help-49

[last slate]

okk

[dreamy lichen]

or if you want to, the x-intercept can be calculated by plugging in y = 0 and then solving for x

[last slate]

do i use that?

and where do i use it

in the one i have rn or the starting equation

[dreamy lichen]

It would work on both

[dreamy lichen]

You can, or you can choose any of the other methods i suggested

they all work equally

[last slate]

soo the x would be (-5,0)?

[dreamy lichen]

yep

and that gives you 2 points

and if you join them (ideally with a ruler), you get a line

[last slate]

ohn okk wait

ill graph it on paper

would it be like this?

[rustic garden]

you have the x and y flipped

x is the horizontal direction

y is vertical

so this is almost right

just fix that and i think its perfect

[last slate]

oops

wait

[rustic garden]

pretty much perfect now

a little bonus thing if you want to help yourself understand this a bit better:

this line, if drawn perfectly, shows all of the points such that if you plug it into the equation, the equation is true

if a point is not on the line, plugging it into the equation would make it false

-2x + 3y = 10
so take your original equation

lets say you want to test (0,0). you can see from your graph that it should be nowhere near the line

plugging in we get 0 + 0 = 10 and it does indeed fail, since its not on the line

from your graph, (-2,2) seems very close if not perfectly on the line

plugging in, we get:
-2(-2) + 3*2 = 10
4 + 6 = 10

and it works!

[last slate]

okayy thanjs

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[lapis slate]

What are the chances that being a front passenger will make you survive and back passenger, differences.

[polar star]

you die either way in the end.

[lapis slate]

Shut up

<@&286206848099549185>

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[keen saddle]

theory will only take you so far

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@lapis slate Has your question been resolved?

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[ionic jasper]

Hi

[ionic jasper]

I’m confused on getting this started

The middle part is confusing

Wait nvm

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[smoky walrus]

@charred fable oops channel closed
yea there's no way to really do that
but you can just say that the original area was 10m^2 for cd_opened = 5, so 20m^2 would be 10

[charred fable]

hi

[charred fable]

the original area of what?

[smoky walrus]

of what they used to decide cd_opened = 5

[charred fable]

oh im sry im new to matlab

are you saying that we can estimate what area orginally used, and based on that multiply that to the cd

[smoky walrus]

yea we can just make up a number that makes sense for the cd_opened to be 5

[charred fable]

alr ok ty

[smoky walrus]

np!

do you have any more questions or should I close this channel?

[charred fable]

you can close it tysm

[smoky walrus]

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[ruby snow]

I'm trying to find the price of a CD (c) and DVD (d) given the contour diagram, but I'm struggling to figure out how to do that 😭

[midnight plank]

@ruby snow Has your question been resolved?

[ruby snow]

here's what i have so far but i don't know if i'm on the right track

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[worn depot]

The speed of cars passing a particular roadblock is normally distributed with a mean of 25 km/h and a standard deviation of 6 km/h. What percentage of the cars go faster than 35 km/h?

[worn depot]

I have z score 1,67

which is 45,25%

im finding just blue one

or the left side too?

blue one is 45,25%

or we finding the pink one?

[polar star]

the pink one

[worn depot]

since MORE THAN 35 km/h

[worn depot]

exact;y

so

50%-45,25%

[polar star]

yes

[worn depot]

,calc 50-45.25

[grand pond]

Result:

4.75

[worn depot]

OMG

IM SO SMART

@polar star

IM SO SMARTT

SEEEE

WOOOOOHOO

!!!

1!1

1

1

1

1

1

IM SO HAPPY

I UNDERSTAND THIS

HAHAHAHAHAHHAAHHAHAHAHAHAHAHHAH

[polar star]

good for you

[worn depot]

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[desert prism]

help it’s my hw

[dawn dagger]

it's an isosceles triangle

and sum of a triangle always is 180°

[desert prism]

agree i think it’s 120

is it correct?

appreciate you

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[desert prism]

please help i don’t get this one and i got a test coming up

.reopen

[midnight plank]

✅

[desert prism]

i got a test coming up and idk how to do this

[lethal path]

If you have M being the midpoint, what information do you know for certain?

By certain this means it comes directly from the definition of the term

[desert prism]

is it D? because i think its SSS

[lethal path]

Yes!

Tricky question cause you really need to pay attention to your givens

And your non givens

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[timber tartan]

Pqrst is a regular pentagon what is the value of x, y, z?

[timber tartan]

Thats y at the bottom left of the triangle

hmm if its regular shouldnt y=z?

And i know each of the pentagons angles should be 108

Does this have anything to do w exterior angles

[dusk pier]

RST and RQP are congruent
Thus angle SRT=angle QRP
Also note that y=z, and that RST=108 degrees, since 180(5-2)/5=540/5=108
Thus, SRT=(180-108)/2=72/2=36, and since y+36=108, y=72.
Wrapping it all up, we have y=z=72, x=108-36-36, so x=36, y=z=72.

[placid spoke]

!nosols

[midnight plank]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[timber tartan]

I dont understand anyth after u said each angle is 108 lmao

[dusk pier]

Do you know the formula for the sum of the angles in a polygon?

[timber tartan]

Yes

[dusk pier]

What is it?

[timber tartan]

(N-2)180 or u can make triangles in the shape and add the measures since eaxh triangle is 180

[dusk pier]

Sure. What should n be?

[timber tartan]

5 sum is 540 each is 108 since there are 5 sides

i know that

I meant after that

[dusk pier]

Oh, ok

Well, I said that SRT=QRP

I guess I was a little ambiguous

Basically, both SRT and QRP are isoceles, since SR=ST and QR=QP

Understand this?

[timber tartan]

Yes

[dusk pier]

Great

So we can deduce that angle STR=angle SRT, right?

By the definition of an isoceles triangle

[timber tartan]

Yes

[dusk pier]

Great

Alright, we know that angle RST=108 degrees.

And that the angles in a triangle add up to 180

[timber tartan]

Oh

I got it

[dusk pier]

Good!

[timber tartan]

108+2k=180

[dusk pier]

Oh, as a fun sidenote, RT and RP trisect angle SRQ. Don't use this, it was just a fun fact I noticed.

[dusk pier]

So k=?

[timber tartan]

36

Right

[dusk pier]

Nice!

You're almost there

[timber tartan]

Yeah lemme check the pic again

[timber tartan]

I havent learnt that

[dusk pier]

So, angle SRT=angle STR=36 degrees. Can you split up angle STP into a y-component and an already known angle?

[timber tartan]

Idk whats a y component but since i know STR is 36 i can do 108=36+y which is 72 so z and y is 72
72+72+x=180 x= 36

Oh so this is what trisect means

Equally divide into three angles

[dusk pier]

Yeah!

[timber tartan]

thanks alot for ur help

[dusk pier]

Yep!

[timber tartan]

btw is there any other way to solve this

[dusk pier]

You're welcome! I'm trying a new, unconventional helping style. Thank you!

[dusk pier]

Well, as I said before, angle SRQ is trisected

So what would that make x automatically?

[timber tartan]

Oh goodluck w that

[timber tartan]

108/3

but how do we know its trisected tho is it cuz its a regular pentagon?

[dusk pier]

Hmm. Well, you have two "diagonals" in a sense, and I think by definition these two "diagonals" trisect the angle.

[timber tartan]

Amd for z and y since the triangle in middle is isosceles we can say 36+2k=180

[dusk pier]

I just had a keen eye for it.

[dusk pier]

Yep!

[timber tartan]

Oh

Makes sense

Thanks again for the help!

[dusk pier]

You're quite welcome!

[timber tartan]

.close

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[jolly roost]

where did i go wrong?

question is asking to solve for a

[dawn dagger]

when you multiply by x(x²+1) you are assuming x != 0

but you plug it in anyway which may lead to contradictions

[jolly roost]

oh yea

but how do i solve a question like this?

[dawn dagger]

well, can you post the original question

[dawn dagger]

the limit doesnt seem to exist other than for a = 1

because if a != 1 you always end up with the form 1/0

[jolly roost]

hmm right

how about this one?

how do we approach solving something of this sort

[dawn dagger]

the only way to get something that may converge here is if you get 0/0 because that's an indeterminate form

[jolly roost]

so we have to make the top = 0 and solve?

[dawn dagger]

yes

well you actually can observe it

[jolly roost]

but isnt (0, 1/27) a point on the graph for the function, so we can plug and solve for a?

[dawn dagger]

well x != 0

so it's undefined

[jolly roost]

oh right

so the only way to solve it would be to get it to indeterminate form?

[dawn dagger]

yes that's your best chances

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[midnight plank]

@vast basin Has your question been resolved?

[midnight plank]

@vast basin Has your question been resolved?

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[exotic laurel]

Can anyone help with e

I thought it would just be a^3

which would be 8

[round parcel]

e has x^3 and f(x).

What's x^3 as you approach x = 2?

What's f(x) as you approach x = 2?

[exotic laurel]

2

[round parcel]

Yes, that's what f(x) is. What about x^3?

[exotic laurel]

8?

[round parcel]

Right, so what's x^3 f(x) as you approach x = 2?

[exotic laurel]

16

[round parcel]

Right.

[exotic laurel]

Thanks! just got confused with the notation

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[round parcel]

You're welcome.

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[wind oxide]

boo

[wind oxide]

integral bash time

hold up i cant find my mse link

wait tf did it get deleted or something

i cant find it...???

well anyhow

It was to find $$\int_0^1 \frac{\ln (x)}{\left(\frac{x+1}{2} \right) \ln \left( \frac{x+1}{2} \right)} dx$$

[grand pond]

rak³en

[wind oxide]

after trying lots of different feynmans that did not want to work

I considered the series expansion

which gives

$$\int_0^1 \sum_{k \geq 1} \frac{(-1)^{k-1} (x-1)^k}{k \left(\frac{x+1}{2}\right) \ln \left( \frac{x+1}{2} \right) } $$

[grand pond]

rak³en

[wind oxide]

can i exchange the sum and intengral here?

i dont think so because that series is unlikely to converge uniformly

but i have no idea how to check uniform convergence

eh eactually

i think it does

can someone confirm

<@&286206848099549185>

[midnight plank]

@wind oxide Has your question been resolved?

[fallow scarab]

Just explain your reasoning

[midnight plank]

@wind oxide Has your question been resolved?

[wind oxide]

without the (-1)^k-1 in uniform convergence, this is simply int_0^1 \frac{ln(2-x)}{(x+1/2) ln(x+1/2) which def converges

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[forest raft]

How am i supposed to solve this limit question?

[forest raft]

couldn't understand anything so have not tried

[modern shard]

alright, is L' Hospital allowed?

[forest raft]

yes

[modern shard]

yeah then just apply it, as it's 0/0 form

[forest raft]

ohh aright

and without the L'Hospital rule, how are we supposed to solve it?

[modern shard]

well we could expand

using taylor series

[forest raft]

alright

[modern shard]

lemme think if we can apply anything else a sec

[forest raft]

sure

[white gate]

LMAO

rare case of LHopital allowed

anyways, multiply and divide by x

[modern shard]

well you could multiply divide with x and log(x+1)/x is 1 when x is tending towards 0
but thats derived by taylor series itself

[forest raft]

our teacher hasn't said that we cannot apply it so..

[white gate]

banger teacher

[modern shard]

fr FR

[forest raft]

fr... he's rly cool

[forest raft]

ok

thanks

.close

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[cloud moon]

Our teacher said if anybody told us how this is solved no homework for while class for a week. The it is a normal pentagon (all sides same angle)

[cloud moon]

Can you explain me how this is solved please?

[midnight plank]

@cloud moon Has your question been resolved?

[cloud moon]

@helper

Oops

<@&286206848099549185>

Somebody save me please 😭

[midnight plank]

@cloud moon Has your question been resolved?

[cloud moon]

Since a lot of time has passed I'll ping I hope it's okay. <@&286206848099549185>

[dusk pier]

What do you need to solve?

[cloud moon]

X

What angle it is

And the Pentagon is regular

[dusk pier]

What are those squiggle symbols?

[cloud moon]

It means those sides are same lenght

If the side has the symbol then the ones that have it are sameenght

So basicly triangle is regular as well

[dusk pier]

Ok

Let me see

[cloud moon]

:D

[cloud moon]

Hey are you still there?

[dusk pier]

I am working right now, sorry

[cloud moon]

Ah no problem if you can look at it when you have the time it would be amazing but np if you can't

[covert kraken]

Is it just a normal pentagon?

[cloud moon]

A regular Pentagon so each side samr length and each angle same

[cloud moon]

That's not true

You should look at the question again it is not the whole angle

If only it was that simple

[covert kraken]

Ah, I see apologies

[cloud moon]

Np I made the same mistake first time :D

[dusk pier]

This is hard

Are C to the point on the pentagon and A to the point on the pentagon parallel? @cloud moon

[cloud moon]

I'm not sure what the question means but this was all the info given

[cloud moon]

<@&286206848099549185>

[thin tide]

It's probably 85⁰

[cloud moon]

Why

[midnight plank]

@cloud moon Has your question been resolved?

[midnight plank]

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[round parcel]

<@&268886789983436800>

[midnight plank]

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[abstract wind]

anyone have an idea ?

[last slate]

hello!

so

lets take this simple

total cars are 4+5+3+3

[abstract wind]

i understand if one time take gonna be

5 / 15 right

[mortal sleet]

Take the propability of drawing a blue card on the first try

[abstract wind]

the probabilty blue card taken one time

[abstract wind]

yeah

[last slate]

yes

5/15

what is the probability it gets drown the second time

[abstract wind]

yes gonna be 5 / 15 but after that there are many other possibilties

it can b x / 14

x can be 5 or 4

[mortal sleet]

Yh but u gotta take the propability of not drawing a blue on the first

[abstract wind]

5 if he got other card, 4 if he got blue card

[mortal sleet]

Or it says within

My bad

[abstract wind]

at the first take

[mortal sleet]

I thought it said at the third try

[abstract wind]

what is within mean anyway ?

[mortal sleet]

1st 2nd or 3rd try

Either one

So like on first try

It would be 1/3

[abstract wind]

so he have three attempt right to take the card ?

[mortal sleet]

Yh

[dusk pier]

Have you tried a method called complimentary counting?

[abstract wind]

i never heard that before

[dusk pier]

You can do P(event)=1-P(event not happening), then calculate P(event not happening). For this, it is fairly simple, as you just calculate the probability of not drawing a blue card in the first 3 cards.

@abstract wind You wanna try the method?

[abstract wind]

yes

[dusk pier]

Ah sorry

[abstract wind]

im still thinking what is event and event not happening mean

[dusk pier]

Yeah go ahead and send the working when you are done

[dusk pier]

Ok

So imagine we roll a dice

[abstract wind]

yeah

[dusk pier]

What is probability of not rolling a 1?

[abstract wind]

5 / 6

[dusk pier]

And what about rolling a 1?

[abstract wind]

the probability rolling one is 1 /6 so not rolling one is the rest

[dusk pier]

Yep

Which is the concept behind P(event)=1-P(event not)

Since probabilites are not greater than 1

[abstract wind]

so event is the probability of the probability of draw a blue card
event not are the probability of not draw blue card

[dusk pier]

Yes.

So calculate that for each turn.

[abstract wind]

hmmmm

is this what u mean ?

[dusk pier]

No.

[dusk pier]

Wait, in a sense, yes.

How did you reach this conclusion?

[abstract wind]

cause it's 3 turn so every turn the total of the card - 1

[dusk pier]

The probability of not drawing a blue card on the first turn is (4+3+3)/(4+5+3+3)=10/15.

This is how I intended you to do it.

[abstract wind]

yeah i already put 10/15 on 1 - 10/15

P(event)=1-P(event not happening)

[mortal sleet]

Why do u need to do the event not happen?

[dusk pier]

I know, but read what I had.

[mortal sleet]

Ur basically calc the same thing

[dusk pier]

Because without it, this problem is extremely difficult to solve. At least, from an analytical sense.

[mortal sleet]

Wouldnt it jus be

5/15 + 5/14 +
5/13

[dusk pier]

Huh?

[mortal sleet]

On the first turn

P(a) = 5/15

[abstract wind]

i think the first turn 5 / 15

[mortal sleet]

Mb 5

Points stands

[dusk pier]

Continuing from here, we can safely assume that the player has not drawn a blue card, since the game has continued. In that case, what would be the probability of not drawing a blue card?

[abstract wind]

yeah

let me think

10/15 probability not drawing a blue card

this is still on the first turn right or two ?

[dusk pier]

Second.

[abstract wind]

owh

there is two possibility i think
10/14 = if on first blue card taken
9/14 = if on first not blue card taken

[mortal sleet]

What grade is this m8?

[abstract wind]

wdym m8 ?

[mortal sleet]

Mate

[dusk pier]

Blue card cannot be taken, since the game has continued. Read the directions.

[abstract wind]

of you are right

[mortal sleet]

I think you are overcomplicating this ngl

[abstract wind]

if blue taken it's over right

so it's 9/14

[dusk pier]

You try it then.

[abstract wind]

idk i think someone posted the question yesterday and i just trying to solve on my own

[dusk pier]

Someone did. This was the exact response I had.

[abstract wind]

ok so
10/15
9/14
8/13

that's the probability of not drawing blue card on turn 1 till turn 3

[mortal sleet]

I still believe it 5/15 + 5/14 + 5/13

Mby not

[dusk pier]

Good! Now apply the formula.

[mortal sleet]

💀

[dusty portal]

This is not how it is applied

[abstract wind]

so how to applied it ?

[mortal sleet]

So

Probability in total should add up to 1

Probability of picking a blue ball is equal to 1- probability of not picking a blue ball

[abstract wind]

yes it is 5 / 15 ?

so ? @mortal sleet @dusty portal @dusk pier

[mortal sleet]

This problem is jus too anoying to deal with

Ive already told u my solution although im not certain if its right or wrong

[dusk pier]

Okay. What I meant for you to do was $1-\frac{5}{15}\cdot\frac{5}{14}\cdot\frac{5}{13}$.

[grand pond]

mathisfun

[abstract wind]

owh ok

[abstract wind]

i try calculate it the prob > 1

[abstract wind]

0.95421245421

[dusk pier]

Yeah.

[abstract wind]

i don't found any option

so answer is E, none of these i think

[dusk pier]

Yes.

[mortal sleet]

My way is wrong

I get what ur meant to do its jus that its too long for me to explain rn

[midnight plank]

@abstract wind Has your question been resolved?

[midnight plank]

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[main current]

Solid logic

[sacred crystal]

it seems fairly direct i just wanted to double check i wasnt missing anything

seems good enough?

[midnight plank]

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[midnight plank]

@sacred crystal Has your question been resolved?

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[tidal turret]

Let ( E = {e_1, e_2, e_3, e_4} ) be the canonical basis of ( \mathbb{R}^4 ) and ( f : \mathbb{R}^4 \rightarrow \mathbb{R}^4 ) a linear transformation such that:

a) ( f(e_1 - e_2) = -e_1 + e_2 )

b) ( f(e_1 + e_2 + e_3) = 2e_1 + ke_2 - 2e_3 + e_4 )

c) ( f(e_1 - 4e_4) = e_1 + e_2 - ke_3 )

d) ( f(e_2 + e_3 + e_4) = -2e_2 - 2e_3 - 2e_4 )

Find all values of ( k ) for which ( f ) is not an isomorphism. Determine ( \dim(\text{Nu}(f)) ) and ( \dim(\text{Im}(f)) ) for the found values of ( k ) and provide a basis for each.

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[prime hornet]

what have you done so far?

[tidal turret]

,w det {{-1,1,0,0},{2,k,-2,1},{1,1,-k,0},{0,-2,-2,-2}} = 0

[grand pond]

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[tidal turret]

k ∈ R - {2,-3}

[tidal turret]

if f is a isomorphism, kernel is trivial and image = codomain = R4

k can be any real number except 2,-3

[prime hornet]

hmm, this matrix looks weird to me

did you put the coordinate vectors as rows instead of columns?

[tidal turret]

is the coordinates wrt E

[prime hornet]

the determinant should be the same regardless, but...

[tidal turret]

det(A) = det(A^t)

[prime hornet]

yes yes, I'm aware

but I would've figured it'd be easier to find A instead, for the other parts

[tidal turret]

with A you mean the matrix representation of f wrt standard basis? (input and output basis) in my notation that is M_E -> E

look, if k is not -3,2 dim(Nu) = 0

and dim(Im) = 4 when k is not -3,2

when k = -3

,w rank {{-1,1,0,0},{2,-3,-2,1},{1,1,3,0},{0,-2,-2,-2}}

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[tidal turret]

when k = -3, dim(Im) = 3

and thus by Rank nully, dim Nu = 1

,w rank {{-1,1,0,0},{2,2,-2,1},{1,1,-2,0},{0,-2,-2,-2}}

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[tidal turret]

when k = 2

dim(Im) = 3, dim(Nu) = 1

K is NOT an isomorphism when K = 2 or when K = -3

[brisk iris]

Why are you doing this ? I sent you the whole proof with the answer ?

[tidal turret]

the basis

[prime hornet]

yes

I agree with what you said, but N(A) is not equal to N(A^T) in general

[brisk iris]

Ah ok, there was more

[prime hornet]

they will have different bases

[tidal turret]

I think they have same dimension

[prime hornet]

but you want a basis later on

they do have the same dim

but you want a basis at the end

[tidal turret]

true

[prime hornet]

which is why I suggested working with A and not A^T

[tidal turret]

yes

[prime hornet]

your values of k seem fine

[tidal turret]

finding Nu(f) is equivalent to finding the nullspace of M_(EE)(f)

[prime hornet]

yes

[tidal turret]

I think, right?

[prime hornet]

yeah

you can honestly go directly to finding a basis and then use that to determine the dimension of the null space/range

it would be more efficient, imo

[tidal turret]

when k = -3

,w nullspace {{-1,1,0,0},{2,-3,-2,1},{1,1,3,0},{0,-2,-2,-2}}^T

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[tidal turret]

Nu(M_EE(f)) = <(6,2,2,1)> when k = -3

,w rref{{-1,1,0,0},{2,-3,-2,1},{1,1,3,0},{0,-2,-2,-2}}^T

[grand pond]

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[tidal turret]

Im(f) = <(-1,1,0,0),(2,-3,-2,1),(1,1,3,0)>

when k = -3

Nu(M_EE(f)) = <(6,2,2,1)> when k = -3
Ker(f) =<(6,2,2,1)> when k = -3

,w rref {{-1,1,0,0},{2,2,-2,1},{1,1,-2,0},{0,-2,-2,-2}}^T

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[tidal turret]

when k = 2

Im(f) = <(-1,1,0,0),(2,2-2,1),(1,1,-2,0)>

,w nullspace {{-1,1,0,0},{2,2,-2,1},{1,1,-2,0},{0,-2,-2,-2}}^T

[grand pond]

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[tidal turret]

when k = 2
ker(f) = <(1,2,-3,1)>

so thats it

in summary, if you want to find the kernel of f, one way is finding the nullspace of M_{EE}(f)

and if you want the Image of f, one way is row reducing the matrix of M_{EE}(f) then taking out the vectors that are linearly independent

[prime hornet]

it sounds like you already know how to solve this problem

[tidal turret]

I didnt

[prime hornet]

well, you're right

[tidal turret]

you made me refresh the matrix representation of linear maps wrt different bases

[prime hornet]

I see

[prime hornet]

but yes, this is correct

[prime hornet]

and this is too

for the image, you can take the linearly indepedent columns of the RREF, or of the original matrix

both are okay

[tidal turret]

lots of this exercises have very similar tricks, only path that I had clear was that maybe I will need to use det = 0 for finding k, because is 4 vectors of R4, so a square matrix

[prime hornet]

indeed!

the isomorphism language it just to dress it up

you just need to find the values of k for which f is not invertible, which is equivalent to finding the values of k for which the matrix rep of f isn't invertible

the det is a very powerful way of doing that

I think you're pretty much set to solve the problem on your own now

you already seem to know exactly what to do

after you find your bases, you can calculate dim(Nu(f)) and dim(Im(f)) easily

[tidal turret]

if MEE(f) is invertible then its bijective, aka isomorphism, idk if invertible matters but for isomorphism we know isomorphism implies is epimorphism and monomorphism at the same time, epimorphism meaning Im = codomain and monomorphism meaning dim ker = 0, idk if its to dress it up or not you can call it bijection or invertible if you see it as MEE(f) but at the start I thought first about f in general not like the matrix representation idk. but for finding the basis of ker and basis of im we needed matrix representation in this case I think, idk

also, the notation my class uses, MEE(f) is non optimal because it doesnt specify which is the entry basis or output basis is idk

[prime hornet]

a linear map (between vector spaces) f is an isomorphism if and only if f is invertible

[tidal turret]

but since both are E, i dont think it matters, idk

[prime hornet]

the thing is like, the language your class is using is too categorical

it makes this concept a little more complicated than it might have to be

[tidal turret]

but man, invertible is only for square matrices what if MEE(f) is no square?

[prime hornet]

for any function f, bijectivity means surjectivity and injectivity

surjective means that the image of f is equal to the codomain

and injectivity of f means that f(x) = f(y) implies x = y

in the case where f is linear, injectivity means that dim ker(f) = 0

the point is, invertibility of f and f being a linear isomorphism are identical when f is linear

[prime hornet]

if f is not square, then the dimension of the domain of f is not equal to the dimension of the codomain of f

in that case, no linear isomorphism is possible

in fact, two vector spaces V and W are isomorphic if and only if their dimensions are equal

so if we want to talk about a linear isomorphism, its matrix representation will always be square

[tidal turret]

oh yeah right, srry

[prime hornet]

no worries

I hope you've been finding linear algebra a bit more tolerable this time around

I know you've been struggling with it for quite a while now

[tidal turret]

XD

[prime hornet]

it's actually really nice to see how much progress you've made

[tidal turret]

just doing what I can

[prime hornet]

keep doing that, then

[tidal turret]

prof doesnt want us to pass

[prime hornet]

you're doing great

[prime hornet]

bleak

[tidal turret]

appreciate it, ty for the clarifications

.solved

[midnight plank]

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[midnight plank]

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[prime swallow]

I require assistance pls and thx

[kindred dagger]

,rccw

[kindred dagger]

gl

[prime swallow]

😭

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[ruby latch]

So, the binomial distribution tells you the odds of k or more trials of p probability succeeding in n attempts.
Is the formula for specific quantities (or more) of trials of different probability succeeding in n attempts somethings similar?

[nova yoke]

something like this? https://en.wikipedia.org/wiki/Poisson_binomial_distribution

[ruby latch]

Not super fast with parsing mathematical formulas, but the initial textual description sounds accurate.

[nova yoke]

the formulas do look grungy

[ruby latch]

Actually, the more I read... This is for when each attempt in the series of attempts has a different probability for success, no?

[nova yoke]

yea

[ruby latch]

Whereas I'm looking for something where the probability of success does not change between attempts, yet there are multiple different types of successes.

[nova yoke]

oh i see

you probably want the multinomial distribution then

https://en.wikipedia.org/wiki/Multinomial_distribution

[ruby latch]

To go with the common marble bag example, say it contains 100 marbles - 1 black, 3 blue, and 96 white, then the odds of pulling at least 3 non-white marble and 2 black marbles in 10 pulls. (putting them back after pulling them)

[midnight plank]

@ruby latch Has your question been resolved?

[ruby latch]

Ehh, it's not quite the formula I was looking for, but I suppose I can work something out.

[midnight plank]

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[sly horizon]

i know how most of the solution works but im kinda confused by the setup. like why do they set x between 0 and 1

[blissful summit]

i really dont understand why they are doing all of this

cuz if u know the riemann sum for k^2, you can already SEE that that thing is in there

[deft sage]

i think 0 and 1 are the bounds

[blissful summit]

ye but where on earth did they come from? nothing given on the question is sugesting that

unless the book they have introduces the riemann sum in a way that implies some form of bound or delta x_i

[sly horizon]

the i in delta x_i is mainly there for if there are unequal subdivisions i think (which there aren't in this case so idk)

actually it might just be for counting a specific subdivision idk

[midnight plank]

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[last slate]

Do both -2 and -6 work?

It seems the answer sheet only has -2, whys that

[sharp wave]

Should work

[midnight plank]

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[last slate]

Where did i make a mistake

[blissful summit]

abs value i think

actually no

where on earth did u get 7?

besides the absolute value thing, assuming its just positive u get

$|W+2|=e^{50t^2+ln9}=e^{50t^2}e^{ln9}=9e^{50t^2}$

[grand pond]

IronVoltage

[blissful summit]

u did the exponeential wrong

@last slate

@shadow scaffold there is a guy posting sus links

[last slate]

Oh right the 9 is multiplied there not added

[last slate]

Huh

[blissful summit]

a guy came in here posting a sus link for a 50$ steam gift card

[last slate]

Lol

.close

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[soft sky]

How do I prove that this right

[blissful summit]

i asked gpt and it said to use 'integral approximation'

hopefully its a useful rabit hole

[soft sky]

No integration allowed

[blissful summit]

i wish i can help but ive never seen it before

sorry chief

[soft sky]

its aight

[blissful summit]

gl tho o7

[soft sky]

ty ❤️

[short gorge]

!no gpt

!nogpt

[midnight plank]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[blissful summit]

boo

[short gorge]

Give me a sec

I'll try to help kevinsnoe

Maybe a first step is to try and find a single fraction that can represent the LHS

[soft sky]

I think this would help

They asked me in sector a to prove this

for every n>1

[short gorge]

this may help

[soft sky]

so using induction would prove it

wait

i think i got it

[short gorge]

oh?

[soft sky]

on the left side of this

[short gorge]

yes

[soft sky]

is basically this without the first 3 fractions

[short gorge]

oh i see where youre going with this

yea yea that makes sense

[soft sky]

so here I would add those 3 fractions on both sides

[blissful summit]

why not just use one term on the LHS?

ignor the terms with the bigger deenominator

like, we know that 1/sqrt(5) is less than that whole sum of fractions

so just set that to be less than the right hand sign

see if its true

and then sandwhich the whole thing

or seomthing

would be easier to work with one term in the LHS anyhow

and the inequality still holds

[soft sky]

well I have to show algerbic way and use induction that I used in the first sector in order to prove

[blissful summit]

hmm

[soft sky]

it's basically RHS + first 3 fractions

[soft sky]

less than this

[short gorge]

yea that works

[soft sky]

lemme do it rq

[short gorge]

now you just need to prove this

[soft sky]

yea its ez

induction for the win

ty @short gorge ❤️

how do i close this lol

[blissful summit]

.close

[soft sky]

.close

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[drifting mason]

Please help w 3f! Im genuinely so confused as to how the teacher did it

[bronze wigeon]

So you get 3e, right?

[drifting mason]

yes

[bronze wigeon]

So 3f is very similar to 3e

[drifting mason]

The teacher got 0.0980 by the way

@bronze wigeon

[bronze wigeon]

The only difference between 3f and 3e is that you also have to consider the case where the first employee speaks neither language and the second speaks both

[drifting mason]

I see

thank you!

.end

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[lapis slate]

Mode is 4
And stand dev is 2

[lapis slate]

Isn’t it 49,38-43,94?

<@&286206848099549185>

[last slate]

Translate the question

[flat veldt]

Is it the area between there

[lapis slate]

Yes

[midnight plank]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[lapis slate]

Hurry up

,calc -3-4

[grand pond]

Result:

-7

[lapis slate]

,Calc -7/2

[grand pond]

Result:

-3.5

[lapis slate]

@flat veldt I

@last slate I

[flat veldt]

Yes

Oops

[lapis slate]

The z score is -3.5

@flat veldt

@flat veldt

[flat veldt]

I'm coming

Gimme a sec

Well the interval is [-1,1]

So $\frac{1-$\mu$}{sigma}$

[lapis slate]

@flat veldt so what’s the answer

[flat veldt]

!no ans

Please don't rush me 😭

!noans

[midnight plank]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[flat veldt]

Find the z scores at the boundaries

@lapis slate

😌

[lapis slate]

2,28%

@flat veldt

@flat veldt 0,135%

Hurry up

I don’t have time

[flat veldt]

The z scores

[lapis slate]

-2

And -3

@flat veldt

[bronze wigeon]

Is this a test?

[lapis slate]

No

[flat veldt]

Then why are you impatient?

[bronze wigeon]

I mean it’s on Canvas, it’s numbered, it has points

It looks like a quiz/homework to me

[flat veldt]

Damn I didn't even realize

[lapis slate]

Is it 6,68%

@flat veldt

@flat veldt

[flat veldt]

Is this an assignment?

[lapis slate]

Yes homework

[flat veldt]

Okay

Calculate your z scores first

[lapis slate]

,calc 2-4/2

[grand pond]

Result:

0

[lapis slate]

,calc -2/2

[grand pond]

Result:

-1

[lapis slate]

-3/2

,calc -3/2

[grand pond]

Result:

-1.5

[lapis slate]

43,32

3,98

,calc 43.31-3.98

[sudden yacht]

Please don't "spam" the chat with these easy calculations @lapis slate

[grand pond]

Result:

39.33

[lapis slate]

Don’t care

[flat veldt]

Stop skipping steps

[lapis slate]

Is this the answer @flat veldt

@flat veldt

[sudden yacht]

<@&268886789983436800>

[lapis slate]

@flat veldt what do I do?

[flat veldt]

Calm down and take your time to solve it.

@lapis slate

[midnight plank]

@lapis slate Has your question been resolved?

[lapis slate]

Just tell me how to do it @flat veldt

I’ll learn that way

[flat veldt]

First step get your z scores which are -1.5 and -2.5

Then find P(-2.5≤X≤-1.5)

[lapis slate]

What is that…

💀💀

[flat veldt]

Probability for normal distribution

[lapis slate]

And what do I do with that

[flat veldt]

That's your answer

[lapis slate]

Where

[flat veldt]

I'm not telling you the answer you're supposed to get it yourself 😭

[lapis slate]

how the fixk

To do that

Brooo

Ur so unclear

<@&286206848099549185> someone else help me

[proud island]

!15min

[midnight plank]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[lapis slate]

Okay great, you can help me.

[vocal frigate]

.close

[midnight plank]

@lapis slate Has your question been resolved?

[urban widget]

Do you have y axis defined

You could use Riemann sums

[midnight plank]

@lapis slate Has your question been resolved?

[proper hedge]

Use normal table

[midnight plank]

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[gusty falcon]

okay so i know how to solve this, but i wanted to make sure if this new way of solving it is pure coincidence or not.

The question is:

The probability of rain on any given day is 1/3. In a four-day period, given that it rains at least twice, what is the probability that it rains on all four days?

[gusty falcon]

there are 11 arrangements which makes sense and of which only 1 is applicable

so 1/11 and then if you multiply by 1/3, you get 1/33

1/3 being the probability of raining

logically doesn't make sense to me lol

so is it just a coincidence?

but i was told this was right so hmm, what am i missing

[dusty portal]

What is the original solution?

[gusty falcon]

i would do:

,w (1/3)^4/ ((1/3)^4 + (4 * (1/3)^3 * (2/3)) + 6 * (1/3)^2 (2/3)^2)

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by mmmm70177

[dusty portal]

[gusty falcon]

well 1/11 * 1/3 is a solution

looks like a lucky result tho

¯_(ツ)_/¯

[dusty portal]

Hmm.

[dusty portal]

This is just binomial, correct?

[gusty falcon]

everything before "i would do" is describing the said solution

[gusty falcon]

no? but also that's my solution

well okay yeah

the denominator is just an application of binomial

[dusty portal]

Ok, was making sure

[dusty portal]

At least the first part makes sense to me, since RRRR=1, RRRN=4C1, and RRNN=4C2. I’m skeptical of the second part, with the probabilities.

[gusty falcon]

yeah well 1/11 part is right

1/11 signifies 1 of the 11 arrangements work

so that part is yeah obvious

but each of these arrangements have different probability

[dusty portal]

That’s what I was thinking too.

[gusty falcon]

yeah but they didn't quote the answer as 1/11

so they know that too

[dusty portal]

I do have a feeling it it coincidental.

However, this might be an application of some obscure formula.

[gusty falcon]

well

if i were to guess it has to do w identify some kind of bijection

maybe i should ask in #probability-statistics

anyway thx, i'll just ask there

.close

[midnight plank]

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[midnight plank]

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[gritty hatch]

sup

#help-18 message

[gritty hatch]

https://cdn.discordapp.com/attachments/903463354654404658/1341129909325795408/1340453306207572073.png?ex=67b588cf&is=67b4374f&hm=449fbf6d8e35c84e49669327fc8f1c5dbc8a0d4e042c08880e8812c2cdd700e4&

1. What does the Remainder divided by the operations/devisor mean and why is it being added to the polynomialpostdivision, i.e., Q(X).

2. Is it related to how if you have a remainder you can represent it as 3(4div2) ? (also whats the notation for a number and a fraction is it 3 4/2?

[sinful topaz]

it comes from the fact that $p(x) = g(x) \cdot q(x) + r(x)$

[grand pond]

eugene_krabs_has_cake

[sinful topaz]

as long as you understand this, you understand the other statement as well

[gritty hatch]

hm...

somewhat?

so you multiplied by the divisor to und oit

but i dont understand R

what is the R term? so far as I have seen it does not exist

as if you have an R term in the first place it would mean the root you were checking was invalid

so it would just be zero

schisms

[sinful topaz]

either r(x) is 0 (the 0 polynomial) or the degree of r(x) is less than the degree of g(x)

but r(x) certainly doesnt just have to be 0

[gritty hatch]

wdym the 0 polynomial and how does R relate to a degree?

and why isnt it guarenteed to be zero unless the root you checked was invalid

[sinful topaz]

by 0 polynomial i mean r(x) = 0

and degree is just something you apply to r(x)

eg degree of 2x^2 + 3x + 4 = 2, degree of 3x^5 + x^2 + 1 = 5

[sinful topaz]

you dont have to apply it to roots

if we had an equation f(x) = 0 and 2 was a root of this equation then sure if you use g(x) = x - 2, then you would have f(x) = (x-2)q(x)

and in this case r(x) would be 0

but it doesnt have to apply to a root

eg take f(x) = x^2 + 3x - 4 and g(x) = x - 2

then f(x) = (x-2)(x+5) + 6

where q(x) = x + 5 and r(x) = 6

[gritty hatch]

what is F of X and is it the same as G?

[sinful topaz]

surely you understand the meaning of p(x) here no?

[gritty hatch]

yes

[sinful topaz]

its just the same thing

[gritty hatch]

aswell as Q (post division polynomial)

alr

[sinful topaz]

p(x) means we have p as some function of x and similarly with f(x) we have f as a function of x

but if your question was if f(x) equals g(x) then no

[gritty hatch]

idk wdym

[sinful topaz]

You know how to multiply polynomials and add them right?

[gritty hatch]

add: idk Mult: (2 + 2) (2 + 2) = (4+4+4+4)

but with X

[sinful topaz]

Would you be able to multiply (x-2) and (x+5)?

[gritty hatch]

(x^2 + x5 - 2x - 10)

[sinful topaz]

Ok now simplify it

[gritty hatch]

(x^2 + 3x - 10)

[sinful topaz]

Ok and now add 6 to that

[gritty hatch]

?

how

[sinful topaz]

It’s just the same as with normal numbers

You take the constant of your polynomial (in this case that’s -10) and add 6 to that

[gritty hatch]

alr

[sinful topaz]

Then replace that as your constant

[gritty hatch]

(x^2 + 3x - 4)

[sinful topaz]

Ok

So you have just calculated (x-2)(x+5) + 6

And you see that you get to the polynomial x^2 + 3x - 4

[sinful topaz]

Which is what I had here

[gritty hatch]

oh

[gritty hatch]

so basically is what your saying that the remainder may be 6 in the case that the coefficient isnt inside the brackets?

and why is the devisor X - 2 and not (X-2) ?

or is it just (x-2)

in such case is it a devisor because theyd multiply by the negative dividing it?

[sinful topaz]

I guess what I’m saying is (you may want to read this a couple of times if you don’t understand, or ask) the remainder isnt necessarily 0 if your two polynomials don’t share a common root

[sinful topaz]

Same thing whether you have the brackets there or not

[gritty hatch]

so why is it like that?

how does that represent division?>

[sinful topaz]

I mean you can put brackets if you want

[gritty hatch]

no just why is the division represented as x-2

[gritty hatch]

but if it isnt common and it isnt zero isnt the division invalid because the divisor didnt have a remainder of zero meaning the polynomial is not accurately described

[sinful topaz]

You can divide any polynomial by any other polynomial, the remainder just changes based of which two polynomials you pick

[gritty hatch]

https://www.youtube.com/watch?v=WRkgMDVIETE&list=PLybg94GvOJ9FoGQeUMFZ4SWZsr30jlUYK&index=38

but then why in the video do they specifically seem to be targeting polynomials that have no remainder to check if its a correct answer?

and how does having a remainder affect things

[sinful topaz]

Because they want to, I guess

Maybe they do multiple videos and in the next one the remainder won’t be 0

[gritty hatch]

this one?

https://www.youtube.com/watch?v=KfrkcGx2Xl8&list=PLybg94GvOJ9FoGQeUMFZ4SWZsr30jlUYK&index=38

[midnight plank]

@gritty hatch Has your question been resolved?

[midnight plank]

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[robust forge]

How can I show that the no of even subsets are equal to no of odd subsets using counting by bijection ?

[lyric charm]

even and odd subsets of what?

and just to be on the same page: when you say "even subset" you mean a subset with an even number of elements (and likewise for odd), yes? @robust forge

[robust forge]

Yup
Subsets of set of first n natural no

[lyric charm]

ok

so subsets of {1, 2, ..., n}

[robust forge]

yes

[round parcel]

Are you sure an even subset isn't where the elements of the subset sum to an even number?

[lyric charm]

does that actually result in equality? cause im not convinced it does

[round parcel]

Yeah, there's a bijection with it.

[robust forge]

well no my teacher clearly stated that even subset refered to subsets with even no of elements and same for odd subsets

[round parcel]

Oh, OK.

[robust forge]

but the problem regarding the sum being even looks interesting too il try it in my spare time

[lyric charm]

ok so in principle the "formally correct" setup is: let E and O be the sets of all subsets of 1:n. aim to construct a bijection E -> O.

[robust forge]

ok

[round parcel]

Hmm, there's a bijection that works for both meanings of even subset.

[lyric charm]

are you thinking of ||xoring with {1}||

[round parcel]

Right.

[lyric charm]

i am deliberately saying this in an obtuse way so as not to give it away

(or minimize the chance thereof if op does open the spoiler)

[robust forge]

what's xoring

[lyric charm]

actually @round parcel i think you can handle this

[lyric charm]

you will find out soon enough. telling you at this point would be spoilers.

[robust forge]

I see

[round parcel]

OK, so let's say you have an even subset.

What are some things you can do to it to make it an odd subset?

[robust forge]

add or remove a element

[round parcel]

OK, and that works in the reverse direction as well.

You can switch an odd subset to an even subset the same way.