help-39

[umbral briar]

Do i use 0.2

[pastel umbra]

yh

[umbral briar]

Ki

Kk*

[pastel umbra]

idk what indcoh is on about, you did the right thing already lol

[pastel umbra]

https://www.physicsandmathstutor.com/maths-revision/a-level-edexcel/hypothesis-testing/ has some model answers you can have a look through

[umbral briar]

These are the 5 questions we have to do. I feel so stupid cause its only AS stats

[umbral briar]

Thank you very very much

[pastel umbra]

1 b and c are very similar to the question you're on there

[umbral briar]

Yes i think that was odd

Pcd

Oh wow i looked at the cheat sheet makes a lot of sense i think

So to work out

[rare field]

https://youtube.com/playlist?list=PLg2tfDG3Ww4sGRzEzvOBE9kDwFvC5UnBP&si=eIIHNEDr8MFDwrKU
I really recommend watching tlmaths' videos on hypothesis testing
they helped me a lot back when I was learning this stuff

bicen maths is also good if youre looking for edexcel specific videos

[umbral briar]

Thank you

What does the 5% significance thing mean

[pastel umbra]

cf. the 10% thing in the PMT example

But the idea is, if your experimentation gives you results that are less than 5% likely to happen, you're gonna

https://klipy.com/gifs/doubt-press-x-3

[rare field]

in a nutshell (its been a while since ive learned this stuff so might be a bit handwavy), if the probability you calculate ends up being lower than the significance, you reject h0. if its higher, you accept h0

[pastel umbra]

So you can, with a 5% error, reject H0 and opt for H1

[umbral briar]

That makes snes but how do i know if they’re less than 5% likely to happen

[pastel umbra]

That's what critical regions are for

[umbral briar]

Oh right ok..

[pastel umbra]

If I used that Q1 with 8, 5 and 5.7 for instance

[umbral briar]

Mhm

[pastel umbra]

P( X <= 5 ) = 0.19 (or 19%)

[umbral briar]

Mhm

[pastel umbra]

That's not below 5%, so if I were testing whether that binomial distribution probability had the right number or whether it should have been lower, I can say "there's not enough information to reject that H0 hypothesis" (because getting a result of 5 or lower for X was still likely enough to occur reasonably)

If we'd calculated that it was 0.03, say, then we can argue that the Bin. Dist. Prob. was probably wrong (i.e. we reject H0), because of how unlikely it was to happen

[umbral briar]

Ohh right ok

So i need to calculate probability of the null hypothesis happening

And if it’s less than 0.05 then i reject h0

[pastel umbra]

To note - it could have still happened, since it had a probability of 0.03 of actually occurring during H0; we call this 0.03 the probability of error

[umbral briar]

Is that the acceptance region

[pastel umbra]

You're calculating the probability of your test result (or anything more extreme) happening, while assuming H0

[umbral briar]

Oh

Ohhhhh

So im finding the probability

[pastel umbra]

-# Video 2 and 4 here

[umbral briar]

That the number of green >= 11

Which is 1 - p(g <= 11)

[pastel umbra]

yes

And then >= 12

[umbral briar]

What

[pastel umbra]

And >= 13, until we find when that probability gets below 5%

[umbral briar]

Parameter?]

[pastel umbra]

ah, we're being asked for the critical region, so we want to see which range gets that probability low enough

[pastel umbra]

(that's what the p technically actually is called)

[umbral briar]

Ohhh

[umbral briar]

Right so i just keep calculating p

Until it gets less than 5%

[pastel umbra]

Though the wording is a little off ikwym, yh

You're trying to find the smallest a so that P(X >= a) < 5% = 0.05

[umbral briar]

Ahhhhh ok ok

And that a is

The p that i use

[pastel umbra]

When we get that answer, "X >= a" is our answer, what we call "the critical region"

(See video 4)

[umbral briar]

Thank you im going to go and watch the videos and try to figure it out

You’ve been so so sooo much help

Thank you again

Have a good day

.close

[pearl pond]

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[pearl pond]

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[deep atlas]

Determine the number of positive integers
N is smaller or equal to 2000 such that N satisfies the following system of congruences simultaneously:

N=1 (mod 3)
N=2 (mod 4)
N=3 (mod 5)
N=4 (mod 6)

[proper nova]

oh what the hell

you claimed 2 channels for that?

[deep atlas]

Whoops

[proper nova]

do you notice anything special with each of the statements?

[deep atlas]

Hi I'm back

[proper nova]

hi back, answer my question

[deep atlas]

Where its not there

[proper nova]

read

[deep atlas]

The congruences are consistent

[proper nova]

hm

tell me

[deep atlas]

Hm

[proper nova]

what if i add 2 to each statement

what does $n + 2 \equiv$ to for each mod

[jolly parrot]

1 divided by 0 equals Infinity

[deep atlas]

Becomes the same

[proper nova]

wdym the same?

[deep atlas]

.close

[pearl pond]

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[pearl pond]

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[inland epoch]

How do I find the domain and range? I’ve got everything else down but I just couldn’t understand how to find the domain and range

[obsidian sentinel]

For what set of x values is the function defined? As in, which values can you put in that give you an output?

[inland epoch]

I’m confused

I really don’t know what that means

[obsidian sentinel]

Notice how g(x) takes some value of x, giving you an output.
In this specific problem, there's one value of x where the function doesn't give you any output (y)

Looking at the graph you drew might help

[inland epoch]

Okay okay

Both are at -5 does that mean something

[obsidian sentinel]

What happens if you try to evaluate g(-5)?

[inland epoch]

Uh

I honestly got zero clue

[obsidian sentinel]

It's undefined at x = -5, because neither arm of the piecewise function includes -5.

[inland epoch]

doesn’t the endpoint have -5 though

I feel like that’s a dumb question

[obsidian sentinel]

No, you actually drew an empty circle, which means it does not include that point

[inland epoch]

Oh

oh

[obsidian sentinel]

just a question, have they taught you this yet?

[inland epoch]

the whole no line = open
line = closed? If that’s what you’re referring to yes

[obsidian sentinel]

Im mean ( vs [

[inland epoch]

ohhhhhhh

Yes they have

[obsidian sentinel]

nice, that might be what your professsor expects here

[inland epoch]

Here’s what the problems that aren’t for homework look like he hasn’t done anything with the [,) so I assume that’s not something he wants here we did these in class and I only have these domain and ranges written down because I copied the board also have mercy on my messy hand writing and organization I just learned this today and I was stressing over the domain and range like the entire class

[obsidian sentinel]

then we should follow the notation he expects

[inland epoch]

Okay

[obsidian sentinel]

so, we now know that g(-5) is undefined, is every other number defined?

[inland epoch]

Umm, yes? I think

[obsidian sentinel]

Yeah

so the domain is every number except -5

[inland epoch]

wait so like

This is correct?

[obsidian sentinel]

yeah

[jolly parrot]

xelph

[obsidian sentinel]

now, the range

which values of y can the function output

you can also see it on the graph

[inland epoch]

Oh my god I feel so good after doing that

Uh

[obsidian sentinel]

good!

you should

[inland epoch]

5 and 3? Or does the open circle make that wrong

[obsidian sentinel]

let me just forward it rq

[obsidian sentinel]

imagine flow coming in from the left of the graph

at which y values does it touch lines or filled dots of your graph?

[inland epoch]

Lines?

[obsidian sentinel]

yeah, like, your graph lines

[inland epoch]

It would touch the lines then

[obsidian sentinel]

actually I just noticed the range for the above problem is wrong

let's start with that range first

[inland epoch]

Oh

[obsidian sentinel]

I'll help more directly, notice how the graph has absolutely no values above y = 3?

[inland epoch]

yeah

[obsidian sentinel]

so the range can't include anything above y = 3
it also excludes y = 3 itself

only everything less than that

[inland epoch]

so it would be y < 3?

[obsidian sentinel]

indeed

[inland epoch]

Ohhh

[obsidian sentinel]

$y \in \mathbb{R} < 3$

[jolly parrot]

xelph

[inland epoch]

Okay thanks I didnt think about that actually

my teacher just helped the class with that problem and only wrote y is all real numbers

so how do I find the second questions range when both circles are open

[obsidian sentinel]

well, it still includes other values, no?

just not those ones

at least one of the lines goes up a lot

[inland epoch]

so would y > 5 be correct

[obsidian sentinel]

almost, there's another line, though it's only at one specific number

[inland epoch]

It’s at 3

[obsidian sentinel]

yes

[inland epoch]

y > 3 u y = 3

That?

[obsidian sentinel]

y > 5 not 3

but yeah

[inland epoch]

Oh yeah my bad there I mix up numbers a lot

I just tried this one on the back by myself is this right?

[obsidian sentinel]

the range of the 9th question is incorrect

[inland epoch]

well at least I got the domain how do I find the range here?

[obsidian sentinel]

remember to look for gaps horizontally

just visually

I don't see any gaps horizontally

[inland epoch]

There’s a gap vertically though

[obsidian sentinel]

that only matters for domain

vertical gaps = gaps in domain

horizontal gaps = gaps in range

[inland epoch]

Since there’s no gaps horizontally then would that make it range is all real numbers?

except for -2?

[obsidian sentinel]

all of them

there's no gap

including -2

sure the line on the left has a hole at y = -2

but the other line fills it in

so it doesn't matter

[inland epoch]

Ohhhh

[inland epoch]

Can you help me with this one as well if you have the time I feel like I’m probably wrong on the last one but the connecting lines are confusing me

[shell nexus]

Domani and range are both wrong if we talking about ex 11

According to the graph

My suggestion to "visualize" the domain Is to think of seeing It from a perspective on top of the graph

[obsidian sentinel]

oh yeah I've been assuming the graphs are correct

hope they are

[shell nexus]

In this case the domain should be for every x but -1, because all the x axis Is "covered"but that gap

While the range Is y<3 united y=5

[inland epoch]

The graphs shouldddd be correct according to how I was taught

[shell nexus]

Wait its wrong

Cause you put 2 in the last equation, the resoult shouldve been 0

You must have not seen the minus sign

But It doesn't affect domain and range

[inland epoch]

Oh you’re right

[inland epoch]

So it would be x > -1?

[shell nexus]

Nope, both x<-1 and x>-1

Cause theres a single gap, so there Is a single Number that Is not in the domain

[inland epoch]

I’m confused aren’t there two gaps vertically

[shell nexus]

Yeah, but One, whe have two empty gaps, the other we have a full One and an empty one

You Will never find 2 full gaps on the same x

With two gaps, neither of them Will have the x associated, but with One full, the full One Will be the associated with x

[inland epoch]

That makes sense

[pearl pond]

@inland epoch Has your question been resolved?

[pearl pond]

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[pearl pond]

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[warm thorn]

I don't have any specific question I'm stuck on. It's more like division as a concept I have to do mentally. I'm studying for the asvab, so I can't use a calculator. The topic I'm stuck on is averaging speeds so you'll get a question like this. Maybe I'm overthinking it but the final equation comes out to 180/4 and idk how to do that mentally. Is there a simple way to do that?

[dense jasper]

break it down into smaller steps (usually, this means trying to divide your numerator and denominator by smaller numbers to reduce their size)

Ex. 180 and 4 are both even (their units digits are both even), so you can divide both by 2

[warm thorn]

As I sent that, a friend of mine said I could also use the answers to do it, but I want to figure out how to do it without answers.. so basically I'd just
180/2=90
4/2= 2
90/2 is 45 answer, does this work everytime?

Different number of course

[dense jasper]

If you have a fraction and divide the numerator and denominator by the same (nonzero) thing, you'll always end up with something equivalent

[warm thorn]

Is it okay if I run another practice quiz rq before I close ticket? With this in mind

[dense jasper]

the rule to open a new channel for new questions mainly applies if the questions aren't rlly related

here it's the same topic, so you can leave this open and keep it all in one place

in fact, that's preferable

[warm thorn]

Tyty, let me run this practice quiz rq to see if I understand this fully

[dense jasper]

[toxic lichen]

as far as mental arithmetic goes, division by 4 is just division by 2 and then by 2 again

[warm thorn]

What about when decimals get involved into the equation like for say
180/4.5

[dense jasper]

delete the decimals

$\frac{180}{4.5}=\frac{180 \cdot 2}{4.5 \cdot 2}=\frac{360}{9}$

[jolly parrot]

Civil Service Pigeon

[dense jasper]

tbf there's some easier subcases you may have learnt off

like "dividing by 0.5 is the same as multiplying by 2"

alternatively you could do $\frac{180}{9/2}=\frac{180 \cdot 2}{9}$

[jolly parrot]

Civil Service Pigeon

[dense jasper]

where was I going with this again

oh yeah

but usually you're gonna either convert the decimal to a fraction and work with the integers

or if it's something "easy" like with .5, you can inspect how to make the decimal "turn into an integer"

if you're going to do mental arithmetic at least

long division exists but like

I haven't used that in forever

[warm thorn]

To be completely honest, maybe I'm just being stupid, but I genuinely have no idea how to do that mentally.. like with 360/9

180x2/9 same thing

[dense jasper]

you can use the same strategy as earlier

and/or note that $\frac{360}{9}=\frac{10 \cdot 36}{9}$

[jolly parrot]

Civil Service Pigeon

[dense jasper]

10 is usually easier to work with when doing arithmetic

that's why a lot of mental arithmetic tricks revolve around getting things in terms of 10

[warm thorn]

So for this to make it simple you could be 360/3
9/3
120/3 = 40?

[dense jasper]

mhm

[warm thorn]

I see I see

[dense jasper]

if you did it like this, you can remember that 4 times 9 is 36

so 36/9 = 4

multiply that by the 10 and you get 40

[warm thorn]

This might hopefully be my last question, I ran another quiz and I'm on one that seems to have no understandable simplification.. 200/3, how would I do that mentally without relying on multiple choice

[dense jasper]

a little rougher since it doesn't reduce to an integer

actually you can't simplify this at all as a fraction

I assume you want a decimal approximation

ig if you have $1/3=0.33 \dots$ and $2/3=0.66 \dots$ learnt off, you can use that

[jolly parrot]

Civil Service Pigeon

[dense jasper]

otherwise, you're kinda stuck with long division or some shortcut mental variant of it

[warm thorn]

Yes the answer is 66.7, but I only know that because it was multiple choice

If we take out multiple choice I'd be completely stuck

[dense jasper]

^^

this is how I'd do it

[warm thorn]

How would I do that? I don't quite understand

[dense jasper]

remember what I said earlier about 10 being easy?

that

[warm thorn]

So 2/3

Then x100

Or

[dense jasper]

yeah

[warm thorn]

Ooooo

That's extremely helpful

[dense jasper]

I'm sure you can look up some more specialised tricks but I always operated on the "good enough" system for arithmetic in my exams

I never cared much for calculation lol

[warm thorn]

I just wanna be able to solve these mentally so I can apply it if need be without multiple choice

So these solutions help a lot

[dense jasper]

[warm thorn]

❤️❤️❤️❤️

Let me run one more practice run see if I got this down

[dense jasper]

[warm thorn]

Ok slightly more issue I don't see where it falls into

180/3.5

This doesn't fall into the 360/7 category because.. no common denominator

That I'm aware of at least

[dense jasper]

yeah this also isn't an integer

idk how many ppl have $1/7=0.\overline{142857}$ learnt off

[jolly parrot]

Civil Service Pigeon

[dense jasper]

actually is that even right

,calc 1/7

[jolly parrot]

Result:

0.14285714285714

[dense jasper]

oh well

you have that ig

here I'd just do something equivalent to long division

but since you have multiple choice options, you can get away with ballparking

||360 is basically 350||

[mystic geode]

||350 is basically 0||

[dense jasper]

ngl some number related to 360/7 is seared into my mind cause I had to bash calculating a floor for arml relays and I was a p1 lol

I just don't remember what division gave that number

why did they have to pick a heptagon bro

ig SHIPYARD was a cute little gimmick but cringe question overall

[warm thorn]

I'm testing a slight theory regarding this actually let me see if it works rq..

[dense jasper]

we love experimentation

[warm thorn]

Nvm 🥀

Yeah idk how to do that one

I think I have the basis down on most of these tho

I appreciate the help

[dense jasper]

[dense jasper]

yeah look up a vid on long division if you have to

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[warm thorn]

.close

[pearl pond]

Channel closed

Closed by @warm thorn

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[pearl pond]

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[autumn fossil]

There are 70 people. Frst we assign colored hats to each of them. We have 10 hats of 7 kinds of colors.

Then we assign the 70 people to groups of 10 randomly.

Whats the probability that there is exactly 1 group woth all colored hats same?

[autumn fossil]

Is it just me or is it ridiculously hard

These are the ans choices

I think neither is correct, though the intended one is apparently c

[oak quiver]

Exactly one or at least one?

[autumn fossil]

Exactlx

[oak quiver]

Because this doesn't say exactly

[autumn fossil]

(I asked the teacher for clarification)

[oak quiver]

Maybe lost in translation

[autumn fossil]

No, there are literally 3 possible interpretations lol

The test was full of pearls

To me, c seems to be the answer to: "If we look at an arbitrary group, what's the probability that it'll have all colors same?"

im thinking inclusion exclusion now

Order the people by groups, so
1, 2, 3, ... 10
11, 12, ..., 20
21, 22, ..., 30
31, 32, ..., 40
...
61, 62, ..., 70

Now we factor out 7 * 10 and suppose that the one group with the same color is the first one, with color A (colors are A, B, C, D, E, F, G)

The probability that they all have color A is (1 / 7)^10

Now we need to calculate the probability that the remaining groups dont have same color...

so time to include-exclude stuff

[pearl pond]

@autumn fossil Has your question been resolved?

[autumn fossil]

Maybeee, but idk

[pearl pond]

@autumn fossil Has your question been resolved?

[autumn fossil]

its incorrect btw

I tried simulating it for lower values

there's too much dependency for me to cook up anything useufl

[pearl pond]

@autumn fossil Has your question been resolved?

[dusty jungle]

why not just no. of divisions with at least 1 group having the same colour - no. of divisions with at least 2 groups having the same colour

[iron basin]

wouldnt the latter term split again

[autumn fossil]

Yeah

[autumn fossil]

And the issue is, how do you even calculate that

[cinder flower]

i think i have a way with egfs, it's kinda complicated though

not something you could do by hand in a test

i just woke up though i will type it up after i brush my teeth and stuff

[dusty jungle]

ok i get it

[autumn fossil]

Yeah, its probably not doable

The test was all wrong

4/7 questions were wrong

[cinder flower]

hm i need to think about it more

[sage igloo]

by "one" does it mean EXACTLY one or AT LEAST one?

[cinder flower]

he said exactly just below it

[sage igloo]

oh nvm 🤦‍♂️

by symmetry we can consider runners by their colors i think

so then it's just the probability that, for the color we choose, it gets its own group and the others don't

the total ways to arrange the groups is (70!)/((10!)^7) by multinomial coefficient

[dusty jungle]

did you mean total ways to make the groups

also you should probably divide by 7!

[sage igloo]

yeah but you get what i mean 😛

[cinder flower]

well since it's a probability problem what's 'total' and 'desirable' is kinda up to you in terms of what should be ordered and treated differently and what not, but i'm not sure that's a good way to look at the 'total'

[sage igloo]

oh yeah i forgot, so it's $\frac{70!}{(10!)^7\cdot7!}$

[jolly parrot]

nadat12

[sage igloo]

for the other problem, fix a color and group them

[cinder flower]

you can proceed however you want, but do consider that you are currently treating it like order assigned within the groups matters

which is not "wrong" but i imagine it to be quite messy

[sage igloo]

wait should we pie bash the ways to order the 60 colored people into groups so that no color has their own group

[cinder flower]

colored people

[sage igloo]

and then i think simplification would follow from binomial theorem or smth like that

[sage igloo]

i could just call them colors tbh

in total, (60!)/((10!)^6*6!) to give groups to the 6 colors

by symmetry choosing a color for pie is the same so i'll just choose one WLOG and multiply by binomial

actually by pie i think the total is [\frac{60!}{(10!)^66!}\binom66-\frac{50!}{(10!)^55!}\binom65+\frac{40!}{(10!)^44!}\binom64\pm\cdots+\frac{0!}{(10!)^00!}\binom60]

[jolly parrot]

nadat12

[autumn fossil]

whats pie

[sage igloo]

Principle of Inclusion-Exclusion

[autumn fossil]

Hmm and what exactly are you calculating?

[dusty jungle]

i thought the coefficient would be 7 choose k (k=1 to 7)

[sage igloo]

i'm counting ways to give 6 groups to 6 colors so no color has their own group

and for the 7th color, yeah just give them a group

[autumn fossil]

I see

[sage igloo]

if i call whatever that stuff is by N

[autumn fossil]

Wait i still dont quite understand how you calculated that

[sage igloo]

ok

so for two objects

so you know venn diagrams?

[autumn fossil]

Yeah

[sage igloo]

so let's say i had 20 artists

40 mathematiceans

[autumn fossil]

i understand standard principle of inclusion exclusion btw, i just dont understand where you took the specific numbers from

[sage igloo]

and 10 of both

oh

[sage igloo]

i got the fraction factor from

[sage igloo]

this stuff

and the binomial coefficient from being lazy and not caring

(i chose WLOG some colors)

[autumn fossil]

Oh so the first 60! thing is the total number of ways to arrange the groups

[sage igloo]

the solution is [\frac{7N\cdot(10!)^77!}{70!}] i think

[jolly parrot]

nadat12

[autumn fossil]

Yeah, okay, i think i kinda get that

[sage igloo]

and the (10!)^7 comes from binomial coefficient logic

and the 7! also arises from that

[autumn fossil]

what even is N

[sage igloo]

this

[autumn fossil]

oh

Alright, i see, thanks

[sage igloo]

ok so the hard part is actually calculating N

[autumn fossil]

ill try to understand it more thoroughly and verify it with some code

It works

very nice

[autumn fossil]

Thanks man!

.close

[pearl pond]

Channel closed

Closed by @autumn fossil

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[pearl pond]

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[royal galleon]

For 17 a. I solved it but I don't understand how it only gets the max from the range of 0 to 10. Yes I understand that you x <5 but if we graph the function the functions value after 10 goes to infintiy I'm pretty sure. Why does solving this purely algorithmically just in the range of 0 to 10 and not touch that. I understand that after the value 10 it isn't really a max per say but the max we found doesn't have the max value

[royal galleon]

Is it just finding me one of the relative maxes?

[sharp vigil]

you need to decide on a domain for your function that makes physical/geometrical sense

that domain should probably not go all the way to infinity

[pearl pond]

@royal galleon Has your question been resolved?

[royal galleon]

I understand that but if I just did it purely mechanically without thinking of bounds I still don't encounter that issue

[sharp vigil]

remember there are two places that a global maximum could occur, at a local maximum or at the endpoints. it so happened that your global max occurred at a local max in this problem

[royal galleon]

I still don't see how it could be a global max. Since the function keeps getting bigger as its endpoint approaches some value. We can't say that is a max since we don't have an actual point but that would make the max we find using the first deirvative not a global max too

[plush bramble]

can you should the math you did referring to "it" in "don't see how it could be a global max"

[royal galleon]

Ok for part a we have f(x) = x(10-2x)^2

f'(x)=(10-2x)^2 - 4x(10-2x)

And the max of that was 10/6

[sharp vigil]

a global max is the maximum point in the function's domain. remember that you need to define the domain sensibly

[royal galleon]

So your saying that if I just did it mechanically with no regards to bound all I can say is that it is a relative max

[plush bramble]

"mechanically" is the wrong word here.

[sharp vigil]

well if you did not define the function's domain then you would not know what counts as a global max. i would argue that part of your "mechanical" solution procedure for these sorts of applied optimization problems should be defining a sensible domain for your functions, though

[royal galleon]

Ok

Thx

.solved

[pearl pond]

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[proud frost]

For z in C, how is the ROC infinity?

[proud frost]

seems they can just "see it", but I don't really see it

[west sapphire]

use the ratio test?

[proud frost]

Its |a_{k+1}| / |a_k| = k right? edit: 1/(k+1)

oh wait

i cannot do algebrea

i see okay

.close

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[viscid tinsel]

hi

[viscid tinsel]

can anyone help me

[minor cloud]

have you started doing it? if you have, where are you stuck at?

if not, what's confusing you?

[viscid tinsel]

in the 3)

[minor cloud]

part 3) alright. have you started solving part 3)?

[viscid tinsel]

not yet

[minor cloud]

what's confusing you about part 3)?

[toxic lichen]

do you know how to do 1) and 2)?

[viscid tinsel]

yeah

[toxic lichen]

the method is pretty much the same the exact same for all three

you boil it down to a quadratic equation

[minor cloud]

I'll defer to Ann, good luck!

[viscid tinsel]

sanks

[minor cloud]

but I will leave you with one hint

the fact that you are asked to do something with points of intersection in all three parts should clue you in as to what to do for all three.

[viscid tinsel]

it was a checking if someone can help me

i know exactly

how to do all

[toxic lichen]

??

[minor cloud]

you want your answer checked?

[viscid tinsel]

sanks

[toxic lichen]

ok so this was a dishonest help request?

[minor cloud]

if so, show your answer

[toxic lichen]

you dont actually need help?

<@&268886789983436800> not sure what to think about this but like, can yall have a look?

[minor cloud]

I'll disengage from this then

[dim linden]

please dont waste peoples time like this

.close

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[hearty fiber]

can someone please help me with graphing graphs

[hearty fiber]

v=(3)(330+acos(theta))/330 where a=50/3 and cos(theta)=(1800-at)/root((1800-at)^2 +100)
i need to plot v/t graph

i need to graph this function

i need to tell if the function would be like graph A or B

[minor cloud]

are you allowed to use a tool?

[hearty fiber]

no, its for a jee problem

[minor cloud]

hm

[pearl pond]

@hearty fiber Has your question been resolved?

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[latent spire]

how do I go about getting a rule for part a

[solar ember]

It says "constant rate"

So you will have something linear like y = mx + c

Where m is your constant growth rate

[latent spire]

okk'

soo..

Ohh do i find m

by doing the formula

18000-16500/6-3?

5000?

[toxic lichen]

missing brackets here btw

[toxic lichen]

also it is not $5k/yr for sure

[latent spire]

y=5000x-1200?

oh

huh

so then what am i supposed to do?

y=50x-12??

[toxic lichen]

do the calculation correctly

don't try to guess

[latent spire]

i did thoo

[toxic lichen]

show work on paper

[latent spire]

1500/3

500

so y=500x+c

[toxic lichen]

ok so five HUNDRED

[latent spire]

ohh mb!

[toxic lichen]

not five thousand

[latent spire]

true

[toxic lichen]

not fifty

[latent spire]

alr thenn

you get 18000=500(6)+c

[toxic lichen]

then you find c, yes

[latent spire]

15000

so y=500x+15000?

[toxic lichen]

yes but replace "x" with "t" and also "y" with "A"

[latent spire]

ohh

to fitthe fomrula ?

A=500t+15000

[toxic lichen]

to fit the variable names that the question gives you

[latent spire]

yepp so is that correct?

[toxic lichen]

yes

[latent spire]

thankss

.close

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[eager jewel]

i got the relation
a(n+2) = a(n+1) + a(n)

now not sure how get each option

[pearl pond]

@eager jewel Has your question been resolved?

[iron basin]

a1 = a2 = 1

from there i think we can prove A) by induction on
a1 = a3 - 1

B and C by some method similar to sum of AGP

[eager jewel]

yes i did agp method itself

say C option

i got 9S/10 = b1/10 + (b2-b1)/10^2 + (b3-b2)/10^3 + ...

[eager jewel]

a3 = a2+a1 from the relation i got ig so we can say from that

but i didnt rlly want to put n=1

to prove it

[iron basin]

why not write bn in terms of an

[eager jewel]

i tried but that kinda didnt help

given b1= 1

b2-b1 = (a1+a3-1)

b3-b2 = (a2+a4 - a1 -a3)

and so on

[iron basin]

that seems a bit more troublesome

[eager jewel]

yeah

[iron basin]

$1+ \sum {n=2}^{\infty} \frac{a{n-1} + a_{n+1}}{10^n}$

[iron basin]

isnt this easy to find from option B

[eager jewel]

this is for option C?

[eager jewel]

we havent proved option B right

[iron basin]

yeah

divide expression by 10 and add to itself

would help defining the summation = S

[eager jewel]

uh sorry im getting a bit confused

which is our summation S here

the summation b(n) or summation a(n)

[iron basin]

sorry i meant option B

[eager jewel]

for option B how did u write a(n) = a(n-1) + a(n+1)?

isnt it a(n) = a(n+1) - a(n-1)

[iron basin]

no uh

[iron basin]

this is for option C

[eager jewel]

oh

[iron basin]

this is for option B since you mentioned it

[eager jewel]

so well get
11S/10 = a1/10 + (a1+a2)/10^2 + (a2+a3)/10^3 + ...

where S is option B

now we use a1+a2 = a3
a2+a3 = a4?

[iron basin]

mhm

[eager jewel]

11S/10 = a1/10 + a3/10^2 + a4/10^3+...
11S/100 = a1/10^2 + a3/10^3 + a4/10^4 + ...+a1/10 + a2/100 - a1/10 -a2/100
11S/100 = S + a1/100 - a1/10 -a2/100

and from this we solve for S ?

[iron basin]

yuh

[eager jewel]

oh yea that worked

damn

now option C?

tbh i can figure it out from here

thank u

.close

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[warm tiger]

Ok so what i did is and found out that for one face diagnol, there will be 3 skew lines

So for 12 diagonols, there should be pair of 36 skew lines

36/12C2 gives me 6/11

[toxic lichen]

wait, aren't you missing at least one

[warm tiger]

Which one?

[toxic lichen]

let me make my own diagram

i count 5 diagonals skew to the blue one here

one in each face other than the bottom (the other is intersecting or parallel in all cases)

[verbal oak]

Any skew line is equivalent to any other by rotation

So you can just consider 1

As your initial choice I mean

And then compare the other options to that one

[toxic lichen]

also yeah you can and should just do (number of black lines)/11 rather than any of that nCk stuff

[warm tiger]

Oh yea i see the 2 i missed

[verbal oak]

The one in the same face intersects ❌
The two touching the same corner intersect ❌

Are you counting parallel as skew?

[warm tiger]

OK so, for every face diagonol there are 5 lines, together which they make skew lines

[warm tiger]

No

[verbal oak]

Oki doki

[warm tiger]

Isnt the definiton of skew line is that they're not parallel

[verbal oak]

I've seen both tbh

[warm tiger]

Do i have to divide by 2 as well

Coz ig i would be overcounting

[verbal oak]

It's been sufficiently long I can't remember which one is more standard

[warm tiger]

I've been taught that they're not parallel

[verbal oak]

Okay so we'll go with that

[toxic lichen]

girl parallel lines are never skew

[verbal oak]

Do you understand my argument?

[verbal oak]

Pretty sure I've seen this at least once but I could be misremembering. It's fine anyway

[warm tiger]

Uhm not really 💀

[verbal oak]

Okay

So let's start with a basic case

Pick 1 line

And find which lines are skew to it

We've done that

We got 5/11

[warm tiger]

Why 11 in denominator

Wouldnt it be 12

Sorry im just bad at probability and combinations

[verbal oak]

We're picking 2 lines

Our first choice is a particular line

[warm tiger]

Oh rest 11 lines we can choose by 11 ways

[verbal oak]

Then we have 11 options for the 2nd choice

[warm tiger]

Got it

Thanks both @verbal oak @toxic lichen

[verbal oak]

So if we choose AC first, then the probability that a random choice is skew to it is 5/11

Now we can consider another possible option for the first choice

Let's say BD

[warm tiger]

AB isnt a diagonol tho

[verbal oak]

Yeah mb

BD

Had to check the diagram again to make a good choice

So BD

We don't need to do the whole calculation again because we can rotate the cube

And if we rotate it so that BD lies along where AB used to be

[warm tiger]

That's.. genius

[verbal oak]

Then it's exactly the same thing as before

So it's 5/11 again

Lol ty

And we can do this for any side of the cube as well

So every option is 5/11

So the answer is 5/11 regardless of initial choice

[warm tiger]

Makes sense

[verbal oak]

And we're done

[warm tiger]

How do i improve myself at combinations

I like so much struggle in it

I dont find any other topic as confusing as this

[verbal oak]

Alternatively, we can say there are 12 initial choices with a probability of 1/12 each, so 12 x 1/12 x 5/11

Hmm

Practice tbh

[warm tiger]

I've been doing for like since last 2 years but whenever a new question appears, i just go blank

[verbal oak]

It's also helpful to look for situations where things are symmetric, and also where they are different

Actually one thing I would say

I've always found the formulas for C and P an unhelpful abstraction

It kinda outsources too much of the thinking process to a formula

But this stuff is a little fiddly so it's easy to pick the wrong one

I prefer to think of this in terms of options at each stage

So if we want to select 2 balls from a bag of 20 balls, and order matters, then we have 20 options first and then 19, so 380 altogether

If order doesn't matter, then every possible ordering of the balls is now equivalent

So we need to divide by the total number of possible orderings

Which in this case is 2

Which is 190

You can check the formulae if you like and they should evaluate the same

That might not help you but it could be worth trying

[warm tiger]

Hmmm 20C2 is indeed 190

[verbal oak]

I prefer to keep my intuition engaged with what I'm doing

Rather than letting it go and relying on a formula

It does require a little intuition training but you won't kinda lose track of what you're doing any more

One useful result to get comfy with is that the number of possible orderings of n objects is n(n-1)(n-2)...(1) = n!

[warm tiger]

factorial n

[verbal oak]

Mmhmm

[warm tiger]

I'll try implementing and see if it works with me lol

[verbal oak]

The reason for that is that you have n options for the 1st choice, then (n - 1) options, etc

I can give you an example if you like

[warm tiger]

yep that i understand, it's mostly these new questions where you have to analyse a lot and my peanut sized brain cant handle it

[verbal oak]

Got another example you wanna look at?

[warm tiger]

How would you do this?

[verbal oak]

Hmm

A few obvious brute force methods spring to mind

But that's not very elegant

[warm tiger]

My brain always thinks of the brute force, lenghty methods first

[verbal oak]

That's fine

Okay so I think I have a nice way

So the first thing to notice (remember my earlier point about looking for what is the same and what is different) is that the first digit is different to the other 6

In that it cannot be 0

So it's probably helpful to do some shenanigans on the last 6 and stick the 1st one on afterwards

So in that vein, we have a few options

Let's say the first digit is a 1

Then the other 6 need to contain either 1 once, or 0 twice

[warm tiger]

or both 1 and 0 twice

[verbal oak]

Sure

I was including that situation in both

[warm tiger]

Oh okayy

[verbal oak]

I suppose we don't have to actually

Let's do them separately

Actually nvm it's painful

Let's include them

So we have a 1

And then 5 0/2s

Each 0/2 has 2 options

So that's 32 options

[warm tiger]

We have to fill 6 places right

7 digit number

[verbal oak]

And then we scramble all 6 digits however we like

So 6!

Actually a useful check is also to find the total number of options as well

Just as a sanity check

So that's 3^6 x 2

,w 3^6 x 2

[jolly parrot]

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[verbal oak]

Oh actually this is worse than I thought

[warm tiger]

Lol ikr

[verbal oak]

Cuz digits are indistinguishable

It's getting too brute force for my liking

[warm tiger]

It's from an exam, and all the solutions online have used principal of inclusion and exclusion

[verbal oak]

I suppose we can just eliminate bad cases

So we have 1458 total options

[warm tiger]

Wouldn't that make it worse, we would have to subtract when 0 is exactly twice, thrice, 4 times

[verbal oak]

The issue is that the amount of rearranging we can do with the last 6 digits is dependent on what they are

Which means we're not significantly improving on just looking at each case

Okay I think I have an approach but we'll see

Let's say 1 is our initial digit as before

And we want exactly 2 0s

We can place those 2 0s in T{5} = 15 ways

Explanation:
If we place the 1st 0 in the first position, then we have 5 options for the 2nd 0

[warm tiger]

Yep i get why it is 15

[verbal oak]

If we place it in the 2nd position, we have 4 options we haven't done already

Oki

So the 5th triangular number

Also 6C2 by the other perspective

[warm tiger]

Yes yes

[verbal oak]

But I'm endeavouring to avoid that to emphasise this approach

Then the other options have to be either 1 or 2

So that's 4 digits with 2 options each

[warm tiger]

16

[verbal oak]

So 2^4 = 16

Mmhmm

So that's 240 ways

[warm tiger]

correct

[verbal oak]

So that's that case done

Then we want exactly 2 1s

Or in this case, 1 1, since we have 1 as the first digit already

So we can place that in 6 ways

And then 2^5 options for the other digits

So 192 options

If I didn't mess up any mental calculations

Agree up to there?

[warm tiger]

Yep

[verbal oak]

But we've double counted some cases

[warm tiger]

Shouldn't we also count when none of the 1 is at first position

[verbal oak]

Hold thy horses, sir

Yes, we'll do that in a sec

Cuz the 2 0s case includes the 1 1 case and vice versa

[warm tiger]

Yeah

Right

[verbal oak]

When they're both true

We do want to count it

But only once

And we've counted it twice

So we need to subtract it once

So now we need 2 0s and 1 1

Which we can arrange in 20 ways

[][]4
[]-[]3
[]--[]2
[]---[]1
10

-[][]3
etc
6

--[][]2
etc
3

---[][]1
1

Does that diagram make sense?

Ahhhhh

Okay this is still useful but we have more fiddling to do

[warm tiger]

i guess

i get it

[verbal oak]

That sounded like a no and then a yes

We can place them so that one of our choices is 1st, then the next is 2nd, and then there are 4 options for the last

Or we can go 1st place, 3rd place, and then 3 options

Or 1st place, 4th place, 2 options

Or 1st place, 5th place, 1 option

That's 10 altogether

Or we can go 2nd place, 3rd place, 3 options

2nd place, 4th place, 2 options

2nd place, 5th place, 1 option

That's 6

You get triangular numbers each time so one you've got the start point you don't need to do all of it generally

Hence the etc

[warm tiger]

yepp

got it

[verbal oak]

And then 3rd place, 4th place, 2 options

Giving 3

And 4th, 5th, 6th

Which gives 20 altogether

Also 6C3

On top of that, we can also rearrange the numbers within these positions

In particular, the 1 can go in 3 different places

So that's a factor of 3

Also 3C1 (placing the 1) or 3C2 (placing the 0s)

So 60 ways to do it

[warm tiger]

seems correct

[verbal oak]

We want 2 0s and 1 1

Which we already have

So now all other digits must be 2s

So that's no extra ways

Or a factor of 1^3 = 1, if you prefer

[warm tiger]

yep makes sense

[verbal oak]

Oh that shows me an easier way to do that lol but nvm

[warm tiger]

ts pmo so much

[verbal oak]

It's fine

Having our intuition engaged allows us to notice these things

Even if sometimes it's too late lol

At least we're not lost

[warm tiger]

yeah true

in exam im supossed to solve this in under 4 minutes somehow kek

[verbal oak]

Welp

I mean I might be able to

Obviously it takes longer to type all my thoughts out

[warm tiger]

yeah typing took the 90% of time

[verbal oak]

Okay so we need to subtract 60

So we have 240 + 192 - 60

So that's 372

Halfway there lol

Now 2 is our first digit

[warm tiger]

aahh

[verbal oak]

Exactly 2 1s:
We can put them in 15 places, as before

They're indistinguishable so no rearranging

[warm tiger]

and 16 for 2/0s

[verbal oak]

Other digits have 2 options each

Yep

So 240 again

Same thing for 0s cuz the situation is symmetric

So 480

And then subtract the both case

So we need 2 0s and 2 1s

Which gives T{3} + T{2} + T{1} = 6 + 3 + 1 = 10

For possible positions

Actually wait

Might need to draw this out more

[warm tiger]

no? ig it should be simple

if first digit is 2 and then 6C2 * 4C2

[verbal oak]

Yeah it's T{3}+T{2}+T{1} + T{2}+T{1} + T{1)

= 15

[][][]3
etc
6

[]-[][]2
etc
3

[]--[][]1
1

-[][][]2
etc
3

-[]-[][]1
1

--[][][]1
1

Which is 6C2

And then we can rearrange within these places

So that's 4! = 24

But 2 of them are indistinguishable so divide by 2

And again because we have another indistinguishable pair

So 6 ways

[warm tiger]

thats 90

[verbal oak]

Which is 4C2

Mmhmm

[warm tiger]

so for this case, 480-90

right

[verbal oak]

Mmhmm

So 390

[warm tiger]

+372

[verbal oak]

Yup

[warm tiger]

762

YESSSS

[verbal oak]

Nice

[iron basin]

jee bro in the wilderness

[verbal oak]

We did it!

[warm tiger]

xD

[verbal oak]

Okay yeah it was a bit painful

[iron basin]

here is a solution if youre still willing to use inclusion exclusion

[verbal oak]

There's a lot of fiddly bits

That's exactly the same method btw

[warm tiger]

here also, the solution by @verbal oak is similar

[verbal oak]

It's exactly the same actually

[warm tiger]

yeah right

[verbal oak]

Just much more condensed

[iron basin]

well a summary then

[verbal oak]

I was doing it on the fly and explaining my thoughts

So it took longer

[warm tiger]

thanks a lot genuinely

[verbal oak]

I guess what I'm trying to say is that if you saw blurple's solution, it's a little intractable unless you understand already

[warm tiger]

are u in college ?

[iron basin]

almost

[verbal oak]

Do you agree or should I have just done that?

I was able to kinda stumble through it because I understood what I was doing, so even though I didn't have the solution ready to go straight off, I made it through to the end and got the answer

[iron basin]

yeah thats why i sent it after the conversation ended

[verbal oak]

👍

[warm tiger]

what does that mean ;-;

[verbal oak]

The formulae for C and P are faster for simple cases. I committed to not using it

If you're happy using it in simple cases then do so

[warm tiger]

in my level, C and P formula are widely used and everywhere almost

[verbal oak]

Yeah but like... do you ever feel unsure what to use when?

[warm tiger]

i just use the C one

and multiply by the factorial whenever i need to arrange as well

[verbal oak]

Cuz for complex cases, I think it can be difficult to know which to use to approach a problem

If you're thinking about what is indistinguishable and what isn't, you can just start the problem and see what happens

I didn't know exactly how that would turn out when I started but I still got there

Which I think is what makes it helpful

There will be questions where you already know the solution. Go for it

But if you don't, then what are you supposed to do?

You're just stuck now

Whereas if you think in this way you can make progress regardless

Do you agree or would you have preferred blurple's version from the start?

[warm tiger]

i guess the blurple version

coz thats easier to think for me? ig

[verbal oak]

😭

Alright

But like... how would you obtain that without being given it?

[iron basin]

im not sure if its that intuitive on first look though

[warm tiger]

your solution?

[iron basin]

kind of takes a few stumbles to land on inclusion exclusion

[warm tiger]

huh? didnt get what u mean

[iron basin]

and not like its the only way

[verbal oak]

Blurple's method is identical to mine, but much more condensed

He's kinda just hit it with the answer, cuz he knew what to do

[iron basin]

this tbh

[verbal oak]

Given you asked about it, I assume you didn't know what to do

So how would you arrive at the answer without having the solution provided?

[iron basin]

wdym

i didnt copy it off

[verbal oak]

That was posed to him

[iron basin]

oh

[verbal oak]

As I mentioned before, 2 ways to get better at this

1. Practice

2. Learn how to stumble!

Given you said you'd been practising for 2 years and still having issues, I thought maybe learning how to stumble would help

You said you preferred blurple's presentation of the method, fair enough. I'm just curious how you would get to that method on your own

Maybe he's gone rip

Nice job anyway Blurple

[iron basin]

yeah nice job blurple

[verbal oak]

😭

[iron basin]

but yeah practice

it takes a bit too many questions

[warm tiger]

arent both your methods like exactly the same, what you did was just counted 2 and 1 being at first position seperately and he counted them togther

[verbal oak]

They are

[warm tiger]

i would have counted them together as well, thats what i meant

[verbal oak]

But Blurple presented it as "Here is the method for this type of problem", whereas I presented it as "Let's imagine we don't know what we're doing. How can we get the answer anyway, by walking through it?"

[verbal oak]

He actually split it into 2 cases as well

For the ones anyway

And the others he did together

So I guess it's two lines extra on my part

[warm tiger]

for the ones, its necessary to split

[verbal oak]

Indeed

[warm tiger]

cuz the 0 makes trouble

[verbal oak]

Was that helpful? Do you think you can solve these in future?

[warm tiger]

yes a lot, thanks ig i wasted a too much of your time xD

[verbal oak]

And do you understand my suggested approach for future problems you're not sure on?

[warm tiger]

yess i got that as well, i'll try implementing that

[verbal oak]

I guess I can write a condensed version:

If two things can be exchanged without making a difference, they're indistinguishable. If they can't, they're different.

If they are indistinguishable, then they don't add extra options. If they aren't, then you need to consider the ways they can be rearranged as well

Oki doki

And it's fine

You didn't waste our time

We helped out of choice

You were a good student tbh

[warm tiger]

and your explanation was too good

thanks genuinely

[verbal oak]

Np

[warm tiger]

.close

[pearl pond]

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[pearl pond]

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[eternal shell]

i honestly have no idea what the question wants me to do, how do i get mean and standard deviation values from using the midpoints?

[eternal shell]

is there some formula im unaware of or am i js being slow

[pearl pond]

@eternal shell Has your question been resolved?

[pearl pond]

@eternal shell Has your question been resolved?

[livid goblet]

for the mean Im assuming because hes using normal distribution, the mean = median

and the median is midpoint

[eternal shell]

the mark scheme says m = 39.4 and s = 10.3 but i’m not sure where they come from

[dense jasper]

... counting boxes ugh

Typically, you'd find the entire shaded area

you know that this entire shaded area is 180 plants, you can then scale this to get the number of plants in each interval

then the weighted average formula gives you the mean

and the standard derivation formula is just the square root of the sample variance

@eternal shell

[eternal shell]

ic, thank you

i didn’t know about the weighted average formula i wasnt taught that

[dense jasper]

tbf it extends relatively easily from the regular average formula

the "weight" is effectively just how many times you're adding the thing

tldr the idea of weighted average is that repeated addition is multiplication

[eternal shell]

could i have also just tabulated the cumulative frequencies and used linear interpolation to find the median height?

or would i have to use that formula

[dense jasper]

depends on what they want ig

you can see if the numbers match tbf

[eternal shell]

as in would this even work? (checked, it gives a slightly different number but very close 👍)

[eternal shell]

aight

ty

.close

[pearl pond]

Channel closed

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