help-33

[soft tangle]

I think that’s what I have written down

[spark otter]

And we correctly wrote this at some point

[soft tangle]

Is this it

[spark otter]

yes!

[soft tangle]

I don't get how we get that in cross multiplication at all ngl

It’s not working

Not sure what to do

[spark otter]

why did you change? this was correct

[soft tangle]

Ok

I’m there now

I finished it

I don’t see it

This part

[spark otter]

ok, now that we finished this, it will be easier to answer each question

[spark otter]

look at the right hand side here

[spark otter]

try to find which right hand side fits the most

[soft tangle]

Middle one

(Cos^2(t) + sin^2(t))/sin(t)

I got it

There is a third part😭

I GOT IT

FINISHED IT

[spark otter]

congrats!

[soft tangle]

I AM SO SORRY

For the hassle

It just wasn’t clicking

And i’m running off of like 3/4 hours of sleep

Can I get your help on another way easier problem

[spark otter]

alright

[spark otter]

@soft tangle bring it on

[soft tangle]

Thank you 🙏

So now i must solve the right sides

[spark otter]

can you take a better pic pls?

[soft tangle]

Yeah my bad

Better?

[spark otter]

great

[soft tangle]

Cool

Alrifht

Alright

So i’m trying to identify an identity i can sue on the first one

[spark otter]

let's start with the first one

[soft tangle]

Ok

[spark otter]

if you write $y = sec(x)$, the first expression writes itself as $\frac{4y^2 + 4y+1}{2y+1}$

[elfin berry]

rafilou2003

[soft tangle]

Ohhhh yeah

[spark otter]

can you factor the numerator?

[soft tangle]

Yes

I’m not sure about the 1 though

I think we exclude that

[spark otter]

you can do a long division

[soft tangle]

No good???

[soft tangle]

Is that where you draw lines and then do something with the x’s

[spark otter]

,rotate

[spark otter]

You know how to do polynomial division right?

[soft tangle]

I am rusty but I can easily remember

Can you help me tho

I don’t really remember it

[spark otter]

Ok

Basically, in it's easiest form it's something like ax^2 + bx + c divided by x+d

We always start by getting rid of the highest term in x

So we have to get rid of ax²

So we take (ax²+bx+c) - ax(x+d) to get rid of all the x²

Then we're left with terms in x, so we have to get rid of them as well, etc...

[soft tangle]

Ohhh ok

Can you show me the whole procedure of the division so I can understand it

[main idol]

https://www.youtube.com/watch?v=_FSXJmESFmQ

[marsh citrus]

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[marsh citrus]

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[wise jackal]

I get how they got this but

why does integrating from 0 to 2pi

and then dividing by 3

result in 2pi / 3?

cuz im thinking u can just find the total area of the 3 petals

and then divide by 3 to get one

[marsh citrus]

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✅

[wise jackal]

<@&286206848099549185>

[marsh citrus]

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[marsh citrus]

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[wise jackal]

.reopen

[marsh citrus]

✅

[wise jackal]

basically all i don’t get is like

why do u integrate from 0 to pi

and not 2pi

OH THE PERIOD FOR ODD PETALS IS PI

.close

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[opaque wolf]

yeah

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[lusty basalt]

Integral of x*e^(x)^2

[lusty basalt]

What form of integration is required and how do I approach

[marsh peak]

Have you tried substitution?

[lusty basalt]

Can you recommend a substitution?

The x is to the power of 3 sorry

[marsh peak]

Yes

[lusty basalt]

What can you recommend

What is the correct substitution

[marsh peak]

Hm? So it's $xe^{x^3}$?

[elfin berry]

A Lonely Bean

[lusty basalt]

Correct

[marsh peak]

That spices things up

One moment, let me think

[lusty basalt]

Me too lol

It would’ve been easy if it was to the power of 2

But 3makes it confusing

[knotty trellis]

do you know gamma function?

[lusty basalt]

No

And I don’t know

Power series integration

Only sub

Reverse chain rule

[knotty trellis]

this doesnt have closed form solution

[lusty basalt]

What doesn’t?

[knotty trellis]

the integral

[lusty basalt]

So you cant integrate it ?

A level integration

[nocturne vale]

You can try by parts apart from that it's not possible to integrate

[novel juniper]

Have you tried wolfram to check if it has a closed form?

[marsh citrus]

@lusty basalt Has your question been resolved?

[marsh citrus]

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[main bolt]

how i do this one?

[stoic saddle]

!status

[marsh citrus]

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

[main bolt]

i forgot how to do limits and im trying to remember how to do them

[stoic saddle]

multiply & divide by the conjugate

[main bolt]

thanks! .close

.close

[marsh citrus]

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[marsh citrus]

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[trim pulsar]

when finding the radius of convergence and interval do u always use the ratio test?

[trim pulsar]

where the lim n-> inf abs for an+1 / an?

and for the case of alternating series

does the absolute term get rid of (-1)^n?

like i can disregard it/consider as 1^n?

[devout mauve]

abs((-1)^n)=1 so yes you can just ignore that

well you use the ratio test when it works

sometimes it will be pretty impossible to calculate the limit you get

or you can do it and it gives 1, then you have to search for another test

[trim pulsar]

so its like the convergence tests where it may become a case of trial and error with the alternating sreies, comparison, etc?

[devout mauve]

ratio test is a convergence test

[trim pulsar]

do u have any tips on recognizing what tests i should use when testing for convergence in general?

ye i get that ratio is one of them

[devout mauve]

well if stuff will cancel from a_(n+1)/a_n then its probably a good start

classic example being eg factorials

[trim pulsar]

ah ok

[devout mauve]

with time you get a bit of a feeling for what tests might work

[trim pulsar]

got it

thank u

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[still temple]

Hi. I actually asked how I could solve this same problem yesterday. However, I'm stuck in how I am supposed to find when does P = -1. The idea is to solve this only using high-school math, so thankfully someone can help me, not using modular arithmetic and that kind of stuff. Someone provided me an explanation that you could match up "mod" and arguments, but technically there is a way to solve these having 0 idea about modular arithmetic, or in general, more advanced math than the one that is taught in high-school. Maybe you need to use implicitly the concept of "mod" without knowing what does that mean? I am not sure. Anyways, as you can see, the explanation and the step by step doesn't have to be that rigorous. Thank you

[still temple]

Probably the exercise could led to the use of complex numbers, as the way taught in high-school (Euler identity and related I guess, not sure tho). Sorry if my English sounds kind of unusual. I am using a translator.

I just realized the multiplication by a vertex and its conjugate is equal to |z|^2, which is always a real positive number. Maybe there's one way to start solving the problem using that fact, somehow matching up the product of the moduli to be equal to 1. By inspection I could say that [ 1/2 times (√2)^2 ] ^ integer have always modulus 1, which would be equivalent to the product of V1, V4, V3 or V6 with (V2 or V5)^2. However, I'm not sure if this is the right way to proceed. I feel somehow you have to use the fact that the hexagon is centrally symmetric

[marsh citrus]

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[still temple]

<@&286206848099549185>

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[orchid oracle]

Differential equations. Im stuck

[orchid oracle]

What step is next? I substituted y’ , but idk if thats right

[proud ice]

Distribute y^2

[orchid oracle]

[proud ice]

I meant in your step 2 here

[proud ice]

Oh you kinda did here. You forgot to square your whole x term

[orchid oracle]

🧐

Wait

[proud ice]

Where did the ^2 go in your y' here?

[orchid oracle]

Ah, forgot it indeed

But im still stuck 😂

After substitution, I shouldnt have any y's left, Only x's and C's

Atleast, thats what my book says

[proud ice]

!show

[marsh citrus]

Show your work, and if possible, explain where you are stuck.

[orchid oracle]

Can't really show much more than this ^
The book takes the derivative of the general solution (like I did), but solves it for y.y', and I solved it for y'

And I dont know when to solve for what

Or do I just guess and see

[proud ice]

Show me your corrections

[orchid oracle]

or do you want me to actually distribute

the y^2

[proud ice]

Distribute the y^2 now, yes

You'll be left with your original equation, which was given

So that suffices your proof

[orchid oracle]

Yep

[orchid oracle]

Ah so its correct too, okay

Much simpler than my book

What about finding a particular solution? Normally you just pick a value for C at random right?

Nvm, all checks out, thx

.close

[marsh citrus]

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[slow dock]

how do i solve this i have no clue

[fathom ridge]

Did you try multiplying and dividing by the conjugate

[slow dock]

can you elaborate

[fathom ridge]

"rationalize the numerator"

Multiply and divide with $\sqrt{x^2 - 2x} + \sqrt{x^2 + 2x}$

[elfin berry]

NEONPerseus

[slow dock]

ah

ill give it a try

i get 4x now what

[fathom ridge]

4x or -4x

In the numerator

[slow dock]

i mean -4x

[fathom ridge]

I'll help you with the positive infinity part you'll be able to do the negative one by yourself

[slow dock]

sure

[fathom ridge]

$\lim_{x \to \infty} \frac{-4x}{\sqrt{x^2 - 2x} + \sqrt{x^2 + 2x}}$

[elfin berry]

NEONPerseus

[fathom ridge]

@slow dock this is what you have right?

I might have to leave in a bit

So I'm just ask you to factor an x^2 out of each square root and cancel it from the numerator

That should land you with an answer

[slow dock]

ah so the answer is four

?

no i caluculted in correctly '

[obsidian shard]

did you do as NEON instructed?

[obsidian shard]

Looks almost done, just divide numerator and denominator by x

[slow dock]

i did sqrt(x^2 -2x) + sqrt(x^2 +2x) = sqrt(x (x -2) + sqrt(x (x +2)

then i factor out sqrt(x)

[obsidian shard]

oh, if want to take common, then take x^2 as common not x

[slow dock]

can you explain

[obsidian shard]

notice that $x = \sqrt{x^2}$

[elfin berry]

numbpy

[obsidian shard]

If you have difficulty understanding taking common, then I suggest just dividing by x above and below

lemme give an example

$\sqrt{x^2 + x} = \sqrt{x^2 (1 + \frac1x)} = x \sqrt{1 + \frac1x}$

[elfin berry]

numbpy

[obsidian shard]

does this help @slow dock

[slow dock]

Yes

Thx for the help

[marsh citrus]

@slow dock Has your question been resolved?

[marsh citrus]

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[shy urchin]

for this problem, i want to use the LCT

what would i compare the reciprocal with

[lucid bridge]

divide numerator and denominator with k^1/3

[shy urchin]

My main issue is with the denim

Denominator**

After simplification, it would be k^(3/2)+k^(1/2)+sqrt(3)

Correct?

[lucid bridge]

they should all be on the same square root instead of seperately

[shy urchin]

oh so u want me to just divide everything by k^1/3

off rip

[lucid bridge]

yes

not pick as bn but outright divide by that

[shy urchin]

so the top goes to 1

the denominator is now

sqrt(k^3+4k+3)/k^(1/3)

is that correct?

[lucid bridge]

yep now get it inside of the sqrt

[shy urchin]

wasnt that this tho

[shy urchin]

.

[lucid bridge]

not even close to that

[shy urchin]

oh

[lucid bridge]

how did you magically pull other stuff out of sqrt

anyway you shouldnt try to divide all of the stuff by k^1/3 , only divide k^3 by it since it has the highest power

[shy urchin]

our bottom is (k^3+4k+3)^1/2

right

[lucid bridge]

distribute 1/k^3 in it

it should become 1/k^6 inside of the square root

my bad

[shy urchin]

im sorry im a bit lost rn

is up to this step

[lucid bridge]

1/k^2/3

[shy urchin]

correct?

[lucid bridge]

yes

multiply the exponent by 2 while getting 1/k^1/3 inside of the sqrt so you get 1/k^2/3

[shy urchin]

dont u mean 1/2

?

cuz i thought sqrt of x = x^(1/2)

so wouldnt we multiply everything in the root by 1/2?

[lucid bridge]

|x|=sqrt(x²)

[lucid bridge]

this is not true for every x

[shy urchin]

right but for now the numerator of the denominator can be simplified as (k^3+4k+3)^1/2

no?

[lucid bridge]

where is k^(1/3) ?

oh i see yeah

well i thought that was obvious

[shy urchin]

right, we can get to that in a sec

oh lol

but thats how i got the k^3/2 solution

[lucid bridge]

there is a shorter way to tell how to pick (bn)

but this one feels more legitimate

[shy urchin]

ah i see

So this is where I’m at rn

[lucid bridge]

so highest variable in terms of k on the numerator is k^1/3 right, and on the numerator its k^3 but its in square root, if you eliminate every other terms it becomes k^3/2

you can just k^(1/3)/k^(3/2) then pick (bn) as that

[shy urchin]

ah i see

ok

[lucid bridge]

that is very cursed you cant eliminate square root like that

[shy urchin]

oh you cant?

[lucid bridge]

you would have pull k^1/3 in, not pull others out

yeah

[shy urchin]

oops so then we can just use k^1/3 / k^3/2

[shy urchin]

ok i follow what you did here

[lucid bridge]

sqrt(a).sqrt(b)= sqrt(a.b) so you are good now

[shy urchin]

mhm so you can multiply it right theough and get sqrt (k^3+4k+1)/sqrt(k^2/6)

[lucid bridge]

2/6 instead of 2/5 and its right

[shy urchin]

oops

but what would you copmare this with

[lucid bridge]

highest term of k, after division inside of sqrt

[lucid bridge]

which will be same as this

[shy urchin]

right

k^3/2 holds more power than k^1/3

so youd compare it with that correct?

huh?

[lucid bridge]

nvm

its right

[shy urchin]

whats wrong with it

lol

[lucid bridge]

my brain was lagging because im looking after my dog at the same time

yeah go on from there

[shy urchin]

lol

[lucid bridge]

so after you evaluate that, take bn as that value

[shy urchin]

so wed multiply everything by the recipricol with the comparison

right

[lucid bridge]

yeah

well if im being honest you dont need to evaluate, you can just take bn as that

and apply LCT

[shy urchin]

bn diverges

so an diverges aswell

by lct

[lucid bridge]

nice

im assuming you found limit to be 1

[shy urchin]

mhm

got it!

thanks so much for your help and patience lol

i appreciate it

.close

[marsh citrus]

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[thorny iris]

how would you solve this

?

[marsh citrus]

@thorny iris Has your question been resolved?

[late geode]

have you drawn a diagram yet?

[marsh citrus]

@thorny iris Has your question been resolved?

[thorny iris]

yeah

[thorny iris]

hello?

[sleek lake]

it should be clear from your diagram

you're looking for the green length, you just find the angle from cosine

there's a right triangle

oh that's arccos

[marsh citrus]

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[hard stratus]

how do i find the area for this

[stark trail]

the surface area?

or the volume?

[severe hound]

surround area or surround and surface?

[hard stratus]

no the area

[stark trail]

That is a 3D object, it doesn't have an area

It has a volume, and it has a surface area

[severe hound]

I think he/she means total area

Surface

[hard stratus]

the surface area

[severe hound]

Should be 2πrh+2πr^2

[hard stratus]

idk what to do

how do i do the circle thing

[stark trail]

Please send a picture of your original question

[hard stratus]

it’s here

[stark trail]

That is just a picture of a shape, it doesn't state a question

[hard stratus]

it said determine the area of the following solids

[stark trail]

It did not say that, because solids don't have area

[stark trail]

Okay well I suppose it means surface area then

Do you know what surface area means?

[hard stratus]

no

[stark trail]

Surface area is just the area that makes up the outsides of a shape

for example if you take a simple cube

the surface area, is the area of all the square faces of that cube

[hard stratus]

ok

[stark trail]

So what are the different faces on the cylinder that you sent

[hard stratus]

1 rectangle?

[stark trail]

Where do you see a rectangle on the cylinder ? :o

[hard stratus]

what idk

[stark trail]

If the surface area, like we said earlier, is the area of the faces that make up the outside of a shape, to find the surface area for the cylinder you have to identify what its faces are

For example, what shape is on top and bottom of the cylinder

those would be two of its faces

[hard stratus]

ok

just 2 circles

[stark trail]

The top and bottom yes, just two circles

and then the only remaining face is the side of the cylinder

do you have any idea how you could find the area of that?

[hard stratus]

1x2.8 and 4x0.2

[stark trail]

Hint: Imagine taking the cylinder and unrolling it so that the side becomes a rectangle, what would the base and height of this rectangle be? The base would be ||the length of the circumference of the top circle that we unrolled|| and the height would be ||the height of the cylinder|| So the area of this side of the cylinder, or the unraveled rectangle, would be ||circumfrence * height||

So now you have identified the 3 sides of the cylinder: 1. top circle 2. bottom circle 3. the side. To find the surface area, find the area of these 3 sides and add them up.

[hard stratus]

0.2x2.8?

[stark trail]

Why are you guessing random numbers?

You should read my previous messages

[hard stratus]

what is circumference * height

[stark trail]

The length of the circumference of the cylinder, multiplied by the height of the cylinder

Do you know what the circumference is?

[hard stratus]

the length is 2.8 and the height is 0.2?

[stark trail]

No

The height of that cylinder is 2.8m, and the radius of that cylinder is 0.2m

Again I'll ask, do you what the circumference is?

[hard stratus]

no

[stark trail]

The circumference is the length around the outside of the circle

Length of the circumference = 2 * pi * length of the radius

[hard stratus]

ok

[stark trail]

So can you tell me what the 3 shapes are that we need to find the area of, in order to find the surface area of this cylinder?

[hard stratus]

what

[stark trail]

The surface area is the area of all the sides of the shape. What are the 3 sides of this cylinder?

(we figured this out already earlier)

[hard stratus]

2 circles and 1 rectangle

[stark trail]

Sort of, the rectangle isn't really a rectangle though right?

You recognize that it is just the curved side of the cylinder, yes or no?

[hard stratus]

yes

[stark trail]

Okay perfect

So we need to find the area of the 2 circles, and the area of the "rectangle"

Do you know the formula for the area of a circle?

[hard stratus]

pie symbol and x2

[elfin berry]

AustinU

[stark trail]

Does that seem familiar?

[hard stratus]

the dot is where the number goes in?

[stark trail]

The dot is multiplication

[hard stratus]

no idk

what do i do next

[stark trail]

What grade are you in?

[hard stratus]

8

[stark trail]

Okay, well I can't continue to help you find the surface area of this cylinder if you are unable to find the area of a circle. It seems like you are missing a lot of background knowledge needed to solve this problem. I can direct you to a video to learn

https://youtu.be/JC2kRM3jTOo

https://youtu.be/dNDbRbHNXFQ

I'd watch both of those in order, and then you should be able to figure this out

[hard stratus]

ughhhh ok

btw i don’t understand the first vid

[stark trail]

It is 6 minutes long, and I sent it 1 minute ago, how have you already watched it?

[hard stratus]

nvm what do i do now

[stark trail]

stop interrupting other people's channels, I directed you to read #❓how-to-get-help and #rules already

[stark trail]

Watch the videos, otherwise I can't continue to help

[hard stratus]

i did

[marsh citrus]

@hard stratus Has your question been resolved?

[marsh citrus]

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[modest root]

Can anyone help me how subspace scalers are converted into matrixes

[marsh citrus]

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[arctic gorge]

ive tried everything but the most i can get is QR = asinB/2
i knw thatnearly all of these triangles are isoceles so BQ and QC have to equal a/2 im just not sure where to go from there or how cos gets squared

[marsh citrus]

@arctic gorge Has your question been resolved?

[marsh citrus]

@arctic gorge Has your question been resolved?

[arctic gorge]

<@&286206848099549185>

sorry for the ping :'')

[steel mountain]

sorry no one answered you

[arctic gorge]

T-T

[steel mountain]

uh

[arctic gorge]

i tried to big brain it out but now i have to go to school lol

[steel mountain]

oof

uhh

well you're working with an icoselies trangle

right?

do you have a diagram??

Ok

I mae it

you need to use the cos rule

and the sin rule

[marsh citrus]

@arctic gorge Has your question been resolved?

[marsh citrus]

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[olive agate]

Hello! this is calculus so i’m not fully sure if i’m allowed to post this here but anyone please just let me know but anyways i need to find this using L’Hôpital’s Rule and i’m stumped on it! Help is appreciated 😊

[marsh citrus]

@olive agate Has your question been resolved?

[marsh citrus]

@olive agate Has your question been resolved?

[olive agate]

.close

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[rotund walrus]

just a quick question

[rotund walrus]

when you have x^2-2x+8/9=0

can u multiply the whole equation by 9 to get rid of 8/9

[bright jay]

Is that $\frac{x^2-2x+8}{9} = 0$ or $x^2-2x+ \frac{8}{9} = 0$?

[rotund walrus]

?

[elfin berry]

dldh06

[rotund walrus]

2nd

quadtratic

[bright jay]

Yes you can multiply by 9 if you wanted to

[rotund walrus]

alr

bet

thx

just wanted to make sure

[twilit arrow]

note there’d also be a 9 in front of the two other terms

[rotund walrus]

yea

lol

so

9x^2-18x+8

right?

[twilit arrow]

yes

[rotund walrus]

ye

alr thanks

[bright jay]

Don't forget the = 0

[rotund walrus]

yea

to lazy

.close

[marsh citrus]

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[twilit quiver]

Can someone help pls

[marsh citrus]

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

[marsh citrus]

Show your work, and if possible, explain where you are stuck.

@twilit quiver Has your question been resolved?

[shadow ibex]

@twilit quiver https://stattrek.com/estimation/confidence-interval-mean

If you wanna show your work I might be able to see where you messed up

Probably something with alpha

[twilit quiver]

I followed this example

@shadow ibex

[marsh citrus]

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[twilit quiver]

.reopen

[marsh citrus]

✅

[marsh citrus]

@twilit quiver Has your question been resolved?

[shadow ibex]

Sorry I'm running errands, haven't looked yet

n = 25, N = 500
xbar = 25.7,
s = 7.8

I guess we have to do a correction factor since n is only 5% of N

fac = 1.5220227022657792307

[twilit quiver]

I don’t get how u got this?

[shadow ibex]

It's the formula on your page

I was just driving so I didn't multiply by the t value yet

p* = .995, df = n-1 =24

Using a t-table or calculator, that value is...

[twilit quiver]

can I write the full working out pls if that’s okay

[shadow ibex]

Oh, using df =24 and alpha = 1 - .99 = .01

[twilit quiver]

yep got that

Can u write the full working out if that’s okay?

[shadow ibex]

I'll do that in a second if thus works

T-Value (right-tailed): 2.351635
T-Value (two-tailed): +/- 2.609228
https://goodcalculators.com/student-t-value-calculator/
© 2015-2023 goodcalculators.com

I think we want the two-tailed?

Multiply that by the fac

[twilit quiver]

yes I think

[shadow ibex]

So the interval should be

25.7+-3.9713042513875346106

[twilit quiver]

lmao but look at the options

[shadow ibex]

...which is not even close

[twilit quiver]

exactly

that’s what I got

But I was like…

so I don’t understand

[shadow ibex]

If thr mean weight is 25.7g, how can we have a confidence interval for the "total weight of the population" that is between like 10 and 15? 🤨🤨🤨

OH!

[twilit quiver]

uh oh

[shadow ibex]

This is orders of magnitude larger

[twilit quiver]

wdym

[shadow ibex]

Multiply by 500/25?

[twilit quiver]

umm

[shadow ibex]

Doesn't seem like enough though

Or by 500

Yeah

Let's see

I don't have my scientific, what's the min if we do that?

[twilit quiver]

I don’t get what u mean haha

[shadow ibex]

Well...if we do that, the min is kinda close to the fourth option

Not very though

Not by scientific standards

[twilit quiver]

haha

I have no idea

[shadow ibex]

Yeah this ain't working

Let's try the other t value

[twilit quiver]

I was stuck on this for 2hrs lmao

Like idk what I was doing wrong

[shadow ibex]

Me neither

Because you don't show your formulas

You have the notes from the professor, amd I assume the values are supposed to match those

[twilit quiver]

Yes

[shadow ibex]

But you don't show the t-value computations

500(25.7+-2.351635)

crunch that really quick

[twilit quiver]

Oh

[shadow ibex]

Too big

On the min

I really don't know

<@&286206848099549185>

[flat raft]

population total weight

not individualweight

[flat raft]

and no

there seems to be a mistake in the question

I don't think it's the sample standard deviation, but rather the standard deviation of the sample mean

[shadow ibex]

Yeah, my vocabulary is trash

For example, "fac" I was using is actually called "Standard Error" or SE

Everything after +- is the Error Margin (Margin of Error), ME

Can you show the critical value computation?

And was your df 499?

[flat raft]

why is this related to a t distribution? am I missing something

[shadow ibex]

Oh!

I was using the population size and not the sample size

df = n-1

Not N-1

Thank you

[flat raft]

what does this have to do with a t-distribution though

[shadow ibex]

It's part of the confidence interval formula?

[flat raft]

not really

that's only if your estimator / test statistic is t-distributed

in this case, the sample mean is approximately normally distributed

and your estimate of the total weight is just the sample mean times 500 is it not?

[twilit quiver]

I’m confused lol

[flat raft]

the sample mean has a standard deviation of approximately 1.6, so if you multiply the whole thing by 500, I guess that gives you a whole thing standard deviation of 780

[twilit quiver]

can someone work it out

[flat raft]

a 99% confidence interval for a normal distribution is everything within 2.575829 standard deviations of the mean

and the mean of the estimate of the total weight is 500 * 25.7 = 12850

[shadow ibex]

Where did you get 1.6?

They gave a sample standard deviation of 7.8

[flat raft]

7.8 divided by sqrt(25)

[shadow ibex]

We got the times 500 thing

[flat raft]

or rather, an unbiased estimate of the variance of the weight of each individual item is 7.8^2, and the variance of the sample mean is 7.8^2/25

[shadow ibex]

Ah.

[flat raft]

hmm maybe you do have to take into account that the standard deviation is not perfectly estimated

but you can't really I think

or rather, it's annoying to do

[shadow ibex]

Well there is that long bit that ^ @shell inlet has here

In the fomrula

The SE calculated differently b.c. the sample is <= 5% of the population size

...I think

[flat raft]

well, suppose you consider the test statistic t = (sample mean - 25.7)/7.8 grams; maybe that's t-distributed

yeah but why would it be a t-distribution

the sample mean is assumed to be normally distributed

[shadow ibex]

I still get the wrong t value with df = 24 and p* = .995

?

[flat raft]

oh there's a finite population correction

[shadow ibex]

Oh...

[flat raft]

okay that makes sense

[shadow ibex]

I used the wrong value

[flat raft]

I get an answer similar to the last choice now

[shadow ibex]

Significance level is 1-0.99 = 0.01

Use that with df= 24

Gets t=2.49216

[flat raft]

the uncorrected sample standard deviation is 7.8 grams, but applying the finite population correction, one gets an estimate of approx 7.61 grams for the population standard deviation

I still don't know why y'all are doing anything with a t distribution

[shadow ibex]

Makes the rest work out as it does in OP's notes

[flat raft]

why is anything t-distributed

[shadow ibex]

🤷

[flat raft]

hmm maybe it's because of the estimated standard deviation

yeah

that gives the second choice

okay it all makes sense now

so basically, we claim that the value (25.7 - true mean) / (7.8*FPC) is t-distributed, and then a 99% confidence interval for this is something; this lets you compute a confidence interval for the sample mean

and then if you multiply it by 500, you get the second answer

[marsh citrus]

@twilit quiver Has your question been resolved?

[twilit quiver]

wait can someone write out the working on paper step by step pls, I’m new to this

[twilit quiver]

wait can someone write out the working on paper step by step pls, I’m new to this

[flat raft]

I think the easiest way to explain this is probably by referencing your prof's slides

find a 99% CI for the sample mean

then just multiply the thing by 500

[twilit quiver]

Can someone help with b)

[flat raft]

so you know the population mean is 0.5 and the population variance can be calculated as p(1-p)

[twilit quiver]

yes

I used the zscore table to find the values and then I calculated it but I didn’t get the answer lol

[flat raft]

can you elaborate more on what you did?

[twilit quiver]

so I found the zscores that correspond to 0.05 since it’s symetrical then it is -1.65 and 1.65

So that one is the lower bound value

is the wrong?

@flat raft

[flat raft]

it's closer to 1.64; it's about 1.644854

other than that, it looks right to me

[twilit quiver]

Yea but the answer is wrong apparently lol

[flat raft]

well it says four decimal places, so maybe you rounded a little early

actually at four decimal places, you might even have to do a continuity correction ew

[twilit quiver]

Yh😂😭

umm

I got this but idk

[flat raft]

I get 0.4418 with more exact arithmetic

but nah there's definitely a continuity correction here

[twilit quiver]

wait what’s that again?

[flat raft]

basically the sample percentage comes in multiples of 1/200 = 0.005

[twilit quiver]

yep

[flat raft]

so when you approximate the sample percentage with a normal

since it's a scaled binomial

you have to do some silly shit where you add 1/400 or something

idk how I would even answer this question without at least saying something about whether the limits are inclusive or exclusive

I'll just assume the limits are inclusive

[twilit quiver]

ahh i see haha,, do u know how I would do the next missing blank

[flat raft]

the next missing blank is just 0.5 - lower limit

unless

[twilit quiver]

why tho?

I may not understand what it’s asking haha

[flat raft]

waitI can't actually see the entire question for that one

[twilit quiver]

Oh sorry hold up

[flat raft]

I think it's just 0.5 - lower limit

but it might also be asking for percentage error, in which case it's (0.5 - lower limit)/0.5

[twilit quiver]

Can u explain why?

[flat raft]

well if you say that it's contained within [lower limit, upper limit]

isn't that just the same thing as saying it's contained within (0.5-lower limit) of 0.5?

[marsh citrus]

@twilit quiver Has your question been resolved?

[marsh citrus]

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[humble walrus]

If a is an integer and 7|4a, then 7|a.
Proof. Suppose a is an integer and 7|4a. That means 4a=7b for some integer b. The only way this is true is if they are both multiples of 28. Thus 4a=28c for some integer c. So a=7c, and 7|a, by definition of divisibility. |||

Is this proof valid?

[open shale]

"The only way this is true is if they are both multiples of 28"
Why?

[glacial hedge]

They must be multiples of their lcm

[wild sinew]

Because they are coprime

[open shale]

That is not an axiom

[glacial hedge]

Arguably that's by definition of the lcm
Otherwise it's an exercice in and of itself

[wild sinew]

You can just use that if p|a*b, then p|a or p|b, if p is a prime

And 7 doesn't divide 4 so it must divide a

[open shale]

yeah

[humble walrus]

so can I use this argument?

[glacial hedge]

Depends on what you already know

[marsh citrus]

@humble walrus Has your question been resolved?

[humble walrus]

i do know that

about the lcm

[glacial hedge]

If you know it you can use it no ?

[marsh citrus]

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[nova quail]

quick question: what do you write when an integral isn't continous betwenn two bornes?

[glacial hedge]

Assuming it's defined (i.e. not going to infinity, otherwise it gets more tricky to say the least) and there's finitely many discontinuities, nothing special as integrals are naturally also defined for such functions

[nova quail]

.close

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[near blade]

hii

[near blade]

i have dirichlet function

and i have to say

if the functions below are continous and in what ranges

[marsh citrus]

@near blade Has your question been resolved?

[marsh citrus]

@near blade Has your question been resolved?

[marsh citrus]

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[eager sapphire]

hi

[eager sapphire]

is there a way to prove these 2 smaller triangles created from that perpendicular are similar?

[leaden monolith]

they share angles

they both have 90° angles

[eager sapphire]

wait i just realized

[leaden monolith]

that perpendicular line splits the original right angle into 2 angles

[eager sapphire]

since that first angle is 90 the other 2 have to add up to 90

[leaden monolith]

and the 3rd angle of each triangle must be the other one

yeah

[eager sapphire]

thanks that makes sense, is this true when you make a perpendicular to any right angle in a triangle?

[leaden monolith]

well...

not always

[eager sapphire]

is it like a theorem

oh i see

like when will it not

[leaden monolith]

this doens't

[eager sapphire]

is that perpendicular going to the right angle

[marsh citrus]

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[willow lagoon]

A husband and wife 6km apart start walking towards each other. The husband walks at a speed of 4km per hour while the wife walks 3km per hour. A fly starts from the husbands forehead at the instant he starts walking and flies towards the wife towards the speed of 5km per hour. when it reaches the wife, it suddenly turns back and flies towards the husband. It continues to fly back and forth until the husband and wife meet. After that, what is the total distance the fly has flown?

[marsh citrus]

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[willow lagoon]

.close

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[vital swan]

what are some examples of trivial isomorphism functions

[vital swan]

like 2 groups are clearly the same, but I can't think of a function

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[vital swan]

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[full venture]

just a question

[full venture]

is cosine neeeded?

to solve a right triangle?

[lucid zenith]

?

[full venture]

i mean i feel like i can solve any sides just with sine and tan

and just add pythagoreas theorem on some

[ruby yarrow]

You need the length of two sides, or the length of one side and the measure of an acute angle

[full venture]

what

[ruby yarrow]

you can use trig ratios to find missing sides

[full venture]

im just asking if cosine is needed

yea

i know that

.close

[marsh citrus]

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[thorny jungle]

I'm not sure how to proceed here

[thorny jungle]

B(n) is bells number

[main idol]

Show equation 5.3

[sterile rock]

ive seen this b4

but forgot how to do it

[main idol]

and try using the formula for binomial coefficient in terms of factorials

[thorny jungle]

me on exams

[sterile rock]

tbh i prolly just didnt understand the solution

tiny brain

[shadow ibex]

@thorny jungle

[main idol]

if you're still stuck, n is a constant with respect to the summation and you can do a substitution on i

[thorny jungle]

ok, ill try that

oh sorry, eqn 5.3 is just this

[elfin berry]

KN

[thorny jungle]

Im not sure how to produce (n+1)!

<@&286206848099549185>

Ok nvm that was so dumb

I think I got it

.close

[marsh citrus]

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[agile yarrow]

where does two concentric circles have radical axis?

[agile yarrow]

need answer fast

.close

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[jade parcel]

Hello, how can I prove that the limit of this function as n goes to infinity is 1/sqrt(e)

[jade parcel]

$cos\left(\frac{1}{n}\right)^{n^2}$

[elfin berry]

Mathu_lmn

[jade parcel]

.close

[marsh citrus]

Channel closed

Closed by @jade parcel

Use .reopen if this was a mistake.

Available help channel!

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

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

[fleet leaf]

x^2+3x-6=0

[marsh citrus]

Please don't occupy multiple help channels.

[fleet leaf]

can someone help me work this out

not sure how to go about this

[lucid zenith]

I told you how to do it

what's the issue

[fleet leaf]

dont know what to do next

[lucid zenith]

what stage are you at

[fleet leaf]

uhm

i got (x+3/2)^2-33/4=0

[lucid zenith]

I told you how to write the second term as a square

[fleet leaf]

yh

(x+3/2)^2(sqrt33/2)

[lucid zenith]

$(x+\frac{3}{2})^2-(\frac{\sqrt{33}}{2})^2=0$

[elfin berry]

kheerii

[lucid zenith]

now, as I have said before, use the difference of squares

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

[elfin berry]

kheerii

[fleet leaf]

yes I told u

wait

so

ohhh

so

(x+3/2)^2-(sqrt33/2)^2 = (x+3/2+sqrt33/2)(x+3/2-sqrt33/2)

[lucid zenith]

correct

[fleet leaf]

how do u do x+3/2+sqrt33/2

[lucid zenith]

you don't need to

[fleet leaf]

i thought i need to solve it

to get a number

[lucid zenith]

you got $(x+\frac{3+\sqrt{33}}{2})(x+\frac{3-\sqrt{33}}{2})=0$

[elfin berry]

kheerii

[lucid zenith]

when a product is equal to 0, you need to consider each term to be equal to 0 one by one

[fleet leaf]

wait

how did u make

[lucid zenith]

I just combined the fractions

[fleet leaf]

3/2 - sqrt 33/2

isnt it a minus

oh wait nvm

so what now

do i leave it at that

[lucid zenith]

no

read everything I wrote

[main idol]

x+c = 0 means x=-c

[fleet leaf]

so i get

the opposite

of each bracket

like for the first one

(-3+sqrt33/2)

and 2nd (3+sqrt33/2)

[lucid zenith]

nope

I think you're missing some brackets

[fleet leaf]

what about that

[lucid zenith]

you just put a bracket around the whole expression

that doesn't help

$-3+\frac{\sqrt{33}}{2}$ is what you have written

[fleet leaf]

brackets where

[elfin berry]

kheerii

[fleet leaf]

oh so

-3+(sqrt33/2)

[lucid zenith]

you just reiterated what you wrote the previous time

(-3+sqrt33)/2

is what I expected

[fleet leaf]

oh ok

[lucid zenith]

but your second answer is wrong

[fleet leaf]

which

[lucid zenith]

.

[fleet leaf]

so

(3+sqrt33)/2

[lucid zenith]

nope

[fleet leaf]

??

thats what u did for the first one

[lucid zenith]

$x+\frac{3+\sqrt{33}}{2}=0\implies x=\frac{-3-\sqrt{33}}{2}$

[elfin berry]

kheerii

[lucid zenith]

$x+\frac{3-\sqrt{33}}{2}=0\implies x=\frac{-3+\sqrt{33}}{2}$

[elfin berry]

kheerii

[fleet leaf]

oh ok

[lucid zenith]

try being easier to work with next time

[fleet leaf]

what have I done

[lucid zenith]

be obnoxious

[fleet leaf]

i just didnt understnad

[main idol]

Entitled, don't answer questions we ask

[fleet leaf]

maybe bcs I dont know the answer

[fleet leaf]

do I leave it like that @lucid zenith

[main idol]

Those aren't excuses for being entitled or ignoring us

Hundreds of people ask questions here without being difficult

[fleet leaf]

ok

ill be easier next time

[lucid zenith]

yes

you can't simplify the answers further

[fleet leaf]

it says

the answers are

-4.4

1.4

on the back of the book

is that wrong

[main idol]

They rounded

Did the instructions say to round?

[fleet leaf]

nope

[main idol]

,calc (3+sqrt(33))/2

[elfin berry]

Result:

4.372281323269

[fleet leaf]

wait so should I round

if it doesnt ask me to*

[main idol]

Textbooks do that

Do whatever the book says or teacher does in class

[fleet leaf]

ok

mb tho for being annoying

i just didnt understand

.close

[marsh citrus]

Channel closed

Closed by @fleet leaf

Use .reopen if this was a mistake.

[fleet leaf]

.reopen

[marsh citrus]

✅

[fleet leaf]

@main idol wait what were the steps again

after I got i got (x+3/2)^2-33/4=0

[main idol]

Just read the history

Kheerii was very generous and patient with you

[fleet leaf]

bro ok chill

i get it

.close

[marsh citrus]

Channel closed

Closed by @fleet leaf

Use .reopen if this was a mistake.

[main idol]

!help

[marsh citrus]

Please read #❓how-to-get-help

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

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

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

[wooden grotto]

Can someone help me with this?

[mystic minnow]

can't you just integrate it? u-sub with u = sin(x)

[wooden grotto]

Its asking for integration?

[mystic minnow]

yeah, you're given y'

general solution is y

[wooden grotto]

To solve the initial value problem, we need to integrate the given differential equation with respect to t:

∫dy = ∫9t dt

y = (9/2)t^2 + C

where C is the constant of integration.

To find the value of C, we use the initial condition given for each problem:

A. y(0) = 6
6 = (9/2)(0)^2 + C
C = 6

Therefore, the solution to the initial value problem is:

y = (9/2)t^2 + 6

When t = 0, y = 6.

B. y(0) = 1
1 = (9/2)(0)^2 + C
C = 1

Therefore, the solution to the initial value problem is:

y = (9/2)t^2 + 1

When t = 0, y = 1.

[mystic minnow]

i agree with that solution

and again, you were given y', the general solution is y. So you integrate y'

[wooden grotto]

y= 6 and y=1?

[mystic minnow]

y is a function of t

y(0) = 6 is evaluating the function y at the value t=0

so once you integrate, you get y = (some function of t) + c
knowing y(0) = 6 let's you solve what that constant of integration c is

[wooden grotto]

I get it. 🙂

[marsh citrus]

@wooden grotto Has your question been resolved?

[wooden grotto]

Okay wait for this problem, I got part A: To solve the initial-value problem, we integrate both sides of the differential equation with respect to t:

∫dy = ∫e^(9t) dt

y = (1/9)e^(9t) + C

where C is the constant of integration.

To find the value of C, we use the initial condition y(0) = 5:

y(0) = (1/9)e^(9*0) + C = 5

Simplifying, we get:

C = 5 - (1/9) = 44/9

Therefore, the solution to the initial-value problem is:

y = (1/9)e^(9t) + 44/9...but im getting b wrong..

[wooden grotto]

To find the time at which y increases to 100, we can set y = 100 in the solution we found:

100 = (1/9)e^(9t) + 44/9

Subtracting 44/9 from both sides and multiplying by 9, we get:

56 = e^(9t)

Taking the natural logarithm of both sides, we get:

ln(56) = 9t

Solving for t, we get:

t = (1/9)ln(56) ≈ 0.7985

Therefore, y increases to 100 at approximately t = 0.7985.

To find the time at which y drops to 1, we can set y = 1 in the solution we found:

1 = (1/9)e^(9t) + 44/9

Subtracting 44/9 from both sides and multiplying by 9, we get:

-35 = e^(9t)

Taking the natural logarithm of both sides, we get:

ln(-35) = 9t, but that's not right answer....

[wooden grotto]

<@&286206848099549185>

[wooden grotto]

.close

[marsh citrus]

Channel closed

Closed by @wooden grotto

Use .reopen if this was a mistake.

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Send your question here to claim the channel.

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

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