help-28

[hidden vault]

thats what i said in simple formula.. also seperate channel pls

and ill help u

[sand badge]

Rahh

[sand badge]

I have exam 2moro

[hidden vault]

okay js go to #help-6 and ask ur question there

that will auto give u a channel

[sand badge]

Thx

[full forum]

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[urban nebula]

This is more of a chemistry doubt.
How do we find the buffer capacity of an acidic buffer with salt conc. A and acid conc. B

[gritty rose]

https://discord.gg/eexdsFw

[urban nebula]

So should I close this help channel?

[gritty rose]

Yes

[urban nebula]

.close

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[edgy ferry]

so i tried to diffrenciate g(x) and applied chain rule to get 3 terms

[edgy ferry]

and when x is greater than 1 , then f (x) can we written as x-1 directly

i tried to put in the values

nvm i got it mb 💀

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[zinc horizon]

Hey I’m loooking for vab
So I did I1 = 7/4 which is 1.75 and it’s supposed to be going into the top node so 1.75-8+x = 0 and it came out as 6.25 so i did -6.25 x 6 which came out as 36 which is wrong why

[zinc horizon]

Or can this only be solved using nodal?

[outer lotus]

You calculated I1 = 1.75
This would only be true if the other side of the 4 Ohm resistor was connected to ground

Ohms law requires voltage drop, not just any voltage

[vernal mural]

@outer lotus what's vab

[outer lotus]

The difference in voltages at nodes a and b relative to ground

[vernal mural]

k i am not close to understanding that. maybe when i get older

[outer lotus]

Learn about the potential difference

It's not that difficult actually

[vernal mural]

yea ik know pd

and calculating the resistance

just the basics tho

[outer lotus]

So we use node b as our reference point

It'll act as ground

V_b = 0

[vernal mural]

what's that arrow
is that flow of current

[outer lotus]

Yes, it's the direction of the current at that source

[outer lotus]

So then we know that the sum of all currents leaving a node must equal zero

We then consider all paths connected to node a

[vernal mural]

there's only one path

is the question incomplete?

[robust slate]

discord.gg/physics

for future reference ^

this is not the right server

[outer lotus]

There are 3 paths connected to node a

Current goes from a to b

[vernal mural]

but all 3 of them are entering/leaving a?

how

[outer lotus]

No, because b is ground

[vernal mural]

that server is inactive and not tailored

[vernal mural]

oh got it

[outer lotus]

Physics is just applied math, so I don't see a problem

[vernal mural]

<t:1768851168:R>

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[zinc horizon]

I don’t get it

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@zinc horizon Has your question been resolved?

[fathom saddle]

Why is I1 = 7/4? You don't know that the resistor is burning 7V

Namely, there's voltage passing through, and getting burned by the other components

@zinc horizon

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[muted token]

Find three consecutive odd integers such that three times the middle one is one more than the sum of the first and the third

[muted token]

How would I go about solving this? So far, I have

$(2n + 1) + (2n + 5) = 3(2n + 3)$

[glossy valve]

Vortac

[muted token]

If I add one to the middle one, I'd get an even number

[gusty sonnet]

let me see

so using your approach

(2n+1) + (2n+5) + 1 = 3 (2n + 3)
4n + 7 = 6n + 9
2n = -2 so n = -1

substituting n back in, 2n + 1 = 2(-1) + 1 = -1
similarly, 2n + 3 = 1, and 2n + 5 = 3

so the answer should be -1, 1, and 3

[muted token]

It is, but why does the + 1 go on the left?

[round sparrow]

!nosols

[full forum]

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.

[muted token]

I originally tried $(2n + 1) + (2n + 5) = 3(2n + 3) + 1$

[glossy valve]

Vortac

[round sparrow]

That’s a misreading of the problem

The triple should be one more than the sum of the others

Thus the +1 is indeed on the other side

[muted token]

ah okay

[gusty sonnet]

mb

didnt realize

[muted token]

I'm doing practice word problems and am trying to wrap my head around the thinking for them

I get most of them right when they're easier, but ones like this I accidentally put one part on the wrong side

"three times the middle one is one more than the sum..."

when I see something like this, I should read it as apply the one more to the opposite side?

[round sparrow]

It’s one more than the other side, so 1+other side

So yes

But I really gtg to sleep bye

[muted token]

Thanks!

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[jovial flare]

Someone help me how to do this

[round sparrow]

Try squaring the equation

[astral sinew]

Honestly you can just solve for x here

You don't have to be smart about it

[real hull]

just square and cube the equations

$x^2 \frac{1}{x^2} = 3^2 = 9$ and $x^3 \frac{1}{x^3} = 3^3 = 27$

[glossy valve]

DM ModMail for new nickname

[real hull]

wait wait mb mb, i thought it was multiplication

it is addition

if you square both equations you get $x^2 + \frac{1}{x^2}+2(\frac{1}{x} \cdot x) = x^2 + \frac{1}{x^2} +2 = 9$ this makes $x^2 + \frac{1}{x^2} = 7$

[glossy valve]

DM ModMail for new nickname

[stiff musk]

!nosols

[full forum]

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.

[real hull]

oh okay

[round sparrow]

Yeah but I don’t think it’s intended

[real hull]

so you can just square and cube the equations and simplify the expression

[round sparrow]

Squaring symmetrical equations is a very common trick

[real hull]

wyaaa,

[jovial flare]

This is a 9th grade question btw idk how to do it

[real hull]

you in ninth grade?

[jovial flare]

Yep

[real hull]

cool, so you got your question answered or you need more explanation

[jovial flare]

Explanation

[real hull]

okay so you have this equation righht $x + \frac{1}{x} = 3$

[glossy valve]

DM ModMail for new nickname

[jovial flare]

Uhh

[real hull]

square both sides

like $(\right x+\frac{1}{x}\left)^2 = 3^2$

[glossy valve]

DM ModMail for new nickname
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[jovial flare]

Whole square lhs?

[real hull]

yea both LHS and RHS, since they are equal their squares must also be equal

[jovial flare]

Ok makes senses

[real hull]

so what do you get now

[jovial flare]

Formula?

[real hull]

yea

[jovial flare]

X² + y² +2xy

[real hull]

yea but y is 1/x

[jovial flare]

So it's like x² + (1/x)² + 2(x)(1/x)

[real hull]

yea, and what does the 2(x)(1/x) term simplify to

[jovial flare]

2x/x

[real hull]

yea and thats two

so at last, $x^2 + \frac{1}{x^2} + 2 = 9$ ,9 is theh RHS squared

[glossy valve]

DM ModMail for new nickname

[jovial flare]

So 7 is the first answer?

[real hull]

yea, that's right, and you can do the same for the second question

[jovial flare]

Same formula righ

[real hull]

i think so, i didt try it

try it and ask me if you get stuck

[jovial flare]

Alright tysm

[wary condor]

for the cubed one?

are you talking about this

[real hull]

yes

[wary condor]

and what are you suggesting ?

[real hull]

cubing both sides

[wary condor]

I dont think OP knows how to do that

[wary condor]

I-

(x+1)^3 is not x^3 + 3x + 1 or smth like that

it is not the same process as squaring 🙁

[real hull]

yea its like (x+1/x)^3 = x^3 + 3(x^2)(1/x) + 3(1/x^2)(x) + 1/x^3

then the middle terms simplify to 6/x

so its like x^3+6/x+1/x^3

[wary condor]

ik how it will go

[real hull]

yea i think he might need to solve for x

[wary condor]

I'm just assuming that sank doesnt know how to cube a bino

[brittle sun]

Huh what no

[brittle sun]

No

[real hull]

then what do you suggest

[brittle sun]

I think rather than assuming, you should wait for them to talk

[brittle sun]

||You don't get 6/x, you get 3x+3/x||

[real hull]

oh yea mb mb

[brittle sun]

Which should look familiar, no solving required

[real hull]

yea its 3(x+1/x)

wc is 9

no need to solve for x

[real hull]

yea thanks

[wary condor]

only asians learn how to cube a binomial before learning about binomial theory /j

[sacred yarrow]

👀

[rare dock]

[split hatch]

[full forum]

@jovial flare Has your question been resolved?

[jovial flare]

I got it

27

[full forum]

Channel closed

Closed by @jovial flare

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[wary condor]

wait

is that even correct

[full forum]

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[jovial flare]

18*

[jovial flare]

Mb

[wary condor]

yes

great job

.close

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[atomic patio]

I got this question and I don't know what i did wrong. It's from further mechanics a levels.
The diagram shows a uniform square lamina of side 4r and density 2p attached to a uniform lamina in the shape of a quarter circle of radius 4r and density p. Find the distance of the centre of mass from the edges OA and OB.

this is roughly what the diagram looks like

[atomic patio]

hit me up if you gotta know where im stuck

[grave elm]

can you use integration?

[atomic patio]

nooo idts

[grave elm]

and do u know what the COM of quarter circle is ?

[atomic patio]

the book told me to find centre of mass and im assuming its using the formula they gave

uh wait

[grave elm]

what was the formula?

[errant tiger]

can u show ur solution

can't tell where ur stuck without seeing ur steps

[atomic patio]

ah sure not sure if you guys can read my handwriting tho

give me a sec

[atomic patio]

there it is

so the funny thing is my answers are (3.24, 2.11) and the correct answer is 0.2 below (3.04, 1.91)

and i have no clue how 😂

i might not survive senior highschool ;-;

[grave elm]

what was the formula u tried to use?

[atomic patio]

umm which one?

[grave elm]

lets start from the first one

[atomic patio]

for centre of mass for circle i followed the formula booklet

that they provide during exams

[grave elm]

what was the book's formula?

[atomic patio]

let me ss

which gave me 16sqrt2/3pi

using r=4r and a=pi/4

am i right up to this point?

[grave elm]

16sqrt2r/3pi

youre just missing the r

[atomic patio]

yes yes my badd forgot the r

[grave elm]

but okay, what next?

[atomic patio]

mhm?

okay so

mass x centre of mass + mass x centre of mass=total mass mutliplied by x

or y but i wanna do x first

so basically i rearranged to get the x with the line on top not sure what its called

i got 32r^2 rho for the mass of the square

4r x 4r x 2rho

then just the midpoint for the centre of mass, 2r

[grave elm]

,w solve for x, 2r * (4r)^2 * 2p + (4r + 16sqrt(2)r/(3pi))(4pir^2 * p) = x * ((4r)^2 * 2p + 4pi*r^2 * p)

[atomic patio]

eh shouldnt it be summed for the total mass?

square+ quarter circle right

[grave elm]

oh right, i accidentally multiplied the sum instead of +ed

[glossy valve]

Results provided by WolframAlpha

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

Requested by methisalwaysright

[grave elm]

still wrong

oh, maybe not

3.24

huh

[atomic patio]

wait so im right?

[grave elm]

probabably

[atomic patio]

the answer sheet is both my coords -0.2

[grave elm]

lemme double check

[atomic patio]

so faulty answer sheet?

because the coincidence is not coincidencing

😂

[grave elm]

Oh wait, this thing looks weird

[atomic patio]

really?

how so

[grave elm]

oh wait, this is the distance from centre?

if so, u gotta multiply it by another 1/sqrt2

[atomic patio]

what why

[grave elm]

because we need the green distance

not the red distance (which is what we have)

[atomic patio]

wait

ohhh

wait i never bothered looking for the horizontal component in the other questions how did i still get it right

[grave elm]

weird

perhaps u made multiple such mistakes and they somehow cancelled out

but this is definitely sth we must do

[atomic patio]

oh the other question was a semicircle so the centre must've already aligned with the axis

[grave elm]

makes sense then

[atomic patio]

i guess this type of thing only shows up when its a quarter circle instead of semicircle?

since usually only those 2 types of circles show up

[grave elm]

yeah, it doesnt show up in semi-circle, bc there the vertical / horizontal component is same as distance from center

[atomic patio]

eh its still wrong-

wait my fault

i got it

i put in sqrt3 instead of 2

aight got it

[atomic patio]

wow thanks man

i almost lost sleep over such a simple detail ;-;

.close

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[atomic patio]

can someone explain how you find the centre of mass in solids, especially in this worked example? im not sure why they did it like that

[atomic patio]

the first image is just an explanation so i guess thats how im supposed to learn it? but i dont understand it at all ;-;

i want to be able to understand the concept behind this

i'm heading to bed, i'd appreciate it if someone could dm me to teach me about this in case the channel closes, thanks!

[turbid badge]

hold on

[atomic patio]

hey hey im not asleep yet

[turbid badge]

$\overline{x}=\frac{\text{total moment}}{\text{total mass}}$, no?

[glossy valve]

Green

[turbid badge]

nice

[atomic patio]

yes it should be, right?

[turbid badge]

yea

[atomic patio]

by the way what do u call the x with the line on top

[turbid badge]

first we can multiply both sides by the total mass

[turbid badge]

overline i guess?

[atomic patio]

wait i got another question

[turbid badge]

we know the total mass is $\frac{1}{3}\rho\pi r^2 h$

[glossy valve]

Green

[atomic patio]

is this the same as the 2d stuff i did up in the channel before? like you take out pieces of the shape so its big shape-small shape??

does it work the same way

[turbid badge]

so $\frac{1}{3}\rho\pi r^2 h\overline{x}=\text{total moment}$

[atomic patio]

but before they were squares and circles

[turbid badge]

sorry i dont know about the other channel

[atomic patio]

its in this channel

but basically it works the same way?

[glossy valve]

Green

[atomic patio]

ah okay

[turbid badge]

hm not too sure sorry

[turbid badge]

total moment is the sum of the moment of those thin slices mentioned in the textbook, right?

[atomic patio]

yes

by the way

does the overline x=30/13 we get start from the bottom of the cone?(the bigger base)

why did they substract by a in the end

[turbid badge]

we can consider the total moment to be $\int_{0}^{h} \rho \pi x y^2 dx$

[turbid badge]

in the examples?

[atomic patio]

yess

mhm

okay

wait

what

[turbid badge]

one sec

I'll complete this first before the example

[atomic patio]

okayokay

[glossy valve]

Green

[turbid badge]

forgot rho 😔

[atomic patio]

why pixy^2

[turbid badge]

moment is mass • distance right

[atomic patio]

yes

wait

force right

[turbid badge]

each slice has mass rho pi y² and the distance from the y - axis is x, so we consider that

[atomic patio]

oh okay

then we just integrate?

[turbid badge]

okay so we equate this with what have from earlier

$\int_{0}^{h} \rho \pi x y^2 dx = \frac{1}{3}\rho\pi r^2 h$

[glossy valve]

Green

[turbid badge]

y=r/h x, no?

you can make the substitution and see how terms cancel out, and then evaluate the integral

[atomic patio]

sorry i was tryn understand the book but failed so im bac😂

[atomic patio]

okayokay

[turbid badge]

if you do that, you get centre or mass as 3h/4

[turbid badge]

oh 😭

[turbid badge]

this is basically the derivation

[atomic patio]

I SEE IT

[atomic patio]

skill issue on my part but should be fine

[turbid badge]

🗣️

[turbid badge]

happens dw

[atomic patio]

💀💀

thanks

.close

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Use .reopen if this was a mistake.

[fading heron]

Hi

[full forum]

Available help channel!

Send your question here to claim the channel.

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• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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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.

[nocturne roost]

In this case where would p be located and how do I do part b

[frank gale]

whats p

I can't find it

[nocturne roost]

Look at the solution

Part b

[frank gale]

oh whoops

[gritty rose]

amazing p = a substitution out of nowhere

[nocturne roost]

Ok can u show me how to solve this

With writing please

[frank gale]

its the value of x at which the volume of revolution between said x and 20 is equal to 360pi

[nocturne roost]

I don’t get it

Can u visualize it by drawing

[frank gale]

sorry I'm on a phone

on a shaky train

basically draw a vertical line between o and 20 in figure 2

the volume you get from spinning the area to the right of the line

up until 20

[nocturne roost]

Ye ye I got it

Am talking about part b

[frank gale]

I am talking about b

k im out of time im entering a tunnel

be back in a bit maybe

[nocturne roost]

Alr

[frank gale]

ok

so if i asked you to find the volume of revolution of y between x is 2 and 20

would you know how to?

[nocturne roost]

Yes

U just put 20 as upper limit

And 2 as Lower limit

And solve

[frank gale]

yes

the p comes in because it is that lower limit

you dont know what it actually is

[nocturne roost]

But wouldn’t that give volume of full - 360pi

If u put 20 and p as limits

[frank gale]

you'd get an expression in terms of p

[nocturne roost]

Wait I’ll give u visualization can u explain

[frank gale]

because you've plugged the p in

[nocturne roost]

If I put 20 and p as limits

What volume would I get

[frank gale]

you'd get the volume in terms of p, since you're plugging it in as the lower limit

[nocturne roost]

Where’s p tho

[frank gale]

like in line 3 of the solution of b

[nocturne roost]

Oh wait

I GOT IT

[frank gale]

oh those are marking guideljnes

[nocturne roost]

Yes

[frank gale]

they might be incorrect on purpose

[nocturne roost]

Ye ye I looked at it the other way around

[nocturne roost]

It’s official Edexcel

They cannot be wrong

[frank gale]

as in they need examples of incorrect answers so that markers know what to give for incorrect answers

[nocturne roost]

Na they always provide correct answers

And tell what to reject

[frank gale]

nah I dont mean the mark scheme is wrong, I mean the scheme has examples of incorrect answers ppl would write

[nocturne roost]

That’s usually examiner report

I didn’t share that

[frank gale]

see the bold text at the top of this

[nocturne roost]

Ye ye

[nocturne roost]

Ur right

Sorry it’s 12 am for me

I act lil abnormal during that time

[frank gale]

yep been there before

[nocturne roost]

Also

Where u from and studying at

[frank gale]

sorry im not comfortable sharing that

hope you understand

[nocturne roost]

Yes I understand

[merry rain]

no

[nocturne roost]

Ok

[full forum]

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[void jay]

if i use 31 * avg / 100k .. i can think of many scenarios for statement 1 and 2

though, theres only one. I need some help with how to set up the proper fraction

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[stone pier]

Is Dij = Dji given by the definition of a diagonal matrix? Dij = 0 when i =/ j, which means j =/ i. Does this entail Dji = 0 --> Dij = Dji ??

[full forum]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[stone pier]

It is regarding diagonal matrices

the ij are subscripts

[sonic stratus]

Yes, it is.

Since all diagonal matrices are also symmetrical

[stone pier]

If i'm still in properties, do I need to prove that?

Or is that really it

My proof question is: If D is a diagonal n x n matrix, then D^t is

I was assuming D^t isn't one then you get Dji =/ 0 which is a contradiction

But i didn't know if that was allowed because it seems simple

[sonic stratus]

By definition symmetrical matrices are those which are the same as its transpose.

[stone pier]

I'm being asked to prove it though

And we have been going over properties of VS so do you think its more formal than that?

[sonic stratus]

Again, if you can appeal to D being symmetrical, then thats kinda it.
If you had to prove it, then you should probably appeal to the fact that, for square matrices, the transpose doesnt alter the diagonal

[stone pier]

Hm

I see

I guess its that simple thanks

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[leaden violet]

where do i go from here
i did integration by parts twice but it just ends up giving the same thing

[paper rain]

let I = integral

from what you've found,

$\int e^{2x} \cos(3x) , dx = \frac{1}{2} e^{2x} \cos(3x) + 3\int \frac{1}{2} e^{2x} \sin(3x) , dx$

[glossy valve]

Flatus

[paper rain]

apply another application of IBP on the RHS

you should then get

$\int e^{2x} \cos(3x) , dx = \frac{1}{2} e^{2x} \cos(3x) + \frac{3}{4} e^{2x} \sin(3x) - \frac{9}{4} \int e^{2x} \cos(3x) , dx$

[glossy valve]

Flatus

[paper rain]

move that integral to the lhs

$$\frac{13}{4} \int e^{2x} \cos(3x) , dx = \frac{1}{2} e^{2x} \cos(3x) + \frac{3}{4} e^{2x} \sin(3x)$$

Hence,

$$\boxed{\therefore \int e^{2x} \cos(3x) , dx = \frac{4}{13} \left[\frac{1}{2} e^{2x} \cos(3x) + \frac{3}{4} e^{2x} \sin(3x)\right]}$$

[glossy valve]

Flatus

[leaden violet]

i figured it out thanks

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[hushed barn]

Does there exist a positive integer pair (m,n) such that (13 * 3^(n-2) - 2^(n+1))/(2^m-3^n) is an integer other than when (m,n) = (2,2) or (2,3) or (3,2) or (6,4)?

[onyx glen]

$\frac{13\cdot 3^{n-2}-2^{n+1}}{2^m - 3^n}\in\bZ$

[glossy valve]

Ann

[onyx glen]

this?

[hushed barn]

Yes

[onyx glen]

hm.

[split hatch]

You've been asking a lot of questions similar to this one

Is there some natural context in which they arise

Just curious

[full forum]

@hushed barn Has your question been resolved?

[hushed barn]

Yes, I am aware

[hushed barn]

I’m working on my own project

[full forum]

@hushed barn Has your question been resolved?

[brittle sun]

Question, before you post them here, do you run a script in python for some reasonable bounds

[hushed barn]

Idk how too

[brittle sun]

Fair

Also no m in the numerator?

Just to confirm

[hushed barn]

Yes

[brittle sun]

Hmmm

I think you can show there's finitely many (m, n)

[hushed barn]

How?

[brittle sun]

Growth rates

Denominator grows much faster than the numerator

For solutions to exist, you get smth like m ≤ n log 3 + O(1)

[hushed barn]

True, but there could be a very large value of n that works

[brittle sun]

I mean the aim is to prove that for large enough n, D_m,n > N_m,n

[brittle sun]

With this constraint on m

[hushed barn]

Yes, that’s mainly the aim

[full forum]

@hushed barn Has your question been resolved?

[delicate torrent]

mainly $2^m - 3^n \mid 13 \cdot 3^{n-2} - 2^{n-1}$

[glossy valve]

1 divided by 0 equals Infinity

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[hushed barn]

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[twin venture]

Yo, I translated this task from german to english with Chat gpt, so i hope it makes sense.
My problem is on a) the last part which says "With what probability does the actual value of X exceed the expected value E(X)?"

[twin venture]

i am not sure what they mean with actual value of x

heres the original question to

[open lava]

i translated it into english fully because i had no idea what it was about

hopefully the translation is accurate

[twin venture]

are u german?

[open lava]

no no, i used the screenshot auto- text translate thing

[twin venture]

omg i just saw i didnt copy pasted the task

thats what i did to

wow

[open lava]

loll

[twin venture]

no wonder why i get no help

thanks tho

[open lava]

anyway, the real value of x (the actual number of red balls in the container) is different from the expected value e(x)

so after you found the standard deviation, in either the positive or negative direction

(which was the second subdiv)

[twin venture]

ye its around 1,13

[open lava]

you now have to find the probability of the deviation being in the positive direction (i.e. more than expected value)

did that answer your doubt?

[twin venture]

wait i needa think

[open lava]

cool cool dw

[twin venture]

do i need the sigma rules for that?

[open lava]

one sec

wdym

oh you call empirical rule sigma rule kk

[twin venture]

prob ye

[open lava]

yeah you do

[twin venture]

ye

just looked it up

[open lava]

yep

[twin venture]

do i just have to do the weird u-1,13. So 1,6-1,13 and thats 0,47 which means 47%?

wait no

that makes no sense 1 sec

[open lava]

same thought process lmao

wait im not sure if you need the sigma rule here

[twin venture]

aight

[open lava]

i think you need to calculate each and every one of the outcomes and do it

but thats so tedious

[twin venture]

wym

[open lava]

there must be an easier way

one sec

OK WAIT

sigma rule required

[twin venture]

aight

[open lava]

and your answer is correct lol

well its close

[twin venture]

the 47%??

[open lava]

yh

[twin venture]

how close

[open lava]

but uh- your initial answer is wrong

by a bit

its 1.6

unless my translation is wrong

and also its 0.497 so 49.7% prob

[twin venture]

u mean the standard deviation?

[open lava]

m-m, e(x)

[twin venture]

what is 1.6

[open lava]

its not 1.13 is 1.6

no the e(x)

[silk bridge]

14. An urn contains 4 red, 6 yellow, and 10 blue balls. n balls are drawn with replacement. The random variable X describes the number of red balls, and the random variable Y describes the number of yellow balls among the drawn balls

looked a bit confusing

[open lava]

yeah thats way better

considering theres no sizes XD

[silk bridge]

lol

[open lava]

english my enemy

jk i love english but yk

[open lava]

is this considered cheating on math

will i get kicked out of this channel

also did your doubt get answered?

@twin venture

[twin venture]

i got confused what is e(x) and what value is the standard deviation

i got for e(x)= 1.6 and the standard 1.13

[silk bridge]

seems right

[twin venture]

what do u mean by that?

[silk bridge]

now you need P(X>1.6)

[twin venture]

do i need to round 1.6 to 2 or 1?

[silk bridge]

what is greater than 1.6

[open lava]

2

[twin venture]

okay 2

[silk bridge]

yes

[twin venture]

from x = 2 to x= 10 right?

[silk bridge]

no, as long as it helps we can fuck english

[twin venture]

because i draw 10 times right?

[open lava]

thats not why we fuck english, but yes

[twin venture]

bruh that takes ages

[silk bridge]

X holds value of red balls how many red balls are there in total?

[twin venture]

4

[open lava]

my heart breaks for you

yes but 2 to 10

[twin venture]

huh

now im confused

[open lava]

right

[silk bridge]

mb 2 to 4

[open lava]

?

now im confused too

[twin venture]

oh now i get it

because (we call it k) is the total of hits u can get, and u only can get 4 hits because theres only 44 red balls

right?

hope that makes sense

[silk bridge]

you are only counting for P red balls they can be from 0 to 4

[twin venture]

ye

[silk bridge]

yh

[open lava]

yes yes

thats right lol istg im running on 9 hrs of sleep in 72 hours and no caffeine

[twin venture]

but wait the balls are getting putted back

so u can get 10 hits

[silk bridge]

have good sleep

[open lava]

take that sentence outta context rn

[twin venture]

why

now i cry

[silk bridge]

oh i forgot replacement

[open lava]

tyy i will try

[silk bridge]

@twin venture

[twin venture]

ye

[silk bridge]

you need P(X>=2)

[twin venture]

to 4 or 10

[silk bridge]

you can do 1 - P(X<=1) = P(X>=2)

[open lava]

that what i did originally

[silk bridge]

oh

[twin venture]

okay let me try

[open lava]

since P(X>=2) = 1-P(X=0)-P(X=1)

ughh

discord you fucker

[silk bridge]

X = 1 as well

[open lava]

yh sorry dc sometimes sends my msgs in the middle of me sending em

[twin venture]

do u guys know how to do the shortcut on an calc

[open lava]

gdc?

[twin venture]

theres an long way but ik theres an shortcut

1sec

heres the long formula

but there was wan easier way on the calc

[open lava]

can you rotate

oh wait

got it nvm

idk the calc part

[twin venture]

aight

[twin venture]

what does the >= means

what do i need to do

only did like >

without the =

[glossy valve]

scoob

[twin venture]

OH

[open lava]

lol

the only reason i know that >= and <= means that is cuz of IT

we had it in microsoft access

for some reason haha

did you get the answer

[twin venture]

i got 0,442

P(X>=2) = P(x = 2) + P(x = 3) + P(x = 4)

or?

[open lava]

no no

P(X>=2) = 1-P(X=0)-P(X=1)

[twin venture]

😭

[twin venture]

oh i didnt saw that

[open lava]

this is way easier to calculate tho

[silk bridge]

i forgot replacement you have to do till x= 8 this way

[open lava]

you can use a sci calc

[twin venture]

wait

wth

why to 8

[twin venture]

i got one i think

[silk bridge]

do this

[silk bridge]

because n= 8

[twin venture]

oh i see

[twin venture]

0.498

[open lava]

after putting into calc i got 49.66≈49.7%

[twin venture]

perfect

wow

im exhausted

[open lava]

great 🔥🔥

[twin venture]

I appreciate all your help

[open lava]

me too lmao

[open lava]

I SAW THAT

YOU THINK WE WERE HEL?

IM OFFENDED

also youre welcome, no problem <33

[twin venture]

.close

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[twin venture]

14. An urn contains 4 red, 6 yellow, and 10 blue balls.
    A total of n balls are drawn with replacement.

The random variable X describes the number of red balls, and the random variable Y the number of yellow balls among the drawn balls.

a) Let n = 8.
Sketch the corresponding binomial distribution of the random variable X.
Calculate the expected value and the standard deviation of X.
With what probability does the actual value of X exceed the expected value E(X)?

b) How many balls must be drawn at least so that the expected value of the random variable Y is greater than 5?
What is the standard deviation of Y in this case?

c) How many balls must be drawn at least so that the expected value of X is at least 1?

d) How many balls must be drawn at least so that with at least 90% probability, at least one red ball is drawn?

[twin venture]

this is the task, its translated from german

i need help on d

[grave elm]

Start by expressing the probability that after drawing n balls, at least one red ball is drawn

[astral sinew]

Balls

(Lol)

[twin venture]

(0,2)^n

[grave elm]

hmm, how did u get this?

[twin venture]

4/20 balls are red so 0,2 and u draw n times

[grave elm]

oh its with replacement

anyway I asked for at least one red ball

[twin venture]

heres the original

[grave elm]

not all red balls

[grave elm]

this calculates the probability that in n draws, all the balls are red

if u want a hint, start by calculating the probability that in n draws, no balls are red (which is the opposite of at least one red ball being drawn, which is the probability we need)

[silk bridge]

You need number of balls such that P(X>=1) >=90%

[twin venture]

(0,8)^n > 0.9 ?

[grave elm]

What's 0.8^n?

what probability does that express?

[twin venture]

so P(x=1) ... P(x=8) >= 0.9 ?

[twin venture]

all balls are red

[grave elm]

then its wrong

[twin venture]

ye but idk any other thing

[grave elm]

Okay, the idea is that instead of calculating this, which is complicated because there are many P's, we will just do 1 - P(x = 0) >= 0.9

1 - P(x = 0) is the probability that x isn't 0, meaning that at least one red ball was drawn

does that make sense?

[twin venture]

ye

[grave elm]

alright, and can you calculate P(x = 0)?

[twin venture]

ye, 1 sec

[grave elm]

it'll be dependent on n

[twin venture]

i got 0.83 but thats wrong, right?

[grave elm]

Yes, that sounds wrong

[twin venture]

1sec

[grave elm]

it should depend on n

P(x = 0) is the probability that no red balls are drawn in n draws

[twin venture]

im at 1- ln(1-0,2)

[grave elm]

U are overcomplicating this massively

[twin venture]

my calc says 1,22

[grave elm]

Yeah, this should be a sign that it's wrong lol

[twin venture]

ye

but idk what i did wrong

[grave elm]

how did u even get ln in?

[twin venture]

ill try to do it in paint 1sec

i tried to ge the n down

hope that makes any seense

[grave elm]

is this 1 - P(x = 0)?

[twin venture]

ye

[grave elm]

then its correct, until that ln thing

you cant just put in ln when its not there

that'd be like having 4 + n^2, but I dont like the square so ill just do square root and say that it's 4 + n

obviously 4 + n^2 isnt the same thing as 4 + n

we cant just apply random functions whenever we want

[twin venture]

xD

oh shit

[red badger]

sup

[twin venture]

yo

[grave elm]

but we can work with the line above that

it should simplify to 1 - (0.8)^n

[twin venture]

but how do i get the n down then

[grave elm]

and we want this to be >= 0.90

[twin venture]

1 - (0.8)^n >= 0.9

[grave elm]

yep, and now that's an inequality and not just an expression. Now we could apply logrithms to both sides and actually make progress

but first, start by some simple manipulations to isolate the (0.8)^n

[twin venture]

wym by isolate?

[grave elm]

make one side (0.8)^n and put everything else on the other side

[twin venture]

oh

ok

1sec

(0.8)^n >= 0,1 ?

[grave elm]

almost

did you multiply both sides by -1 at some point?

[twin venture]

ye+

wait

[grave elm]

When you do that, you have to flip >= to <=

[twin venture]

i did -1 first on both sides and then :(-1)

tahts what u mean right?

[grave elm]

yeah

[twin venture]

oh

[grave elm]

but when we multiply or divide by negative number, we have to flip the >= sign

[twin venture]

(0.8)^n <= 0,1 ?

[grave elm]

5 > 3; -5 < -3

[grave elm]

alright, now we can either solve it by guessing n, or use logarithms (and then round it appropriately)

if you wanna do logarithms, do log base 0.8 on both sides

[grave elm]

but if you do this, keep in mind that log base 0.8 is decreasing and so you'll have to flip the <= sign again

[twin venture]

ill need an minute on that one

okay ill just wrote everything else down and im not quite sure how to write that down, because i didnt use ln that much

if i ln(0,8) in my calc its -0,223

should i write only that instead of the (0,8)^n ?

[grave elm]

when you ln both sides you should get $\ln\left(\left(0.8\right)^{n}\right)\le\ln\left(0.1\right)$

[glossy valve]

MathIsAlwaysRight

[grave elm]

no

we dont ln just 0.8, we ln (0.8)^n

[twin venture]

but how do il write that in the calc

theres no n

[grave elm]

the inequality isnt solved yet

[grave elm]

now the reason why this is useful is that by log laws
ln((0.8)^n) = n * ln(0.8)

[twin venture]

oh

that makes sense

[grave elm]

this is waht's meant by "logarithms bring exponent down"

they dont make it magically disappear, they just put it down

anyway, now we have $n\cdot\ln\left(0.8\right)\le\ln\left(0.1\right)$

[glossy valve]

MathIsAlwaysRight

[grave elm]

one last step is remaining

[twin venture]

wait i thought we have to flip the <= again

[grave elm]

not when we apply ln. ln is increasing

if we applied log base 0.8 though, we would have to flip it

logarithms with base < 1 are decreasing, so we'd have to flip it

ln is base e, which is around 2.7 which is >1 so its increasing and we do no flipping

[twin venture]

wow

[grave elm]

so if we apply logs with base > 1 we dont flip, if the base is < 1, we flip

[grave elm]

anyway, do you know how to solve this now?

[twin venture]

im not sure with the > thing everytime i put something on the other side

but i would divide both sides by ln(0.8)

but i would assume we flip

because its negativ

[grave elm]

If we add / subtract a number / expression -> we do not flip
If we multiply / divide by positive number -> we do not flip
If we multiply / divide by negative number -> we do flip
If we apply logarithm with base > 1 -> we do not flip
If we apply logarithm with base < 1 -> we do flip

[grave elm]

exactly

ln(0.8) is negative, so when we divide by it we do flip

[twin venture]

thats smth i gotta save

[grave elm]

$n\ge\frac{\ln\left(0.1\right)}{\ln\left(0.8\right)}$

[glossy valve]

MathIsAlwaysRight

[grave elm]

so we should have this

[twin venture]

yes

[grave elm]

,calc ln(0.1) / ln(0.8)

[glossy valve]

The following error occured while calculating:
Error: Undefined function ln

[grave elm]

,w ln(0.1) / ln(0.8)

[glossy valve]

Results provided by WolframAlpha

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

Requested by methisalwaysright

[grave elm]

so what do u think the answer to the question would be?

[twin venture]

11

holy

Thanks alot dude

[grave elm]

np

[twin venture]

i appreciate it

i will close it now

.close

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[sullen flame]

Hi. Can any one prove me how does
aCosB+bcosA=c in a tringle??

[rough tundra]

do you know the law of cosines?

[outer lotus]

Draw an altitude down to the side c

Use the definition of cosine in a right angled triangle

Pretty straightforward

[sullen flame]

Fs

[sullen flame]

a2+b2-c2=2abcosC?

[rough tundra]

<@&268886789983436800>

[rough tundra]

Yep! But more importanly for this proof it also holds when a^2=b^2+c^2-2bccos(A) and b^2=a^2+c^2-2accos(B), you agree?

hello chartbit

[outer lotus]

There you go

[rough tundra]

if you do, then rearrange for the cosine terms, plug into a*cos(B)+b*cos(A) and you'll see the computation simplifies to c and that the identity you've proposed it simply a corollary of the Law of Cosines

[outer lotus]

You can see that the yellow bit is b•cosA, blue bit is a•cosB, and they add up to c

[full forum]

@sullen flame Has your question been resolved?

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[marsh gorge]

Could someone help me with this question i dont know where to start

[queen crater]

You are supposed to write down two equations and solve them as a system

[marsh gorge]

Idk where to get the equations from

[queen crater]

From the bold parts of the text

[marsh gorge]

Is it (L x W) = 50 and L = 3 +2W

[queen crater]

No, it says perimeter not area

[gritty rose]

second eqn is right tho

[marsh gorge]

So the first one is 2L + 2W = 50

[queen crater]

Yes

[marsh gorge]

Ok

What do I do with the equations

OHHH

[gritty rose]

last part

[marsh gorge]

I havew

W*

And L is

[gritty rose]

find L and plug it in to see if it fits the two starting eqns

side question: who's this in your search bar

[marsh gorge]

Huh

Oh idk

Microsoft just puts things there

Nelson Mandela im guessing

Oh

The equations didint work

[queen crater]

That doesn't seem right

[gritty rose]

mistake here

you interpreted your 50 as 56

write neater

[marsh gorge]

I had to throw the 6 over

Oh so it's 44

?

[gritty rose]

oh then you should be subtracting 6, not adding

6-6 = 0, not 6+6 = 0

[marsh gorge]

Fixed

[gritty rose]

do this when you get L

[marsh gorge]

Got L im check now

Works

Thx guys

.close

[full forum]

Channel closed

Closed by @marsh gorge

Use .reopen if this was a mistake.

[full forum]

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[tribal marlin]

hi I'm just tryna figure out to find the volume of a right pyramid and trying to find the volume and surface area of objects with multiple shapes

[subtle marten]

,rccw

[worldly matrix]

Hello!! What're you getting confused on?

[tribal marlin]

I have no clue where to start :(

[worldly matrix]

Well, would you like to tackle question 7 or 6 first?

[tribal marlin]

6 first

[worldly matrix]

ok

do you have on-hand the formulae for the surface areas of a sphere and cone?

[tribal marlin]

I do yes

[worldly matrix]

and do you know how to find the area of a circle?

[tribal marlin]

yes

[worldly matrix]

The shape is constructed with a cone with its circular face covered by a hemisphere

what surfaces on those objects are exposed, whose surface area we are to measure?

[tribal marlin]

Both the hemisphere and the cone are exposed but their base isn't

[worldly matrix]

correct!

so how are we gonna evaluate that?

[tribal marlin]

do we find the surface area of the hemisphere first? or is it the cone?

[worldly matrix]

According to the commutative property of addition (that a+b=b+a), it doesn't matter which one we do first so long as we do both

will the surface areas of the individual objects change if we calculate each of them in a different order? no

[tribal marlin]

it wouldn't? right?

[worldly matrix]

it would not

[tribal marlin]

ok

so then I'd just do the regular 3πr² for the hemisphere and πrs+πr² for the cone

and then after that I'd add both surface areas?

[worldly matrix]

not quite

you still have to consider the two circular faces being occluded

[tribal marlin]

oh yeah

so then would I find the surface area for the circle?

I'm confusing myself sorry

so after finding both of the areas of the hemisphere and cone we still have to find the surface area of the "base" (none exposed circle)

or am I skipping something

[worldly matrix]

yes, you're on the right track

what's the area of that circle? πr²

occluded on each shape

[tribal marlin]

it's 113.0973

[worldly matrix]

I usually prefer putting all the variables together algebraically then evaluating

it makes for better practice

so you've got the hemisphere (minus the circle) plus the cone (minus the circle)

[worldly matrix]

so do this but then subtract 2πr²

make sense?

[tribal marlin]

ohhh okay

so I subtract the unexposed circle

[worldly matrix]

exactly! you don't count the surface area of the part that's covered!

[tribal marlin]

okay thank you

that makes sense

[worldly matrix]

did you get the answer?

I'm gonna be honest, I can't really make out the numbers on example 6 in the photo

[tribal marlin]

the radius is 6 and the height of the cone is 12

I'm almost done with the question

[torn jolt]

Did you solve the question?

You can close if solved using . solved

[subtle marten]

or .close

[worldly matrix]

there

s

3 more parts

no need to rush

haste breeds error

[torn jolt]

I'm not rushing, just asking

[tribal marlin]

so my final answer was 565.0cm²

and I think that's right

[torn jolt]

Example 7?

[tribal marlin]

example 6

[worldly matrix]

example 6; surface area

radius of 6; height of 12

[torn jolt]

The text is very hazy

r = 6cm and h = 12cm?

[worldly matrix]

ye

I got 479

[torn jolt]

Oh that's easy
Combination of solids, you can make the calculations easier by taking radius squared as common

[worldly matrix]

lemme recalc

[torn jolt]

Okay I'll try

[tribal marlin]

i problem did something wrong

[worldly matrix]

so did you evaluate 3πr²+[(πr)sqrt(r²+h²)+πr²]-2πr²

[torn jolt]

479.2808340256903

evaluate

[worldly matrix]

I combined like terms to make 2πr²+(πr)sqrt(r²+h²)

[torn jolt]

this expression will be too difficult, first solve for slant height

[worldly matrix]

which is sqrt 180 yes

[torn jolt]

Now onto volume, did yall find volume?

Do you need help to find volume or you know the formulas?
@tribal marlin

[tribal marlin]

I'm trying to figure out my mistake with the area

[torn jolt]

Hmm, could you show your work?

If you can show your work, we can help you find out your mistake

[tribal marlin]

yeah I can show you my work

[torn jolt]

Okay show

[worldly matrix]

how'd you get 18 for the slant?

[torn jolt]

,bccw

[torn jolt]

,rccw

[tribal marlin]

cause √144=12 and √36=6 and I added the two which gave me 18

[worldly matrix]

you can't do that with roots

[worldly matrix]

18 squared is 324

[tribal marlin]

oh. my teacher told the class to do that..

[torn jolt]

I think you've tried to split sqaure roots, which is not done like that
It is only when two terms are multiplied, given that neither of the terms are less than one
root ( a x b) = root(a) x root(b) {a, b > 0}

[worldly matrix]

to square root a number you just squared is to get the number you started with

at that rate you're just summing the numbers, which is incorrect