer

Thank you!

.close

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[misty temple]

How does this converge at .75

[misty temple]

I understand that the limit is 0 and therefore the numbers added get smaller

But I’m lost on how the numbers sum to 3/4

[obtuse pebble]

@misty temple Has your question been resolved?

[tawdry pewter]

Thats really strange, I dont know if thats right.

but then again I'm not the brightest in math so

[frosty trench]

I’m guessing that the summation series goes:

1/3 + 1/8 + 1/15 + 1/24 …

Right?

[potent dagger]

u can split that expression and express in terms of two partial fractions and proceed from that stage, it'll become more clear to you

[misty temple]

I got to here

[misty temple]

But I’m not sure what to do next

[potent dagger]

my handwriting sucks so I can't send a pic

[misty temple]

Aw :/

[novel knoll]

\begin{align*}\sum_{n=2}^\infty\frac{1}{n^2-1}&=\frac{1}{2}\sum_{n=1}^\infty \frac{1}{n}-\frac{1}{n+2} \
&=\frac{1}{2}\sum_{n=1}^\infty \left(\frac{1}{n}+\frac{1}{n+1}\right)-\left(\frac{1}{n+1}+\frac{1}{n+2}\right)\end{align*}

[misty temple]

How did you get the 1/n - 1/n+2? Thank you btw

[warm shale]

ScapeProf

[novel knoll]

You have 1/(n-1)-1/(n+2)

Index shift

[novel knoll]

Telescoping now

[misty temple]

Oh hmmm very interesting

Ive never seen that but it makes sense!

Thanks

[misty temple]

What changes when we do this?

Other than n

I’m not sure where the 1/n is from

[novel knoll]

If you plug in n=2 in 1/(n-1) it is the same as plugging in n=1 in 1/n?

[misty temple]

Ohhhh yes thanks

[obtuse pebble]

@misty temple Has your question been resolved?

[obtuse pebble]

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[timid silo]

Can someone help me? How can I write the Euler-Lagrange equation in terms of generalized coordinates and velocities?

[brave bramble]

Question is very general

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[timid silo]

How should I do this?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[tardy epoch]

.close

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[lost radish]

Hi

[lost radish]

Here

In the point of inflection

@timid silo ????

<@&286206848099549185>

[lost radish]

In the point of inflection I

it has the abscissa x = -2

and what is y?

Does it f''(-2) ???

[fervent cradle]

if it wants the point on the graph

then the point will be just f(2)

[lost radish]

But it is point of inflection

Which is f''(x)

@fervent cradle

[fervent cradle]

yes

let the point where the abcissa is be (x, y)

then at that point, f''(x) = 0

then, for that x, y = f(x)

[lost radish]

ok

@fervent cradle

What about this?

[fervent cradle]

nope

should be f(2), not f''(x)

[lost radish]

So it should be f(-2) not f''(-2)

[lost radish]

@fervent cradle is this true?

[fervent cradle]

that's what i said

[lost radish]

Ok thank you @fervent cradle

How to give an feedback? @fervent cradle

t!rep @fervent cradle

[sinful light]

Mahdi has given @fervent cradle a reputation point!

[lost radish]

Ok

.close

[obtuse pebble]

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[timid silo]

Good afternoon folks

[past ore]

Hello guys

growth of a spherical tumor modeled as a function of the population of germs (g in Million) in tumor, v(g) = 1.5 g?, When the diameter of the tumor is 2V15 , and population of germs is 20m and it is growing at a rate of 2m /day. How fast is the diameter of the tumor changing at that time?

My brain froze

[timid silo]

1: Why did we need to calculate all those derivatives when we could just insert a = 1 into the taylor summation formula.
2: How did we go from the left side of the summation equation to the series?

[past ore]

Any help?

[timid silo]

Sir this is wendy's

[fervent cradle]

bruh

get outta here, someone else got here first

one question per channel at a time

[past ore]

K thx

[timid silo]

Yes, I believe this channel has been assigned to me.

No worries tho @past ore

[fervent cradle]

what do you think the taylor summation formula is

[timid silo]

you sum up all the derivatives?

[fervent cradle]

??

[fervent cradle]

yeah so you need the derivatives to calculate it

[timid silo]

How can you capture the range (0 to infinity) with just 4 derivatives tho?

[fervent cradle]

?

basically they noticed a pattern

[timid silo]

^

[fervent cradle]

the zeroth derivative was 1
the first derivative was -1
the second derivative was 2!
the third derivative was -3!
the fourth derivative was 4!

so they say that the nth derivative is gonna be (-1)^n * n!

if you have the nth derivative, the general derivative for any n, then you can get all of them

[timid silo]

Ok, so the point of derivatives is to see what values they return at a given x = a point

[fervent cradle]

sure

[timid silo]

then you identify a pattern, write that into a series form

is that what they did?

[fervent cradle]

you want to work out this

and you know this

so if you just substitute that into the first thing

then you get

[timid silo]

so doing the derivatives lets you find out this component of the taylor series, correct?

[fervent cradle]

yes

[timid silo]

@fervent cradle sorry im jut lost on where each component fits

[fervent cradle]

?

i don't understand

[timid silo]

my line of thinking is

Given a fn and x =a, find the first nth derivative to find the circled component in blue, and use that to find the Taylor series of the fn.

[fervent cradle]

yes

[timid silo]

Ok

but how do we systematically figure out what those series of numbers equal?

[fervent cradle]

??

i mean

you've found the taylor series

[timid silo]

like you figured this out > (-1)^n * n!

[fervent cradle]

you're done

like

[timid silo]

no i don't understand the process rn

[fervent cradle]

so this is the end result

this is still a series

the end result is a series

you're not finding an answer that doesn't have a sigma in it

like, there isn't gonna be any nice answer to 'what does this series of numbers equal'

there's not gonna be a more nice form than this last series

[timid silo]

I guess my question is how the dude on the left became the dude on the right

I don't follow how the blue line happened

How are you able to just come up with an algebraic form like that

[fervent cradle]

ok well if you ignore the minus signs

then it's just 1, 1, 2!, 3!, 4!

so it's just gonna be n!, right

now, considering the minus signs

it goes like +, -, +, -, +...

so that's gonna be (-1)^n

[timid silo]

right

oh wait

how did u know to add ^n

[fervent cradle]

because (-1)^0 = 1, positive

[timid silo]

I get the -1 part

[fervent cradle]

(-1)^1 = -1, negative

[timid silo]

but why ^n?

[fervent cradle]

(-1)^2 = 1, positive

(-1)^3 = -1, negative

[timid silo]

ohhh

OH WOW

ok

[fervent cradle]

so yeah it's just a thing that works

[timid silo]

right

this is why I get hooked on calc

the dopamine kick I just got from figuring out (-1)^n is better than cocaine caffeine

[timid silo]

What about n!?

Who says that's true for all n?

[fervent cradle]

they just took it for granted bc the pattern is so obvious

they just assumed

you can probs prove it explicitly

induction

[timid silo]

Right but I'm a noob so I don't understand why it's taken for granted

like, I wouldn't be able to figure this out on an exam

[fervent cradle]

well it's taken for granted bc it fits lol

[timid silo]

[fervent cradle]

because the 0th one is 0!

the 1st one is 1!

the 2nd one is 2!

the 3rd one is 3!

so they just assume it's gonna be like that always

[timid silo]

because of the numerator?

[fervent cradle]

yes

they just evaluate those at 1

they put 1 in, and it comes out as
0!
-1!
2!
-3!
4!

so they just assume it's gonna be (-1)^n * n! the whole time

[timid silo]

ohh

hm alright

im gonna go give this a try

thanks

@fervent cradlety

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[humble wind]

sin(180-2x) times 7.5 = 9 times sin (x)

[humble wind]

how would i solve that?

[obtuse pebble]

@humble wind Has your question been resolved?

[obtuse pebble]

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[timid silo]

i got up to:

2^2x - 18^x = -8

wait im stupid its just a quadratic nvm

sorry

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[urban glen]

How would you draw the point of, on a graph, something like..

Y is bigger than or equal to -2

× > 3

Y > 3x + 1

[true rain]

It’s not a point l

It’s an area

Draw a line and shade the inequality

[gloomy valve]

Also, because x > 3 and y > 3x + 1, you automatically get y > 10, which makes the y >= -2 condition redundant

[obtuse pebble]

@urban glen Has your question been resolved?

[urban glen]

I'm confused lol

I think I get it nvm

[obtuse pebble]

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[left needle]

Find solutions for 11x+7y=200 in set of natural numbers, i got:
(x,y)=(400+7k,-600-11k);keZ
but how do i make it be in set of natural numbers

[compact shadow]

11 and 7 are coprime

There is a result about solutions of ax+by=c where a and b are relatively prime

if you find one solution (x_0,y_0) then all solutions are (x_0+bt,y_0-at) for all integers t. Use this

[left needle]

yes

i did that

but for just natural numbers/ whole numbers >0

i tried this

[compact shadow]

Just solve range of t then

k

You used k

[left needle]

400+7k>0 => k>-57
-600-11k>0 => k<-54
k e {-56,-55}

is that correct

[compact shadow]

400+k>=0,-600-11k>=0

[left needle]

no just >0

0 isnt a natural number

[compact shadow]

I see, so 0 isn’t in your version

[left needle]

wdym my version?

[compact shadow]

Wrong, k>=-57

k>-57,1428… right? So k>=-57

[left needle]

oh yes correct, i tought that it was -7k my bad

[compact shadow]

So -57,-56,-55

[left needle]

so that it would be negative

[compact shadow]

Yeah negative

[left needle]

thanks

[compact shadow]

Np

[left needle]

could you also help me with this (4x+10y)(8x+7y)=86, i have no idea how to do it

i can only think of writing 86 as a product of whole numbers

but thats too much work, there has to be a "smarter" way to do it

[compact shadow]

Thought the same

But don’t worry for each linear system there will be only one solution (x,y)

Because
(4 10
8 7)
is invertible

But yeah.,, still so much work😂

So 4 solutions

[left needle]

i solved it, you get 8 systems of equations but since its a diophantine equation only 2 valid solutions, rest arent integers

i wrote it in paint

[obtuse pebble]

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[compact shadow]

You know sum of C(n,k) k from 0 to n is 2^n right?

The first equation, the left hand side, is half of the total sum

2. is trivially just solve it

No

2^26

[compact shadow]

(1+x)^n=ΣC(n,k)x^k we just let x=1

Yeah

Just by definition I think, it holds for any c actually

d>=x^2+42-c right

You can always find such d, no matter what x and c are

Make d larger than the right hand side

Yeah

👍

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[coarse tapir]

Just wanted to check my method and answer

[coarse tapir]

I've calculated the perpendicular distance to be 4sin(3) x 80N

and the other force should just be x100N

and equating the two together

I have x = 1.6

am I missing something?, and also which distance have I just calculated?, the distance from wall to x or from x to the point at the end of the rod

seems to simple to be true 🚟

[brave bramble]

Perpendicular distance? Perpendicular to what?

There's an N in your distance

You are confused about whether your equation represents a force, or a distance. 4sin(30) is the height of the wall section, why are we multiplying that by 80N?

Wait, no it isn't even that. What is 4sin(30) here?

Make sure to draw an FBD and split it into components

[obtuse pebble]

@coarse tapir Has your question been resolved?

[coarse tapir]

Sorry should have clarified

The moment of the tension force is perpendicular distance x force

Distance is 4sin(30) and force is 80N

[coarse tapir]

The 4sin(30) is also the perpendicular distance if you make another triangle

Sorry im on phone but I'll try to draw it

[brave bramble]

That blue line is 4sin(30) indeed

[coarse tapir]

Mhm

Aka, the perpendicular distance to the tension

[brave bramble]

Why is that important?

Seeing as gravity acts directly downward, I don't see what you gain from picking a weird frame of reference

[coarse tapir]

Its about the tension... When I'm resolving forces tension acts anti clockwise while gravity acts clockwise

And it says they are in equilibrium

[coarse tapir]

How would you solve it?

[brave bramble]

There's forces holding up the left and right side of the rod. Having the tension of the cable is enough to get the force on the right using an FBD. Then check the moment at the left side.

[obtuse pebble]

@coarse tapir Has your question been resolved?

[obtuse pebble]

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[glossy wharf]

suppose you have n dice. every time you roll all of them, you remove every die that shows a 6. for example, if 2 show up as 6s, you remove them and then roll n-2 dice, etc. (the original problem was with 25 dice)
what is the probability that it will take N rolls to eliminate all the dice?

[glossy wharf]

i started by just looking at n=2 but i wasnt able to figure out a pattern

[river tree]

well maybe the first step to solving this problem could be observing each individual probability in the first trial/roll.
let's say you roll all N dice, with an even 1/6 chance of rolling a six for each individual die. Then the probability of rolling x sixes is equal to the binomial pdf of x, given that p=1/6 and n=N

in the second trial, the probability of rolling x sixes is equal to the binomial pdf of x, given that p=1/6 and n= the amount of dice remaining from the first trial

i can't think of where to go from here though; my method might lead to nowhere. i'm thinking about it

[obtuse pebble]

@glossy wharf Has your question been resolved?

[drifting wraith]

it's that thing where you go in circles

[river tree]

?

[drifting wraith]

what if n = 1

[river tree]

then P=1-(5/6)^N), where N is the amount of trials/rolls

[drifting wraith]

but what is the probability

N = 10, n = 1

[river tree]

yea sorry for the confusing variables im tired

[drifting wraith]

it's 5/6 × [same problem with N = 9]

[river tree]

probability is equal to 1-(5/6)^N, N rolls for n=1 die

[drifting wraith]

we're probably not seeing the same problem

[river tree]

how so

[drifting wraith]

you did something like N or less

or maybe N or more

i read exactly N

[river tree]

are you saying i'm thinking about a cumulative probability?

[drifting wraith]

not really

i don't know

1-(5/6)^N is not(die lives for N rolls)

so N or less

[river tree]

For one die, P(rolling a six in N rolls) = 1-(5/6)^N, correct?

[drifting wraith]

yes

[fervent cradle]

it's a branching process except no branching

[drifting wraith]

that's N or less

[fervent cradle]

gimme a sec, this is semi-advanced probability theory

[drifting wraith]

yeah

[river tree]

it is a difficult problem but i believe it can probably be solved with simple probability math

[fervent cradle]

nope

[river tree]

hopes=crushed

[fervent cradle]

it's a branching process, you want to bring in the heavyweight stuff, gimme a sec

[drifting wraith]

you;re answering a different question

it asks not more and not less than N

and you're only doing not more

[fervent cradle]

well

hmm

[river tree]

you're right, i'm doing it cumulatively when it should probably be considered individually

[fervent cradle]

wait i'm a fool

yeah it's simple enough that you can do it with basic prob

ok so

[drifting wraith]

oh?

[fervent cradle]

probability that it lives through 24 rolls is (5/6)^24
probability that it dies on the 25th roll is 1/6
so probability that it survives 24 rolls but dies on the 25th is just gonna be (1/6) * (5/6)^24

[drifting wraith]

ok

that's 1 die

[fervent cradle]

we know that at least one is like this

[drifting wraith]

tru

[fervent cradle]

the rest can be anything, they can live to 25 or less

ok so really we want the probability that none live to 26

minus the probability that none live to 25

[drifting wraith]

damn

💯

[fervent cradle]

no i'm erratic

i'm changing it up

this is not what i was originally saying

i'm starting over

probability that none live to 26 is (1 - P)^n

where n is the number of dice, P is the probability that any single one makes it to 26

P is (5/6)^26

so it's (1 - (5/6)^26)^n

then the probability that none live to 25 is just

(1 - (5/6)^25)^n

so the probability that at least one lives to 25 is 1 - (1 - (5/6)^25)^n

wait, bugger

conditionals

end me

[river tree]

this is killing me

so the probability that it ends after N trials is equal to (1-(5/6)^N)-(1-(5/6)^(N-1)) or is that not true

[fervent cradle]

ok it's fine

[river tree]

have you solved the problem o great statistical problabilical genius

[fervent cradle]

the probability that none make it to 26 is (1 - (5/6)^26)^n

the probability that at least one lives to 25 is 1 - (1 - (5/6)^25)^n

[drifting wraith]

and you can do it for every number

[fervent cradle]

so the probability that at least one makes it to 25 but none make it to 26 is the second minus the first

[river tree]

substitute 26 for N and 25 for N-1 so that we can get a true representation of the variables initially presented in the problem

[drifting wraith]

and kill them off at the last moment

[fervent cradle]

1 - (1 - (5/6)^25)^n - (1 - (5/6)^26)^n

final answer

maybe

[drifting wraith]

it rings tru

[fervent cradle]

no

[river tree]

maybe

[fervent cradle]

it's gibberish

[river tree]

yea

[fervent cradle]

there's no way it rings true

it's madness

[river tree]

i put it in desmos

[fervent cradle]

it's too complicated

[river tree]

it fucking sucks

how is this problem so difficult

[drifting wraith]

(1 - (5/6)^26)^n − (1 - (5/6)^25)^n

just an extra 1 -

all die minus all die too early

[fervent cradle]

P(A) = P(a single one reaches 25) = (5/6)^25
P(B) = P(a single one reaches 26) = (5/6)^26

P(C) = P(none of the n dice reach 25) = (1 - P(A))^n
P(D) = P(none of the n dice reach 26) = (1 - P(B))^n

||P(C' and D) = P(at least one die reaches 25 but none reach 26) = P(D) - P(C and D) = P(D) - P(C) = (1 - P(B))^n - (1 - P(A))^n = (1 - (5/6)^26)^n - (1 - (5/6)^25)^n

ok done

that looks more reasonable now

[drifting wraith]

but 26 should be before

[fervent cradle]

why

oh no yeah

[drifting wraith]

oh

[fervent cradle]

i goofed it

[river tree]

other than that, though, would it be correct

[fervent cradle]

ok fixed

wait

ok so 26 > 25

so (5/6)^26 < (5/6)^25

so 1 - (5/6)^26 > 1 - (5/6)^25

[drifting wraith]

no, that's larger

[fervent cradle]

fixed already

so (1 - (5/6)^26)^n > (1 - (5/6)^25)^n

so (1 - (5/6)^26)^n - (1 - (5/6)^25)^n > 0

which is what we want to see

done

[river tree]

yea i believe that would be the correct solution

considering the integral of f(n) on the interval zero to infinity would obviously be equal to one i don't see anything wrong with these numbers

[drifting wraith]

awesome

[river tree]

finally i can rest in peace knowing the answer to a silly little probability problem

[river tree]

does this satisfy the answer you needed?

[glossy wharf]

still trying to understand what was done

so it sort of looks like instead of viewing it as one big system, we look at each die individually since each roll is independent.

[glossy wharf]

im a bit confused here though. the original problem was starting with 25 dice (a particular case of n=25 dice), and seeing if it takes N rolls to eliminate all of them, but it looks like this is solving for n dice and 25 rolls.

[fervent cradle]

am i the biggest fool in existence

sorry

i'm sleep deprived

P(A) = P(a single one reaches n) = (5/6)^n
P(B) = P(a single one reaches n+1) = (5/6)^(n+1)

P(C) = P(none of the 25 dice reach n) = (1 - P(A))^25
P(D) = P(none of the 25 dice reach n+1) = (1 - P(B))^25

P(C' and D) = P(at least one die reaches 25 but none reach 26) = P(D) - P(C and D) = P(D) - P(C) = (1 - P(B))^25 - (1 - P(A))^25 = (1 - (5/6)^(n+1))^25 - (1 - (5/6)^n)^25

swapped everything around

[glossy wharf]

im guessing it should be N instead of n here?

[river tree]

arent we all

[fervent cradle]

sure whatever

[river tree]

yea in place of n, they use 25 and 26 to represent the probability at n=26

[drifting wraith]

yeah

[glossy wharf]

P(A) = P(a single one reaches N) = (5/6)^N
P(B) = P(a single one reaches N+1) = (5/6)^(N+1)

P(C) = P(none of the 25 dice reach N) = (1 - P(A))^25
P(D) = P(none of the 25 dice reach N+1) = (1 - P(B))^25

P(C' and D) = P(at least one die reaches N but none reach N+1) = P(D) - P(C and D) = P(D) - P(C) = (1 - P(B))^25 - (1 - P(A))^25 = (1 - (5/6)^(N+1))^25 - (1 - (5/6)^N)^25

[fervent cradle]

same difference

[glossy wharf]

just making sure

[river tree]

also wouldnt you want to use (n-1) and (n) instead of (n) and (n+1), respectively?

[fervent cradle]

no

we want it to reach n but not n+1

n-1 is irrelevant?

wait

fuck

[river tree]

oh never mind we have a different interpretation of what n means

[glossy wharf]

wait is that last line actually correct?

[fervent cradle]

no i think you're right

[drifting wraith]

wouldn't you do minus 1?

[fervent cradle]

oh i copypasted something without changing it

[river tree]

n is the amount of rolls where there are no more dice

[fervent cradle]

ok lemme rego

P(A) = P(a single one reaches n-1) = (5/6)^(n-1)
P(B) = P(a single one reaches n) = (5/6)^n

P(C) = P(none of the 25 dice reach n-1) = (1 - P(A))^25
P(D) = P(none of the 25 dice reach n) = (1 - P(B))^25

P(C' and D) = P(at least one die reaches n-1 but none reach n) = P(D) - P(C and D) = P(D) - P(C) = (1 - P(B))^25 - (1 - P(A))^25 = (1 - (5/6)^n)^25 - (1 - (5/6)^(n-1))^25

_ _
third time lucky

[drifting wraith]

(5/6)^N → a die lives for N rolls
1 - (5/6)^N → a die dies at some point before N+1
(1 - (5/6)^N)^25 → they all die before N+1

(1 - (5/6)^(N-1))^25 → they all die before N

you want them all to die before N+1 but not too early

[glossy wharf]

okay that makes a lot of sense actually

[river tree]

yea im pretty sure this is correct. like 95% sure

[fervent cradle]

the amount of time it took me to do that

i am unfit for any task beyond making simple gruel

[river tree]

well look on the positive side. you can feel satisfied now that you've synthesized a coherent and logical answer to the problem you've been analyzing for about half an hour at least

[high tiger]

is this a geometric distribution?

[glossy wharf]

its been a few months since i took probability so i need to check

but i think so

[fervent cradle]

surely not

[glossy wharf]

close to at least

[fervent cradle]

it's the difference between geometric distributions

the difference between n and n-1

[glossy wharf]

i was kinda of thinking a geometric distribution might come up here since it deals with how many failures before a success

[fervent cradle]

wait, no

i'm blind

it's not

[fervent cradle]

for each individual die it's clearly geometric

overall, no

more complicated

[glossy wharf]

yeah

thats what i meant

alrighty neat

thank you guys so much for the help. that was a doozy to be sure.

.close

[obtuse pebble]

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[void plover]

Is the homework program wrong?

I've even tried getting it with the LCD

[drifting wraith]

it's 3

oh it says that

[void plover]

How did you get to 4(13-x)=5(11-x)?

[timid silo]

U crossed multiply

[void plover]

ahhhhhh ok gotcha thank you very much!

.close

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[low fossil]

Can anyone help me to show this?

I checked in the Wikipedia, they used the young’s inequality

But I didn’t learn that in my courses

Is there any ways to start the proof?

Originally I want to try the mathematical induction

But since it is not proving for integers only, so I am not sure about that

Can I just use basic knowledge of vectors and Cauchy Schwartz inequality to prove it?

[novel knoll]

Can be done with Jensen’s also

[low fossil]

Umm sorry I didn’t learnt that as well

Can I develop the inequality only by Cauchy?

I know it is quite impossible since Cauchy is the special case of holder

[novel knoll]

Uh

Sounds weird you are asked to prove it then to me

https://www.emis.de/journals/JIPAM/images/299_05_JIPAM/299_05.pdf

Claims to only use CS (haven’t checked its valid)

[low fossil]

Umm sorry for the confusion, but when I search in the web, I always find the integration version of the Holder’s inequality

And idk how to proceed

[novel knoll]

Sums are just integrals

[low fossil]

Oh

So they use the integral sign to replace the sigma

Seems we have to use Jensen’s anyway

[low fossil]

Ok thanks for your help, I will try

.close

[obtuse pebble]

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[robust sleet]

partial decompisition

[timid silo]

um

this isnt for me

[short spruce]

what

[timid silo]

my friends been stuck on this question for awhile

[short spruce]

tell your friend to google partial fraction decomposition

[timid silo]

ok

thanks

.close

[obtuse pebble]

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[thin torrent]

would i substitute x = au, y=bv, z=cw. Get boundaries for z between -1 and 1. Get the first jacobian to be abc

then substitute further into polar

and get another boundary

0 to sqrt(1+w^2) for w

and 0 to 2pi for theta

[obtuse pebble]

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[gloomy tusk]

anyone help? how to find Θ

[gloomy tusk]

idk why i find the slant height but my goal is to find Θ

[timid silo]

angle formula

[gloomy tusk]

which one?

can we like determine the Θ at the 3d cone? not the one that's been deform

maybe that'll help

[obtuse pebble]

@gloomy tusk Has your question been resolved?

[gloomy tusk]

<@&286206848099549185>

[obtuse pebble]

@gloomy tusk Has your question been resolved?

[high lily]

theta is defined to be the angle of the sector of the unfolded cone, dunno why you're trying to get it from the 3d form

consider the which parts of the 3d cone correspond to the unfolded cone

[obtuse pebble]

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[gloomy tusk]

.close

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[desert sparrow]

How do I show $X_{n} \overset{L^{r}}{\longrightarrow} X$ and $Y_{n} \overset{L^{r}}{\longrightarrow} Y$ implies $X_{n} + Y_{n} \overset{L^{r}}{\longrightarrow} X + Y$? I am confused when $0 < r < 1$. For $r\geq 1$, Minkowski inequality works

[warm shale]

pepper

[royal basin]

is the case 0<r<1 even a thing worth talking about

might not even hold true then??

[desert sparrow]

Doesn't seem to be qualified in my homework

[royal basin]

might at least be worthwhile to consider convergence towards zero to have less moving parts

[obtuse pebble]

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@desert sparrow Has your question been resolved?

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[autumn patrol]

https://i.imgur.com/dzwVGhJ.png

[autumn patrol]

After reducing the matrix to row ech form and doing the diagonal product, why isnt the determinant -2?

I dont understand the last step of (-1)^1 * -2 = 2

Why interchange rows? Which rows are being interchanged? and why does it negate the determinant

[obtuse pebble]

@autumn patrol Has your question been resolved?

[robust sleet]

it doesn't matter which rows are interchanged

you can find out why it negates by doing a trivial example

[obtuse pebble]

@autumn patrol Has your question been resolved?

[autumn patrol]

im not following

[robust sleet]

take a trivial 2x2 matrix

solve its determinant

then change the rows

what do u get

[autumn patrol]

1 2
3 4

4-6
6-4

hmm ok makes sense now

but why interchange rows anyways? what does it achieve?

[robust sleet]

I think when making the matrix into diagonal form and taking its det

u have to swap 2 rows then, to get the det of the 'actual' matrix

[autumn patrol]

alright

thanks

.close

[obtuse pebble]

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[scarlet silo]

Hi, how to find the cumulative frequency for Q2 and Q1?

[obtuse pebble]

@scarlet silo Has your question been resolved?

[scarlet silo]

<@&286206848099549185>

[obtuse pebble]

@scarlet silo Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[obtuse pebble]

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[toxic canopy]

Prove that the diagonal of the 3D parelelogram ( not a native english speaker ) ABCDA1B1C1D1 cuts through the center of the triangle BDA1

[fringe cedar]

Hm

Let's see

[toxic canopy]

we are working with vectors btw!

[fringe cedar]

The triangle inscribed inside is an isosceles triangle

As far as I can tell because all 3 sides r equal

[toxic canopy]

i just drew the sketch to make it easier for people to visualize

[fringe cedar]

The center of the tringle is also the midpoint

[fringe cedar]

What grade math is this

[toxic canopy]

college math

[fringe cedar]

the incenter

[toxic canopy]

analytical geometry

[fringe cedar]

im in middle school

thats way but i'll try to help

[toxic canopy]

i need to prove it tho

[fringe cedar]

yes

[toxic canopy]

u learn this stuff in middle school?

[fringe cedar]

no

[toxic canopy]

i know what vectors are?

u**

[fringe cedar]

i think you should start by proving its an isosceles triangle

[fringe cedar]

yes at least i think

uh

hm

because

if u slice the thing in half

u would have like a trianglular thing

2

no not like that

[toxic canopy]

i mean

i can visualize it

and see why that is the case u know

[fringe cedar]

go for it

[toxic canopy]

but i need to write it on the paper

[fringe cedar]

i-my brain's stuck with colege math

[toxic canopy]

yeah yeah ahahahaha

[fringe cedar]

im 7th grade

and im doing college math

[toxic canopy]

yeah you probably should avoid doing that

[fringe cedar]

mk doing these

i want challenges cus middle school stuff easy for me

[toxic canopy]

i wish i was thinking like that during my highschool days

i went to the easiest highschool i could find cause why not

and then i went on to go to the hardest university in serbia lmao

Prove that the diagonal of the 3D parelelogram ( not a native english speaker ) ABCDA1B1C1D1 cuts through the center of the triangle BDA1

[fringe cedar]

k

[obtuse pebble]

@toxic canopy Has your question been resolved?

[toxic canopy]

<@&286206848099549185>

[compact shadow]

You let A=0

You have three vectors

a=AB

b=AD

c=AA_1

So AC_1=a+b+c

On the other hand

The center is P, AP=(1/3)(a+b+c)

So P is on the line AC_1

[obtuse pebble]

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[toxic canopy]

thanks!!!

[obtuse pebble]

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[runic arrow]

You randomly take a card from a well-shuffled deck of cards and then put them back.

1. What is the probability that after 100 times you have not drawn an ace of hearts.
2. What is the probability that after 20 times you drew an ace at least once.

[runic arrow]

I don’t really have any work bc I don’t know how to start and we use this formula called la place or whatever

[drowsy girder]

Well in theory if you got 1/54 chance to draw ace of hearts you should draw at least 1 in 100 drawns. But thats not true

[runic arrow]

Well yeah obviously, but that’s the whole point of this exercise it’s to go beyond that logic

[drowsy girder]

@runic arrow ok after a bit of search

I found out the formula
$$(1 - p)^n$$ to be true

[warm shale]

Pluton

[drowsy girder]

p being probability n being tries

[runic arrow]

Yea

[drowsy girder]

Well you just plug it in and thats question 1 done

questio 2 uses same formula

[runic arrow]

Idk what to put in tho

bc if u say like

[drowsy girder]

Well a deck has 54 cards

And 4 aces

Probability it wont happen
$$(1 - p)^n$$
Probability it will happen
$$1 - (1-p)^n$$

[warm shale]

Pluton

[drowsy girder]

Just plug in second formula

[runic arrow]

But then I get one as the answer

oh wait I got it

But the n is supposed to be only for p

Not one

[drowsy girder]

,w 1 - (1 - 4/54)^20

[warm shale]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by Pluton#6301

[drowsy girder]

Seems correct to me

[runic arrow]

Not if u do the first one lol

[drowsy girder]

What do you mean

[runic arrow]

If u put in

1-(1-53/54)^100

My calculator says one

[drowsy girder]

Bc that would be approx 1

,w (1/54)^100

[warm shale]

Results provided by WolframAlpha

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Requested by Pluton#6301

[drowsy girder]

Basically 0

[obtuse pebble]

@runic arrow Has your question been resolved?

[obtuse pebble]

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[gritty dragon]

Hey i am in 8th grade and I tried everything i just don’t get the question

[fallen mantle]

solve for x and y first

do you know how to solve?

[gritty dragon]

no sorry

[fallen mantle]

have you learnt solving simultaneous linear equations?

[gritty dragon]

yes but we got a really bad teacher so I asked about it be she couldn't really help me so i thought I would aks here

[fallen mantle]

okay

do you know what are the methods to solve simultaneous linear equations?

[gritty dragon]

I think you should trace

[fallen mantle]

For the equation 3y - 2x = 12, express y as the subject in terms of x

[fallen mantle]

i hope you know how to do this

[gritty dragon]

yeah i think so

[fallen mantle]

try it now, and let's see what you get

[gritty dragon]

i will

hey i am back

i dont have the answer yet

I tried but I just get stuck

<@&286206848099549185> can you help me i don't understand the question

[drowsy girder]

I dont think thats how its supposed to be solved

[gritty dragon]

oh

[drowsy girder]

Normally when i see this type of questions you need to manipulate somehow both equations

[trim oasis]

I think for an 8th grader, the most straightforward solution would be to solve for x and y

[fallen mantle]

I agree with you

[drowsy girder]

@gritty dragon do you know quadratic formula

[gritty dragon]

yeah I think so but i am not relly good at math

[fallen mantle]

have you tried this?