help-10

[mystic fiber]

he said u were able to make an expression of sum sort and basically simplify that whole lot of 31 equations as some variables can cancel each other out

[obtuse pebble]

@mystic fiber Has your question been resolved?

[obtuse pebble]

@mystic fiber Has your question been resolved?

[obtuse pebble]

@mystic fiber Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

@median skiff Has your question been resolved?

[median skiff]

.close

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[glass junco]

I dont know how to go further with this question. The second picture shows what i have already tried. I am stuck on the integral part or all the steps i did are also wrong idk.

[final thunder]

I’m not sure what you’re trying to do but I would use integration factor here

Ah that’s what you’re doing,

Should get $(e^{-6x}y)’ = -3e^{-6x}(\frac{6}{x} + \frac{12}{x^2})$ just by using the formula

[warm shale]

Pure

[final thunder]

Which I guess is what y ou wrote down $$ e^{-6x}y = -3 \int e^{-6x} (\frac{6}{x} +\frac{1}{x^2}) dx$$

[warm shale]

Pure

[final thunder]

Okay now you have to do integration by parts
Split into two sums like you did

For the first one differentiate 3e^-6x and integrate 1/x^2

Then see if you notice something

[obtuse pebble]

@glass junco Has your question been resolved?

[glass junco]

Aha i asked my teacher, integration by part will not work on this!

You need to guess a value and thats why its really hard

but thanks!

[final thunder]

It will work

The integrals will cancel out

[glass junco]

that will not happen because its x^-1 and x^-2

[final thunder]

Have you tried it

[glass junco]

yes

[final thunder]

Well WA agrees

[glass junco]

with you?

[final thunder]

Yea

[glass junco]

o damn

But i dont see how one of the two x will dissapear

[final thunder]

For the first one differentiate 3e^-6x and integrate 1/x^2

Do this

Leave the second integral alone

[glass junco]

do you try to remove the first or second one?

i guess the second right?

[final thunder]

Second one will cancel with the first one after you perform IBP on first

[glass junco]

ok lets try

1/x^2 => -1/x^-1

you will need to again take the anti derrivative of that

so -1/x^-1 => -ln(x)

[final thunder]

?

[glass junco]

but yea i dont see it dissapear

[final thunder]

Don’t

Why are you doing it twice

[glass junco]

because after the first time the x will not go awa y

[final thunder]

Sure as I said just leave it as it is

And write down the integral you get after first round of IBP

[obtuse pebble]

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[final thunder]

? Did you get it?

[final thunder]

.close

[obtuse pebble]

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[timid silo]

Yo

[timid silo]

So I have this exercise Im failing to solve

Basically I have a cilinder whose diameter is known, d = 2a

A person cuts this cilinder in half following the y axis and then makes another cut 45 degrees on the right, see pic

The question asks to find the volume of the piece that has been cut off from the cylinder

My reasoning is that we can find the area of a section (which is a half circle), and then we find a function that represents how that area changes between different x

So since the degree is 45 we know both lines are of length a

And also we know that depending on the x, the radius of the circle is (a-x) because y = x

So…

Since radius = r = a, we can make a function out of the half circle’s area

Which becomes A(x), because the radius keeps getting smaller

And then we simply use the integral to calculate the volume through the new found A(x)

But this turns out to be wrong… the solution is 2/3 * a^3 — no pi whatsoever

Anyone understands what I did wrong with my reasoning?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

😭

[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?

[forest sinew]

i wonder if switching to planar somewhere would make the problem easier

having a really hard time visualizing the problem though

[timid silo]

I solved it in the end

The areas werent these of half a circle, but rather the areas of circular sectors… silly me

.close

[obtuse pebble]

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[visual dirge]

pls help as fast as you can
Lucca claims that in figure n there are n + n + n + 4 red squares.
5.4 You must explain why Lucca has re

picture:

[obtuse pebble]

Please don't occupy multiple help channels.

[visual dirge]

?

.close

[obtuse pebble]

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[maiden epoch]

this question has our high school math teacher stuck

We're not sure where the error is

Answer is listed as 300/17 (not 100% on this but I remember it was a rational number)

[nova bane]

seems legit

let me just check it

nvm im stuck here too

[maiden epoch]

Makes sense

Our teacher got like a masters in mathematics and he's confused as hell

[spice citrus]

A = s^2 + 2w^2, w = (150 - 4s)/6
I think this right

[maiden epoch]

Yea

[spice citrus]

A = s^2 + (150-4s)^2/18 = s^2 + 150^2/18 - 2 * 150 * 4s/18 + 16s^2/18

Which is a quadratic in s you should minimize

17/9 * s^2 - 200/3 * s + 1250

s = -b / 2a = 200/3/(17/9)/2 = 17.65

Approximately

I think

Nvm

Edited it

[maiden epoch]

He left now but I'll take a screenshot of this and show it to him

Thanks

(I have no idea what I'm looking at)

[spice citrus]

Sry for not using latex I'm on mobile

[maiden epoch]

Yea the unicode writing is a lil confusing

.close

[obtuse pebble]

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[lucid flume]

How can I show that this function is continuous at x=1 but undifferentiated?

[kind hawk]

well clearly the left part is not integrable

it's bigger than the function thats constant 1

[sage geode]

Show that the limit of (f(x) - f(1))/(x - 1) doesn't exist

As x approaches 1

[lucid flume]

Oh, fuck

I think I get it

[obtuse pebble]

@lucid flume Has your question been resolved?

[lucid flume]

.close

[obtuse pebble]

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[fierce mountain]

Using the induction method , show that 3^2n+1 + 2^n+1 | 7 , any n ∈ N

[eternal pond]

For part e why is it a true statement?

[untold flax]

this chnanel is occupied now, go to an avaiable one

[untold flax]

uhh it isn't

I think you wroiei t wrong

wrote it wrong*

[fierce mountain]

I got it as homework

And i have indication

Somehow my professor gets it to

3^3n+2 + 2^n+4 | 5 , any n ∈ N

And i really dont get it

[eternal pond]

whoops sorry

[untold flax]

3^(2n) +1 + 2^(n) +1 do you mean this?

[fierce mountain]

no

[untold flax]

it's just wrong though

[fierce mountain]

Wait i'll send a photo

Maybe i did write it wrong

[untold flax]

oh you did write it wrong

ok well the base case is true, right

now second step of induction is to assume that the general nth case is correct

so now we just have to show that the n + 1th case holds

[fierce mountain]

oh bet i got it

i think

wait

Yes i got it

Thanks

.close

[obtuse pebble]

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[deep steeple]

Reall unsure on how to solve this using parameters. Using Gauss-Jordan elimination I get that the second equation is just 0 = 0 and the first is the same. If I let y = s and z = t the first equation becomes -4x-7s-3t = 13, giving $x = -\frac{13}{4}-\frac{7s}{4}-\frac{3t}{4}$

[warm shale]

Ippo

[deep steeple]

M friend and I both arrived at the same answer, however, I'm wondering, is there any conditions for the values for s and t?

Textbook isn't really clear about this

[obtuse pebble]

@deep steeple Has your question been resolved?

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[stone phoenix]

Not sure what the question is asking or how to approach it

[hallow mesa]

If the question

Was just

Find the domain where

X^2-4 has an inverse

Could you do this

@stone phoenix

[stone phoenix]

Let me try

but what does it mean by that it has an inverse

are you supposed to find the domain of the inverse?

[hallow mesa]

I think it means

[hallow mesa]

This

[stone phoenix]

Ah okay

so the inverse of y = x^2 - 4 would be

let me do it rq

y = sqrt x + 2

i think

[hallow mesa]

Close

[stone phoenix]

sqrt x + 4?

[hallow mesa]

Yea

Sqrt is for both

[stone phoenix]

hm i was thinking but wouldnt the square root make the 4 into a 2 anyways?

[hallow mesa]

No

Because it is

Sqrt(x+4)

[stone phoenix]

Ah

[hallow mesa]

If they where getting

Multiplex

Then you could do that

[stone phoenix]

mhm

[hallow mesa]

And take the 2 out of the sqrt

[stone phoenix]

yeah

[hallow mesa]

Now just need to find a range for this

[stone phoenix]

so wouldnt it be

[hallow mesa]

Not the domain

[stone phoenix]

x > (equal to) - 4

[hallow mesa]

Sorry

One min

[stone phoenix]

ah ok

[hallow mesa]

Sqrt(x+4)

Will give 2 answer

[stone phoenix]

yes

[hallow mesa]

So you need to restrict the output

So it only give

[stone phoenix]

one answer

[hallow mesa]

The positive root answers

For example

If you have sqrt(x)

You can set that y

Is always greater then 0

[stone phoenix]

ah

Find a domain that includes the vertex for which the parabola y = x2 - 4 has an inverse.

[hallow mesa]

Then x^2 has an inverse

[stone phoenix]

ah

otherwise it is not

so it will be

[hallow mesa]

There are

Many domains that work

[stone phoenix]

x > equal to 0

or x < equal to 0

[hallow mesa]

Or 1

Not for the less

But for the grater is works

[stone phoenix]

yea

[hallow mesa]

So for

Sqrr(x+4)

-4 is when it is 0

[stone phoenix]

ye

[hallow mesa]

Do anything below -4 will have

Sorry I meant 4

[stone phoenix]

but sqrt 4 is 2

[hallow mesa]

When x is less the 4

[stone phoenix]

so it will be 2?

[hallow mesa]

Sqrr(x-4)

Will be 0

[stone phoenix]

yeah

[hallow mesa]

When x is 4

[stone phoenix]

yeah

[hallow mesa]

Below that

It won’t be defined

[stone phoenix]

yes

because you cannot sqrt negatives

[hallow mesa]

So it does not have an inverse

Below

[stone phoenix]

undefined numbers imaginary

[hallow mesa]

-4

[stone phoenix]

ah

[hallow mesa]

-4 is an inverse

[stone phoenix]

yes

[hallow mesa]

But below it no

[hallow mesa]

Yes

[stone phoenix]

yea

[stone phoenix]

so would this answer be correct

[hallow mesa]

Yes

[stone phoenix]

for the domain of inverse of y=x^2 - 4

but what does it mean by includes the vertex

because in the textbook the answer for this question is

[hallow mesa]

Wait

You

[stone phoenix]

this is why i am confused

[hallow mesa]

Still have not

[hallow mesa]

Answers the mini problem

[stone phoenix]

ah

[hallow mesa]

Remember

[stone phoenix]

turning pont of parabola

[hallow mesa]

The inverse really

[stone phoenix]

flip y=x

[hallow mesa]

Should be in terms of y

Sqrt(y-4)

[stone phoenix]

yes

oh yeah

because inverse

[hallow mesa]

And if y is greater then or equal

To to 4

Then x would be

What

[stone phoenix]

hm

https://cdn.discordapp.com/attachments/903430332324405288/1035168399401177158/unknown.png

when it is asking for domain that includes the vertex

[hallow mesa]

That just means

The turning point

Of the graph

Is included

Like you said

In the domain

[stone phoenix]

ah okay

so we should find turning point of y= x^2 - 4

as well?

[hallow mesa]

Well

If you include all

Numbers that are inverses

[stone phoenix]

Vertex = (0,-4)

[hallow mesa]

Yes

But if you just include all

[stone phoenix]

yea

[hallow mesa]

Numbers that

Are Inverses

Then the vertex

Will be one of them

[stone phoenix]

yeah

[hallow mesa]

Also

You know

Sqrt(y-4)=x

[stone phoenix]

yea

[hallow mesa]

And y is greater then or equal to

4

Not 5

My bad

So x is

Greater then or equal to what?

[stone phoenix]

-4

no

its less than?

nvm

4

[hallow mesa]

Yes

[stone phoenix]

because less than 4 is negative

*undefined

[hallow mesa]

Yes

But

[stone phoenix]

$sqrt(y-4) = x$

[warm shale]

Gr00by

[hallow mesa]

In the terms of the original function

If you plug in values

Less then 4

Then you get y values

[stone phoenix]

yes

[hallow mesa]

That you can get

With plugging values

Above 4

[stone phoenix]

yeah

[hallow mesa]

Hopefully that made sense

[stone phoenix]

a bit

not rlly tbh

[hallow mesa]

I think

[stone phoenix]

but if u plug less than 4

[hallow mesa]

I can show it better with a graph

[stone phoenix]

wouldnt it be

negative

[hallow mesa]

On Desmos

[stone phoenix]

yea

unless u mean with

y = sqrt(x+4)

because then

ah

i get now

OH

BECAUSE

X > equal to 0

[stone phoenix]

for this function

[hallow mesa]

Yea

[stone phoenix]

and then it is x < equal to 0

[stone phoenix]

for this function

https://cdn.discordapp.com/attachments/903430332324405288/1035170259998625792/304051441964679168.png

so thats why there is two answers

[hallow mesa]

Yes

[stone phoenix]

x ≥ 0 or x ≤ 0

[hallow mesa]

Yes

[stone phoenix]

because of inverse and original function

[hallow mesa]

Ya

I’ll show with a graph

[stone phoenix]

that would be nice

[hallow mesa]

One min

See how when x is greater then 0

[stone phoenix]

AH

It is half of parabola graph

[hallow mesa]

You get one solution

Also

[stone phoenix]

because the inverse of parabola is always only one side

[hallow mesa]

In the future

You don’t need to do all the work

[stone phoenix]

yeah

[hallow mesa]

You just need to split it in half

[stone phoenix]

ah okay

[hallow mesa]

And that will be

The vertex

[stone phoenix]

so you can graph the parabola

or find the Vertex

[hallow mesa]

Yes

[stone phoenix]

through finding two x roots and then divided by two

then sub back into original equation

[hallow mesa]

If you don’t need to include the vertex

Then you get

x>1

Works to

[stone phoenix]

ah okay

[hallow mesa]

Cause it still

Only

On one half

[stone phoenix]

so the reason why this one was -

was because the vertex was in -4

[hallow mesa]

Of the parabola

[hallow mesa]

Yes

Sorry

[stone phoenix]

so if the vertex was (0,0)

[hallow mesa]

For the confusion

[stone phoenix]

then the domain would be

x > equal to 0?

or no

[hallow mesa]

That is correct

[stone phoenix]

its okay

because now I learnt the idea

so I can like use it later too and will remember it

[hallow mesa]

Graphing

[stone phoenix]

yeah

[hallow mesa]

Can always help to

[stone phoenix]

yeah

so if vertex is in negatvies

basically

it is two

x > equal to 0

and x < equal to 0

[hallow mesa]

That’s only true

If vertex

[stone phoenix]

has no y axis

[hallow mesa]

Is on the y axis

[stone phoenix]

yea

[hallow mesa]

This

[stone phoenix]

but if vertex is like (-4,-4)

then what?

you also need range?

[hallow mesa]

No not really

Just x value

[stone phoenix]

hmm

[hallow mesa]

Should I show a graph?

[stone phoenix]

yes please

[hallow mesa]

Ok

One min

Ok one minute

It’s uploading

Okay it’s done

[stone phoenix]

Thanks

[hallow mesa]

See how

When is x is greater the -4

It splits the graph

[stone phoenix]

yeah

[hallow mesa]

In two

Nothing to do with y

[stone phoenix]

yeah

[hallow mesa]

Just the x value

Of the vertex

[stone phoenix]

yeah

so

would the answer still be

x > equal to 0

x < equal to 0

or it depends

on the inverse of the equation

and what the equation is

[hallow mesa]

It would be

x>= -4

For this one

Last one it would be 0

[stone phoenix]

ah okay

[hallow mesa]

And in general

It depends on the x value of the vertex

[stone phoenix]

oh ok

so the x value of the vertex is what decides the domain

[hallow mesa]

Yes

Cause see in the graph

[stone phoenix]

https://cdn.discordapp.com/attachments/903430332324405288/1035173666863992872/B18B7CD8-8FB7-4162-B32F-05C78B2DBA02.png

[hallow mesa]

The vertex x value

Splits the graph in two zones

Hear wait

I’ll show what I mean

[stone phoenix]

thanks

[hallow mesa]

Okay it’s uploading

See how x=-4

[stone phoenix]

yea

[hallow mesa]

Splits it into two

[stone phoenix]

mhm

[hallow mesa]

Different areas

So one side

[stone phoenix]

the left and right

[hallow mesa]

Is one of the inverse

And the other is the other inverse

[stone phoenix]

yeah

because inverse has to do horizontal line test

so parabola must be in hal

halkf

[hallow mesa]

Good job

I got to go

[stone phoenix]

All good

[hallow mesa]

In a minute

[stone phoenix]

thanks

that was good enough

[hallow mesa]

Sorry

Hopefully

[stone phoenix]

Dw i got it, and i can study a bit more on my own

[hallow mesa]

But you got the concept

Right

[stone phoenix]

yeah

[hallow mesa]

Ok

See ya

[stone phoenix]

cya

[obtuse pebble]

@stone phoenix Has your question been resolved?

[obtuse pebble]

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[stable rain]

Let f(z) be holomorphic in a domain containing the annulus 1 ≤ |z| ≤ 2. Suppose there exists g(z) holomorphic in the domain {z : |z| < 2} and h(z) holomorphic and bounded in the domain {z : |z| > 1} satisfying

f(z) = g(z) − h(z).

Prove that g'(z) is determined uniquely by f(z).

[drowsy grail]

The answer is 37 dumbass

Tantalizing

[wild swallow]

👽

[drowsy grail]

Heck

[stable rain]

so i tried letting F, G, H be the analytical continuation of f, g, h respectively on C and they all satisfy F(z) = G(z) - H(z) due to theorem of isolated zeros in some domain that contains a point of accumulation

not sure what else to do :c

[wild swallow]

what do you think

theres about 1 theorem

in total

[stable rain]

no i wasnt thinking

[wild swallow]

that relates derivatives to the functions themselves

[stable rain]

i was doing jordans lemma

so um

how do i specify contours

like i did

C_R = {Rt - R(t-1): t in [0,1]}

...

i q is being auto modded

S_R = (Re^(qi): q in [0,pi]}

then i proceded to use em as contours

does that work?

[wild swallow]

thats

too much work bruh

[stable rain]

residue theorem?

[stable rain]

oh

[stable rain]

lol

ok but essentially they r sets

contour integrals is like

integrating over a contour which is represented by a set of vals

ok but the prev issue

[wild swallow]

well

not exactly

theyre curves with an orientation

[stable rain]

with jordans lemma

i got that its upper bounded by

$\frac{ 2 \pi R (R^2)}{(R-1)(R- \sqrt 2)( R- \sqrt2 )}$

but this is not vanishing

[stable rain]

its ~2pi

hm

i can just work with that right

use it as 2pi rather than 0

as R-> inf

[stable rain]

whats it tho

[wild swallow]

uh

didnt check but

that seems like you have too many powers of R

should be R^2/R^3

anyway

[wild swallow]

literally

cauchy's integral formula

[stable rain]

in num i got R^2 from x^2 and the last R from the contour

[wild swallow]

there is no R from the contour

[stable rain]

wasnt this residue theorem

[wild swallow]

you're using jordan's lemma

[stable rain]

wait isnt ML bound M from contour like

2pi R

[wild swallow]

its not ML bound

its jordan

jordan is stronger than ML bound here

[stable rain]

but we use ml bound right

[wild swallow]

no

you use jordan

[stable rain]

to show it vanishes

[wild swallow]

no

you cant use ML bound here

its not strong enough

[wild swallow]

residue theorem is like

[stable rain]

:c

[wild swallow]

cauchy's integral formula

omega boosted x9001

oh wait

hold on

i think i quoted it wrong lol

[stable rain]

this?

hais wait

how do i jordans lemma here

[wild swallow]

literally

apply the statement of the theorem

[stable rain]

this is like when we use residue theorem with a pole of order 2 right

[wild swallow]

its cauchy's integral formula

[stable rain]

isnt this the same

[wild swallow]

residue theorem is proved using cauchy's integral formula

[stable rain]

hm

ok the other 1

with jordans lemma we got

$(1 - e^{- \pi R}) \max_{z \in S_R} |g(z)|$

[stable rain]

so as R-> inf, e^-piR vanishes

so we get O(R^-1)

[wild swallow]

there should be a pi prefactor but it doesnt really matter

g vanishes

so you're done

[stable rain]

its pi/pi = 1

alpha = pi

[wild swallow]

oh alpha = pi

ok

[stable rain]

ok imma

come back after thinking about the other Q

imma

BREAK rn

or some would even say

Breakthrough

lmao

[wild swallow]

[stable rain]

Level up Imperial Sky Layer 1

thanks

❄️

[wild swallow]

❄️

[stable rain]

.close

[obtuse pebble]

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[drowsy grail]

You're still trying to solve this simple problem? You're dumb

[obtuse pebble]

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[humble dock]

Idk how to solve this?

[cedar lichen]

What have you tried

[obtuse pebble]

@humble dock Has your question been resolved?

[obtuse pebble]

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[versed drum]

Can someone please help me with this problem here?

[versed drum]

<@&286206848099549185>

[obtuse pebble]

@versed drum Has your question been resolved?

[marble junco]

try to decompose each term

[obtuse pebble]

@versed drum Has your question been resolved?

[versed drum]

How?

Can you please elaborate?

Send a solution perhaps?

[marble junco]

for example, try to make 1/(sqrt(0.2) + sqrt(0.4)) something like a/sqrt(0.2) - b/sqrt(0.4)

you need to obtain a and b values

[versed drum]

.close

[obtuse pebble]

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[obtuse pebble]

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[nocturne karma]

Hello I have a question over a trig substitution problem

[nocturne karma]

The problem is the integral of 9x^3 / sqrt of (1+x^2) dx

That would be an improper fraction right?

Pls help

Yes that

This is what I have so far

Not sure what to do now though as I have a 9x^3 up top

[warm shale]

luckydongdong

[marble junco]

you need integral?

[marble junco]

try to define $u=x^2$

[warm shale]

luckydongdong

[nocturne karma]

wdym

I have to use trig sub to solve it

[marble junco]

oh ok

sorry

[nocturne karma]

Do you know how?

[marble junco]

your figure looks correct

change every x to $\theta$

[warm shale]

luckydongdong

[marble junco]

including dx

[spice citrus]

Try to just sub it in

I think you'll get sin x * tan^2(x)

[nocturne karma]

ok

So the top would be tan^3(x)?

[spice citrus]

Then use a trig identity for the tan^2

And bottom?

[nocturne karma]

Top would be tan^3x(sec^2x)

And bottom sec (theta)

So it cancels

And now it’s tan^3theta (sec(theta)

[spice citrus]

Yes, I was wrong

[nocturne karma]

So I can write tangent as sin/cos and sec as 1/cos

So now I have sin^3(theta)/cos^3(theta)*1/cos(theta)

[spice citrus]

Try a substitution again

[nocturne karma]

Wdym

[spice citrus]

Rewriting tangent and secant isn't necessary

[nocturne karma]

It’s also a IBP problem now

[spice citrus]

No, u = tan x works out I think

[nocturne karma]

oh ok

So u^3(sec(theta))?

Now I could do IBP?

[spice citrus]

du = sec^2(x)dx right?

[nocturne karma]

Let me try

[nocturne karma]

Yes

[spice citrus]

Wouldn't the integrand be u^3du?

[nocturne karma]

Wait what

[spice citrus]

Oh, nvm thought it was tan^3 * sec^2

[nocturne karma]

I’m so confused

[spice citrus]

Okay, we have integral of tan^3 * sec right?

[nocturne karma]

yes

Thetas

[spice citrus]

Yes

u = sec theta

This should work if you write tan^2 as sec^2 - 1

[nocturne karma]

What

[spice citrus]

du = sec(theta) * tan(theta) dtheta

[nocturne karma]

Can you write it on paper so I can see it?

Easier than writing thetas

Or on online notes

[spice citrus]

Maybe online

[nocturne karma]

ok

[spice citrus]

Hopefully this is correct, I have to go now

[nocturne karma]

I meant like an online sketch pad lol

[spice citrus]

I'm on mobile lol

[nocturne karma]

Oh

[obtuse pebble]

@nocturne karma Has your question been resolved?

[obtuse pebble]

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[frail karma]

Hello guys i need help with this question

[frail karma]

Null hypothesis is people cant tell the difference between butter and margarine

expected number is 100

I dont know about 2b tho

[plain stag]

an alternate form of the null hypothesis is that the proportion of people that can correctly identify butter is 0.5

[frail karma]

5%is 10 people

Idk now

[plain stag]

true but not relevant

you want to test what is the probability of getting 120 people (or more) out of a sample of 200, given that the proportion is 0.5

that's where the 5% significance comes into play

[frail karma]

hmm

I am trying to think about it

80 people

Can not differentiate

Welp i am still confused

[frail karma]

Do you mind explaining a little more

[plain stag]

What doesn't make sense

[frail karma]

I am not sure how to think about it still

[plain stag]

The null hypothesis is that the proportion of ppl who can correctly identify butter is 50%

[frail karma]

Yeah i know that

[plain stag]

Under the assumption that the null hypothesis is true, your goal is to compute the probability of obtaining a sample where at least 120 out of 200 people correctly identify butter

[frail karma]

I just dont know 2b

[plain stag]

To reject the null hypothesis at 5% significance is to do the above computation and get a probability of less than 5%

[frail karma]

How do i compute that?

[plain stag]

You have a sample of 200, and a constant proportion of success of 0.5

What distribution sounds appropriate?

[frail karma]

Binomial

[plain stag]

Indeed

[frail karma]

0.003

That is less that 0.05

So

We accept the null?

[plain stag]

You don't want =, you want >=

[frail karma]

Hmm

[plain stag]

I suggest you do a little more reading on hypothesis tests, you want to compute the probability of getting a result at least as extreme

[plain stag]

With n=200, every point will have tiny probability

[frail karma]

I will!

[plain stag]

Good

So if we assume the null hypothesis is true, the probability of getting 120 or more people is 0.3%

So do you think the null is true?

[frail karma]

P(X>=120) is 0.00284

[plain stag]

=

Unless your screenshots are saying different things

[frail karma]

Its the same

[plain stag]

P(X = 120) = 0.001 according to your first screenshot

[frail karma]

Ye

I edited

[frail karma]

0.3 is less than 0.5

Which is not = to 0.5

So we reject?

[plain stag]

What is 0.5

[frail karma]

50% can identify correctly

[plain stag]

You're getting your percentages mixed up

,calc 120/200

[warm shale]

Result:

0.6

[plain stag]

The proportion who got it right is 0.6, which is greater than 0.5

However the probability of getting a sample proportion of at least 0.6 given that the null proportion is 0.5 comes out to about 0.003

[frail karma]

oHH

At 0.05 significance level we cant reject

We are comparing 0.003 to 0.05

[plain stag]

Yes

If the probability of getting a sample at least as extreme as what we got is less than 5%, we reject

[frail karma]

Yeah its so small

Compared to 5%

[plain stag]

So what does that tell you about the null?

[frail karma]

We can reject it

[plain stag]

Not just 'can'

[frail karma]

We reject our null hypothesis

Wording matters

[plain stag]

Indeed we do

[frail karma]

Thank you for helping me

<33333

[plain stag]

No worries, good luck moving forward

[frail karma]

Oh we are not testing for 120

[plain stag]

What does that mean

[obtuse pebble]

@frail karma Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[smoky zenith]

If cos and sec has the inverse restriction of -pi over 2 to pi over 2 and SIn and cos sec has the inverse restriction of pi to 0 whats the inverse restriction of tan and cot?

[daring rock]

To be clear, you're referring to the range of the inverse functions, not the domain, right?

And you have them backwards

arcsin ranges from -pi/2 to pi/2
arccos ranges from 0 to pi

And arctan ranges from -pi/2 to pi/2

[smoky zenith]

oh ya my bad

[daring rock]

Oh, and arccot ranges from 0 to pi apparently

That makes sense, it avoids having a horizontal asymptote in the middle

But I've literally never used the inverse of cot, sec, or csc. There's no reason you'd ever need to

[smoky zenith]

k ty

[obtuse pebble]

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[fringe ingot]

What is this equation for partial fraction decomposition called

[haughty coyote]

Just looks like the regular PFD in C, with arguably debatable notations with A_l^(j)

[fringe ingot]

How is PFD in C any different than in R

[haughty coyote]

The irreducibles are different

x^2 + 1 is irreducible in R, not in C

[fringe ingot]

Is that all?

My proff said he wants it written with limits and everything tho

[haughty coyote]

There's different methods to find the coefficients in practice, but for the general theorem ? You don't have a given way of just giving a nice formula for every coefficient

[obtuse pebble]

@fringe ingot Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[covert girder]

Help

[rigid lintel]

solve 6x - 12 = 18

[covert girder]

Oh

That's easier than I thought thanks

[rigid lintel]

yeah math be like that sometimes haha

[obtuse pebble]

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[stone raptor]

brain isnt working help me

[knotty crow]

what is the part a

[stone raptor]

a) 1/(10+x)^2 = series ( n/100 (-x/10)^(n-1) )

and i used that series i set it up in second line of part b

i dont know how to take derivative of series

[knotty crow]

maybe use the fact that

$$\frac{\text{d}^2}{\text{d}x^2}\Big( \frac{\frac{1}{2}}{10+x}\Big)=\frac{1}{(10+x)^3}$$

[warm shale]

Modus

[knotty crow]

series of (1/2)/(10+x) is easier

[stone raptor]

ye thats what i had

its literally the same thing as what i wrote for part b second line

but i dont know how to take derivative of n/100 (-x/10)^(n+1)

oh sh i think i know the problem

[knotty crow]

aa, you've found the series of 1/2 / (10+x) in a

or?

$$

[stone raptor]

ye i found it in a

[knotty crow]

but it's wrong then

how 'n' there

[stone raptor]

nvm i got it

my brain is broken today

[knotty crow]

ah ok it's 1/(10+x^2)

np.

I was thinking about (1/2) / (10+x) 😄

good job

[obtuse pebble]

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[cinder quest]

Hi

[velvet peak]

Can someone explain why this is wrong?

[cinder quest]

hey

my channel :p

can someone help me with this

[velvet peak]

Well 4.5 percent of interest per annum gained $83.75 with $460 extra in the initial investment so the question would be what amount of money would get you 83$ on 4.5percent interest

[warm shale]

sarahsokool
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[velvet peak]

And deduct the 460 at the end

[cinder quest]

oooh ok ok

how do i write this down

[unkempt galleon]

isnt the 83.75 from both the 1.5% and 4.5%

or am i reading the question wrong

[cinder quest]

hi @unkempt galleon

[unkempt galleon]

hi

[fast cliff]

write system

[cinder quest]

how

[fast cliff]

i think you would use the money in each for varaibles

x can be lower

x + 460 = y

[velvet peak]

You don’t have to worry about the 1.5 percent

It’s a different savings account

[cinder quest]

ok ok

[fast cliff]

(1.015)x + (1.045)y = 83.75

[fast cliff]

i think thats the system

x + 460 = y

(0.015)x + (0.045)y = 83.75

wait

there lol

[cinder quest]

ok ok

what do i do now

[fast cliff]

do you know how to solve a system

[cinder quest]

No

[fast cliff]

rip

[cinder quest]

I mean

[fast cliff]

so you already have y in terms of x

[cinder quest]

okay

[fast cliff]

you can substiute y = 460 + x into the second equation

[cinder quest]

how does that look

[fast cliff]

(0.015)x + (0.045)(x+460) = 83.75

[unkempt galleon]

substitue the 0.045y into 0.045(460+x)

[fast cliff]

you know how to solve this?

[cinder quest]

i multiply

0.0045 with x+460 ?

[fast cliff]

you should multiply everything by 1000 first to make the numbers easier

[unkempt galleon]

other way around

yea

[fast cliff]

15x + 45(x+460) = 83750

no more decimals

[cinder quest]

ok ok

[unkempt galleon]

then distribute the 45 over the x and the 460

[fast cliff]

yea

[fast cliff]

that doesnt make sense, multiplication is communative

[unkempt galleon]

its how my brain works

i realized after i sent

[cinder quest]

@fast cliff

[fast cliff]

lol

yea

[cinder quest]

Wdym

[fast cliff]

ignore he messed up what he meant

[cinder quest]

I have 15x+45(x+460)=83750 RN

[fast cliff]

yea

multiple x by 45 and 460 by 45

[unkempt galleon]

mhm

[cinder quest]

!calc

[fast cliff]

15x + 45(x) + 45(460)

[cinder quest]

460x45

[fast cliff]

!calc 460x45

lol rip

[fast cliff]

20700

[cinder quest]

ok ok

[fast cliff]

can you

simplifiy a little

[cinder quest]

do i do

[fast cliff]

15x + 45x + 20700 = 83750

[cinder quest]

60x+20700=83750

[fast cliff]

yea

you should be able to solve for x

i believe in you lmao

[cinder quest]

I minus 20700 correct

and put it on 83

[fast cliff]

yea

[fast cliff]

what does this mean

[unkempt galleon]

from both sides

[cinder quest]

83750

[fast cliff]

yea subtract 20700 from that

[cinder quest]

I got

x=1050.833

[fast cliff]

yea good

[cinder quest]

So A

[fast cliff]

nope

[cinder quest]

WHAT

):

[fast cliff]

think about the variables we used

[cinder quest]

X

ok i see what to do i think

[fast cliff]

lol what did x mean/represent

[cinder quest]

variable X

[fast cliff]

yea no

[fast cliff]

these 2 equations are what we started with

[cinder quest]

do i multiply .015 with 1050

[fast cliff]

if x had a