help-10

[keen plover]

0.2CH=4

CH=20

AK=CH+4=24

[urban quartz]

wait wait

alr I'm there

and then how do i get AB BC and AC

[keen plover]

cause BC=AC

[urban quartz]

yes

[keen plover]

so H is the midpoint of AB

[urban quartz]

yes

[keen plover]

which means

(0.5AB)^2+(CH)^2=CB^2

AB=1.2BC

therefore

(0.6BC)^2+(CH)^2=BC^2

therefore CH^2=BC^2-0.36BC^2=0.64BC^2

get it?

[urban quartz]

wait I'm processing it

[urban quartz]

i didn't get this

[keen plover]

AB=1.2BC

[urban quartz]

why 2CH?

[keen plover]

so you replace AB to 1.2BC

[urban quartz]

alr

[keen plover]

There's no 2CH I think

[urban quartz]

u wrote CH^2

[keen plover]

this is Pythagorean theorem

[keen plover]

yes

[urban quartz]

ohh yeah

i forgot that

[keen plover]

so you get it

[keen plover]

therefore CH=0.8BC

and as we did, CH=20

[urban quartz]

wait how did u come here

[keen plover]

OK I do it again

(0.5AB)^2+(CH)^2=CB^2

this is pht theorem

[urban quartz]

yes

[keen plover]

and you replace AB to 1.2BC

[urban quartz]

alr

and we divide it with 2

0.6

[keen plover]

right

(0.6BC)^2+(CH)^2=BC^2

it is this

then

(CH)^2=BC^2-(0.6BC)^2

[urban quartz]

so

[keen plover]

and add like term together

[urban quartz]

0.64BC²

[keen plover]

YES

and then root(CH^2)=CH

root(0.64BC²)=0.8BC

[urban quartz]

20= 0.8BC

OMGGG

[keen plover]

yes

then BC=25 and AB=30

[urban quartz]

bro i love u

[keen plover]

and C=25+25+30=80

[urban quartz]

like I've been stuck here for hours

[keen plover]

DM me if you have any ques bro

[urban quartz]

u're the goat bro

thanks

[obtuse pebble]

@urban quartz Has your question been resolved?

[obtuse pebble]

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[icy latch]

Need help with these qs

[obtuse pebble]

@icy latch Has your question been resolved?

[novel knoll]

With what in 9?

In 10 convert to z-scores

[icy latch]

A

[novel knoll]

P(-5<B-G<5)=P(B-G<5)-P(B-G<-5)

Can you solve it now?

[icy latch]

Yeah

[icy latch]

And for b I have done this

When I sub them in I get 0.292 which isn't right

[obtuse pebble]

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[slim summit]

For this integration question, would I include the area of days 5 and 8, or stop at day 5 and 8

[raven badger]

,rotate

[slim summit]

It’s asking for how much water is lost during 5 days, would I Include the fifth day’s area or not

My gut is telling me to stop at the fifth day and not include up until day 6

[obtuse pebble]

@slim summit Has your question been resolved?

[obtuse pebble]

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[sand agate]

hi

[obtuse pebble]

Please don't occupy multiple help channels.

[sand agate]

how do you get the equation of this?

i need to get the other factor

[thick gyro]

assuming that the curve is a quadratic

well for this, we know that its in the form ax^2 + bx + c = y

you can find out what c is

looking at the grpah

[sand agate]

so sub in lets say -2 in for x and 0 for y? @thick gyro

pllllllllzzzzzzzzzzzzzzzz help me

i beg

please help me

😢

i got the answer and its -x^2+2x+8

[minor pilot]

Did you google it 🤨

[sand agate]

no trial and error 😢

[minor pilot]

Hm

Here's the first step

When x=0, we get y=8 correct?

[minor pilot]

(We know this from the graph)

[sand agate]

okay

[minor pilot]

We're going to do some algebra

So far we've got the ax^2 + bx + c part down

[sand agate]

yeah

[minor pilot]

We can say c=8 from this so it saves time

However

We have 2 other points to use

[sand agate]

yeahhh

[minor pilot]

(-2,0)

(3,5)

Let's do the first one here

0 = a(-2)^2 + b(-2) + 8

Just subbing x in

[sand agate]

yeahhh

i did that

[minor pilot]

What did it simplify to?

[sand agate]

4a-2b+8=0

[minor pilot]

Good

[minor pilot]

We have another point

[sand agate]

3a+3b+8=0?

[minor pilot]

Close

[sand agate]

oh

+5

=5

i forgot

sey

[sand agate]

sry

[minor pilot]

Rmb it's a(3)^2

So it'd be 9a + 3b + 8 = 5

[sand agate]

im actually special

[minor pilot]

We have two equations, do you think we can find both variables with just two equations?

[sand agate]

yeah?

[minor pilot]

How would we do that?

[sand agate]

cancel bs outr

out

[minor pilot]

Show me

Please

[sand agate]

1 sec

[minor pilot]

Hm I think you should take another look at your working

You are on the right track though

[sand agate]

yeah i seen

im doing it again

a is -1

[minor pilot]

Correct

[sand agate]

b is -2

omg thank you so much

i was just making stupid mistakes

i owe you so much

[minor pilot]

No problem

[sand agate]

can i help you with anything?

[minor pilot]

Haha issok

Thanks though

[sand agate]

tysm

[obtuse pebble]

@sand agate Has your question been resolved?

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[strong crane]

Its probably easy and I tried so many answers but none of them worked , I have unlimited tries to get it right.

[timid silo]

write the values for the frequency to the left

[strong crane]

yh ik it goes up in 0.1

but when i times by the lengths

and put answers

it said im worng

wrong*

[timid silo]

Maybe it does not go up by 0.1

[strong crane]

been on this question for like 2hrs

so confused

i done 16 divided by 10

got 1.6

then counted and its 0.1 per box

[timid silo]

hmm

ok well if you call a box 5 lines

[mental solstice]

I dunno where you're getting 0.1, i count 16 boxes of height 5 small boxes in the second column

[timid silo]

yeah even then it doesn't work

Just write the values on the frequency axis

[strong crane]

ok

so do i write in 0.1

[timid silo]

i suggest you go 2 by 2

0,2...,16

[strong crane]

aren't you meant to divide frequency by the length

[timid silo]

I think this is just about reading the graph

[strong crane]

so for the first frequency what would I do

[timid silo]

Did you put some values on your frequency density axes?

[strong crane]

idk how to

is there a digital

whiteboard tool or something

[timid silo]

You can screenshot it and do it with paint

Well since you have the image you can open it with paint directly

[strong crane]

the task provided me with a video and the video does it a completely different way

its like an example

thats what the example shows how to do it

[timid silo]

Yeah you multiply the frequency density with the lenght to get the density

[strong crane]

im confused about how much to go up by

for the frequency density

i went up by 0.1 before

each time

[timid silo]

ok i see where the 1.6 comes from

[strong crane]

yes

[timid silo]

so more like go 0.2 by 0.2 from 0 to 1.6

[strong crane]

so instead of going by 0.1 per box i go by 0.2 per 2 boxes?

[timid silo]

forget about that

You know it's 0.1 per box

[strong crane]

yh so idk why im getting it wrong tbh

[timid silo]

Now let's look at our first value a

We see it takes less than 2 boxes and more than 1

We need to be more accurate

So instead of looking at boxes we need to look at lines

[strong crane]

0.16?

[timid silo]

Wait

You said a box is 0.1 right?

[strong crane]

i mean like per line

these lines

black ones

[timid silo]

Yeah

[strong crane]

per one is 0.1

[timid silo]

Yeah what about the very small lines in between

[strong crane]

0.02

[timid silo]

That's what i call a line

ok 0.02 per line

[strong crane]

yes

[timid silo]

so how many lines do we have for length between 0 and 50

[strong crane]

8

so

0.16

[timid silo]

ok then we multiply by the total length which is 50

Did you try that for all already?

[strong crane]

got it right tysm

yeah i tried it before

but

i got the last and first ones wrong

[timid silo]

Ho ok

[strong crane]

.close

[obtuse pebble]

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[dusk dagger]

hi?

[obtuse pebble]

Please don't occupy multiple help channels.

[dusk dagger]

?

[wise talon]

go back

.close

[obtuse pebble]

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[wise talon]

#help-21｜아리스킨충1

[obtuse pebble]

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[fierce glen]

I am doing Kumon math tutoring and have no idea how they went from one step to another. I have a picture of the question.

[timid silo]

Which step did you not understand?

[timid silo]

Sure, I’ll expand it, gimme a sec

[fierce glen]

thanks!

@timid silo u still here bro

[timid silo]

Yah sorry, I’m getting sqrt(3 +sqrt(5))

[fierce glen]

no problem

[timid silo]

Maybe someone else can answer

[fierce glen]

this shit is confusing 😭

[fierce glen]

should I ping helpers?

[timid silo]

Sure

Oh u have to do completing the square on the numerator

[fierce glen]

huh

[fierce glen]

can u show me?

[timid silo]

Sorry for the delay lol

[fierce glen]

no problem

[timid silo]

(a+b)^2 = a^2 +2ab + b^2

That’s the formula to use

To simplify the numerator

[fierce glen]

wow

tysm bro

i was so confused

[timid silo]

No problem

[fierce glen]

thankss

[obtuse pebble]

@fierce glen Has your question been resolved?

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[grand lily]

How would you arrange this in order to apply inverse Laplace?

[obtuse pebble]

@grand lily Has your question been resolved?

[obtuse pebble]

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[drifting nova]

can someone please help me with rolle's theorem for the problem:

x(x-3)^2 ; [0,3]

[drifting nova]

i took the derivative and got x=1 and x=3 but it says that the answer is only x =1

[gilded needle]

rolle's theorem guarantees a point in the open interval (0,3) where the derivative has the appropriate value

how was the question worded?

[rigid pine]

I'll include the theorem. I think the problem only wants the value of x where the theorem is applicable. It's not saying there is no others.

[drifting nova]

​Rolle's Theorem applies and the​ point(s) guaranteed to exist​ is/are x=

[drifting nova]

why isnt it applicable on x=3

[gilded needle]

rolle's theorem doesn't even require the function to be differentiable at the endpoints, just continuous

[rigid pine]

Notice the inequalities < are not <=.

[gilded needle]

so it certainly can't promise anything about the derivative at the endpoints

[drifting nova]

so ur saying that its because it was [] instead of ()

[rigid pine]

The theorems conclusion only guarantee f(x) = 0 for some x inside x = a and x = b but says nothing about it at the end points.

[drifting nova]

by endpoints are you talking about the end of the interval?

nvm

that makes sense

thanks guys

[rigid pine]

Yes.

[obtuse pebble]

@drifting nova Has your question been resolved?

[obtuse pebble]

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[supple stream]

how to do completing the square?

[supple stream]

don’t look at bottom half

what am I doing wrong I know I am

Standard form to vertex form

<@&286206848099549185>

[obtuse pebble]

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

[unreal musk]

,w expand 4(x-3/2)^2+3.75

[warm shale]

Results provided by WolframAlpha

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

Requested by chartbit#3171

[unreal musk]

Oh actually I see what you did

[unreal musk]

Should be multiplied by 4

As you basically get 4(x^2-3x+2.25-2.25)+6

[supple stream]

wait

I do 6-2.25 right

nvm

is that right

[obtuse pebble]

@supple stream Has your question been resolved?

[unreal musk]

6-4(2.25)

[unreal musk]

When you split this you keep the (x^2-3x+2.25) in and take the -2.25 out so you multiply that by 4

[supple stream]

Thx

[obtuse pebble]

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[tribal hazel]

never really understood exponent that are fractions.
i always assumed that it multiplying the numbers half.
so say 5^1/2 is the same as 5x2.5.
but so far its wrong. can someone explain how exponent fraction works?

[scarlet locust]

5^1/2 is the same as sqrt(5)

generally $a^\frac{1}{n}$ is the same as nth root of a

[warm shale]

Swas.py

[tribal hazel]

what?

[vast pendant]

The definition is $a^\frac{b}{c}=\sqrt[c]{a^b}$

[tribal hazel]

yeah i get that

[warm shale]

Waffle

[tribal hazel]

i just dont understand how do u multiply a number 1/2 times

[vast pendant]

You don't. It's just that it's a natural extension.
n/2 is half of n "addition-wise", but sqrt(n) is half of n "multiplication-wise"

[tribal hazel]

oh i see

its just a concept that doesnt really make sense

i guess it dont matter

.close

[obtuse pebble]

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[cerulean tartan]

how do i find the covariance

and how do i find the variance of this distribution

[obtuse pebble]

@cerulean tartan Has your question been resolved?

[frosty spoke]

write out the definition of covariance

[obtuse pebble]

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[open nimbus]

having trouble with part b

if it was just r = 6, ik the equation would be x^2 + y^2 = 36

but idk what to do with that cos in there

[lethal sand]

you can start with multiplying both sides by r

[supple granite]

part b as in graphing (x-3)^2+y^2=9 ?

[open nimbus]

ohhh wait i think i remember now

we eventually have to complete the square right

[lethal sand]

sure, if you want to reach to the same form as the answer

[open nimbus]

hm ok

so im at r^2 = r6cos(theta)

i cant remember what we do next tho 😦 @lethal sand

[lethal sand]

how do you represent x in terms of polar coords?

[open nimbus]

x = rcos(theta)

wait so r^2 = 6x? or am i thinking of it wrong

[lethal sand]

yes

and at the same time what would x^2+y^2 be represented as in polar coords?

[open nimbus]

thats just r^2 i think

[lethal sand]

yes

[open nimbus]

ohhhh

[lethal sand]

so what is the final equation in terms of x and y?

[open nimbus]

x^2 + y^2 = 6x...?

[lethal sand]

yes

[open nimbus]

and how do u complete the square again?

is it just the (number / 2) ^2

to find that middle value

if that makes any sense

[lethal sand]

you can look that up

[open nimbus]

ohhhhhhh yeah it is

ok

that makes a lot more sense now

thank you!!!!!!

.close

[obtuse pebble]

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[timid silo]

hi

[grizzled shore]

!status

[obtuse pebble]

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

[timid silo]

<@&286206848099549185>

[grizzled shore]

I’m literally here to help you

[timid silo]

oh

yep

I know how to solve it but I dont know what to do first

[grizzled shore]

Have you tried anything

[timid silo]

so whats the problem

[west imp]

try solving the brackets first

use the BODMAS rule

[timid silo]

Brackets

[grizzled shore]

I want to see what they have tried/know

Before pointing them in a direction

[timid silo]

operations

[west imp]

brackets over division, multiplication, addition, subtraction

[timid silo]

bodmas

[west imp]

yup thats what i said

[timid silo]

so I'll solve the brackets first?

[grizzled shore]

Have you tried anything yet

[west imp]

answer is 97

[timid silo]

solve whats in the brackets first (answer)

hi

[west imp]

yo

@timid silo should i explain how i got the answer?

[timid silo]

yeah..

[grizzled shore]

<@&268886789983436800>

[west imp]

alr so after solving the brackets

[timid silo]

@icy geyser ur the real deal

[west imp]

you will get 13 (13) - 12(6)

cause 6+7 = 13

and 12/2 = 6

so 13x13 is 169 and 12x6 = 72

[timid silo]

You just got your acc today

[west imp]

then 169-72

which is equal to 97

[timid silo]

oh

Now I get it

Thankyou

[west imp]

so in these questions always follow the bodmas rule

no probs

[timid silo]

I already a suber

you can close it now..

is it really you 😲

[west imp]

.close

[obtuse pebble]

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[timid silo]

!status

[obtuse pebble]

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

[west imp]

6

[obtuse pebble]

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[bleak eagle]

For a set to be a sub space of a known vector space, does the set have to contain the same zero vector that the known vector space has? Or can it have its own distinct zero vector that is not the zero vector for the vector space?

[ocean ingot]

I'm not sure if it's possible for a subset U of V to have a different zero vector than V in the first place, is it?

[royal basin]

it has to contain the parent space's zero vector yes

[bleak eagle]

I’m not sure lol probably not!

[bleak eagle]

Oh okay thank you!

[royal basin]

it is impossible for another vector to play the role of 0 given that the operations of addition and scaling by definition are inherited from the parent space

[ocean ingot]

Yeah that makes sense

I prefer calling it the identity element or additive identity personally

[bleak eagle]

Ahhh okay that makes sense. I didn’t think I had found an example of one, just wanted clarification because I was wondering about it!

[bleak eagle]

That’s fair, sounds much better!

Thank you for the help

.close

[obtuse pebble]

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[icy cape]

hello

I dont know where to start

please help

btw i might respond a bit late since i'm doing textbook stil

l

[obtuse pebble]

@icy cape Has your question been resolved?

[icy cape]

<@&286206848099549185>

[dense imp]

you can use 30/60/90 triangle on the right to get ratios between the x side and the triangle's legs

then the small leg of that 30/60/90 triangle is the same length as the 2nd leg of the 45/45/90 triangle (it's isosceles)

if you remember 30/60/90 triangle ratios, it is helpful since they come up a lot, or you can just use trig

[icy cape]

huh!!!!

what is the 30/60/90 ratio?

[dense imp]

the measures of the triangle's angles

[icy cape]

ik

but whats the relation

[dense imp]

the short side will be half the hypotenuse

x/2

you can just see this if you do sin(30) = y / x

y = sin(30) * x

y = x/2

[icy cape]

Oh ok

[icy cape]

how would it work for the A leg

[dense imp]

and the other leg is xsqrt(3)/2

but since in your problem you have the hypotenuse as x, just divide the other ones by 2

[icy cape]

wait I dont get this

can you please draw

it

pls

[dense imp]

draw what?

[icy cape]

what u are saying

[dense imp]

i just sent a diagram

just divide everything in that diagram by 2 to match with your problem

[icy cape]

ok so we have established that small side of right triangle = 1/2 of x

how about the base/

?

x square root 3

damn

(x square root 3) / 2?

[dense imp]

yep

[icy cape]

ok

so then 0.5x

= the smallers sides

smaller*

of left triangle

right?

[dense imp]

yep

so how would you get length of AB which is the original question?

[icy cape]

basically add 0.5x and x square root of 3/2

[dense imp]

exactly

$\frac{x}{2}+\frac{x\sqrt{3}}{2}$

[warm shale]

Sooshon

[icy cape]

Yessir!

wait how to add x and x square root 3

since they are diff

can u do 2x square root 3?

nvm u cant

[dense imp]

no

[icy cape]

since the inside of root

isnt same

[dense imp]

you can't really add them

[icy cape]

true that

[dense imp]

you can optionally factor out the x, which is what the desired answer did

[icy cape]

anyway how do we get square root 3 + 1

/2

from that

[icy cape]

oh

[dense imp]

well they have same denominator so write as

[icy cape]

OK

[dense imp]

$\frac{x+x\sqrt{3}}{2}$

[warm shale]

Sooshon

[dense imp]

and now just factor out the x:

[icy cape]

wait to factor would that be x ( 1 + square root 3)/2

[dense imp]

$\frac{x(1+\sqrt{3}}{2}$

[warm shale]

Sooshon

[icy cape]

damn isnt it 1/2 tho

NVM!

I forgor

[dense imp]

we just wrote with common denominator already

[timid silo]

You forgor

[dense imp]

so when factoring, just working with the numerator, nothing going on with denominator

[icy cape]

Yessir! we up 💯

[icy cape]

Alright!

damn never heard of 30 60 90 ratio

ngl

thanks tho

🙂

goodbye

[dense imp]

it just follows automatically from sin/ cos of 30./60 degrees

you can just use that if you dont happen to remember

[icy cape]

ok

.close

[obtuse pebble]

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[grizzled shore]

!status

[obtuse pebble]

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

[grizzled shore]

!show

[obtuse pebble]

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

[prime summit]

i tried finding the width in terms of x

then use w x w x l

to find volume

but i can't seem to eliminate the denominator of 2

for the width

and since it is an expression I can't multiply by a factor to eliminate the 2

[dense imp]

how come you are working with the assumption that the other 2 sides of the box are equal?

[prime summit]

?
there are two variable the width and length
where did I do that

[dense imp]

as far as i can tell, rectangular box just means it has right angles, there would be a width, a length, and a height. i don't see anything to indicate two of those are equal

[prime summit]

ah i see

use depth for the 3rd dimension

ah okay

so how to you go about it then?

[dense imp]

hm not sure, i was trying something, but any way to get y in terms of x from this?:
96 + x^2 + y^2 +xy -18x -18y = 0

[prime summit]

No I don't think we do that

[dense imp]

i think i got something, it's quite a lot of algebra

yes I think i have an equation that's just in terms of V and x : )

no idea if it's supposed to be the right way to approach it, seems kinda obscure

[prime summit]

Maybe work backwards from the answer

Might help

[dense imp]

so let x, y, z be the sides, we can write the equations for side length, area and volume:
4x + 4y + 4z = 72
2xy+2xz+2zy = 192
xyz = V
but lets simplify the first two so we have:

x + y + z = 18
xy + xz + zy = 96
xyz = V

good so far?

[prime summit]

Mhm

I got that

[dense imp]

write z in terms of x, y: z = 18 - x - y

[prime summit]

I see

[dense imp]

now plug into the area equation and we get:
xy + x(18 - x - y) + y(18 - x - y) = 96

this will simplify to:
96+x^2 + y^2 +xy -18x - 18y = 0 if you work out the algebra

good so far?

[prime summit]

Yeah

[dense imp]

actually lemme rearrange terms for that last equation (youll see why later):
(y^2 + xy - 18y) + 96 + x^2 - 18x = 0
just grouped the ones with y in front together

now take the V = xyz and substitute the (18 - x - y) for z:
V = xy(18 - x - y) = 18xy - x^2y -xy^2

divide both sides by x and we get:
V/x = 18y - xy - y^2 or i'll move a negative sign over to the other side and get:
-V/x = y^2 + xy - 18y
does the right side of that look familiar?

[prime summit]

Some what

Yes

[dense imp]

it's exactly the grouped y terms in the previous equation : )

(y^2 + xy - 18y) + 96 + x^2 - 18x = 0

so we take -V/x and substitute for that parenthesis

-V/x + 96 + x^2 - 18 x = 0

[prime summit]

Oh I see

[dense imp]

V/x = 96+x^2 - 18x, now just multiply through by x

V = x^3 -18x^2 +96x

[prime summit]

Mhm
Wow thx a bunch

I didn't expect it to be so convuluted

[dense imp]

that was a hard one, i didnt spot the idea of dividiving that volume expression by x for a while, not sure if there is an easier way

fun problem though, i enjoyed it so thanks : )

[prime summit]

Np

I was able to make an equation a+b=12

and tried the quoitient rule

[obtuse pebble]

@prime summit Has your question been resolved?

[obtuse pebble]

@prime summit Has your question been resolved?

[prime summit]

<@&286206848099549185>

[obtuse pebble]

@prime summit Has your question been resolved?

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[nova nebula]

I’m lost does anyone know this?

[delicate junco]

graph theory?
are you generally familiar with the terms edges and vertices?

otherwise you may have to 'try'

[obtuse pebble]

@nova nebula Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[uncut girder]

At a trainstation, a new train arrives every 5 minutes. Every 7 minutes a passanger comes to the trainstation to catch a train.
How long on average does a passanger need to wait for a train?

[obtuse pebble]

@uncut girder Has your question been resolved?

[obtuse pebble]

@uncut girder Has your question been resolved?

[uncut girder]

<@&286206848099549185>

[uncut girder]

<@&286206848099549185>

[uncut girder]

<@&286206848099549185> at least tell me if the ping works xD

[timid silo]

Hello

what do you need help with

[dusk night]

Yes, the ping works, to solve this I would choose a sample size (ex: 10) and I am not sure from here however, I would just calculate the times manually although as mentioned below, matrices will probably work quite well.

[uncut girder]

hello guys

[quaint violet]

hii

[timid silo]

if she can begin and end in the same position

by passing only once through each line

it means there is a cycle

cycle only exists if the number of points that are conected to a pair number of points is pair or has 2 that are impair

is the opposite of odd

pair*

i dont know how to say it in english

i can solve it for you in vc

@uncut girder

[uncut girder]

What is vc?

[timid silo]

voice chat

[uncut girder]

thanks for taking the time

Maybe I need clarify something which will be easier in text

[timid silo]

u can call me through discord

i can stream photoshop

and solve it for u

[uncut girder]

nice thank you, I might take you up on that offer

[timid silo]

nvm i was answering another quesiton

i just saw what your quesiton is

[timid silo]

.

At a trainstation, a new train arrives every 5 minutes. Every 7 minutes a passanger comes to the trainstation to catch a train.
How long on average does a passanger need to wait for a train?

lets start from time 0

5 minutes a train arrives and goes

passenger gets in at 7 min

another train arrives at 10 minutes

[uncut girder]

wait 1 sec

the example I gave I would solve like you are attempting now

going one by one until it comes full circle

then deviding to get average

I'm looking more for a formula where I can just input different numbers to get the result without too much work

[tawny sail]

Trial and error isn’t very useful here. Calculate the probabilities

[uncut girder]

yeah because what if it never goes full circle?

or we have numbers with a comma

[timid silo]

it will

if its a series

sequence

[tawny sail]

It does at the LCM of 5 and 7

[timid silo]

its a nice start

[uncut girder]

it is but for some example you will need to do very many steps no?

[tawny sail]

I can’t picture a general solution, but can see how to solve this one

[uncut girder]

how would you solve?

[tawny sail]

I would write out all the possibilities and take the average, it isn’t clean but it works.

[timid silo]

first one waits 3

second one 1

third one 4

fourth one 2 minutes

fifth one waits 0

and then it repeats

3+1+4+2+0

[tawny sail]

As long as you are getting questions with clean numbers its okay, if they are going to give you like 5 minutes 3 seconds and 12 minutes 7 seconds for example, that would be very different

[timid silo]

10

10/5

2 minutes in average

@uncut girder

[uncut girder]

Thanks cece

[timid silo]

you are welcome

[uncut girder]

That is how I solved it too but it's not what I'm looking for

I posted a simple scenario to see if there is an easy one step solution

[timid silo]

ur looking the average time

someone has to wait

there is an easier way

u know every 5 min a new train comes in

5x, with x being how many times u wait

will give u the min

for example to find the first one we did

7x1=7

but to have a 7 u need 2 times 5

10-7=3

so u have to use modules

did u do modules in school or not yet @uncut girder

[uncut girder]

Ok I need to clarify the real purpose of my question

[timid silo]

5x-7a>0

[tawny sail]

I understand - you are looking for a general solution/formula for it with any two times

[uncut girder]

YES

[timid silo]

5x-7a>0

first equation

need to find a second one

because u have 2 variables

[tawny sail]

Trying to work it out, but I’m laying in bed at 3am and haven’t done this kind of probability for a while

[uncut girder]

hehe no worries

thanks for helping anyway

[timid silo]

x=2 and a=1 verifies it

5x2-7x1= 3

3>0

3 is a time to wait

so 2 trains for the first person

[verbal swan]

Can someone help me with division please

[timid silo]

need to find the second equation now

[tawny sail]

Read #❓how-to-get-help

[tawny sail]

I have an idea, don’t know if it works in all cases. It isn’t exactly what you want but it appears to work for the couple examples I tried

It’s easier than doing everything but still slow/tedious

Basically you add up all the waiting “options” and divide by the number of options

Wait sorry it doesn’t work in all cases I found an exception

Might be able to find an answer for you tomorrow mate, probably don’t have time tonight sorry. There are a few answers on the web for similar scenarios (average time waiting if you appear at a random time, for example). Maybe you can get an idea for this from reading those. I’ll see if it’s still unsolved tomorrow, good luck

[uncut girder]

I'll write what I've come up with so far as a generic formula:
a=7min (passanger)
b=5min (train)

b-((b//a)*a)
2b-((2b//a)*a)
3b-((3b//a)*a)
........

the // means divide and round down to whole number

and then continue this process until cycle is complete, and then divide by the number of results

Does that make sense a little bit?

[timid silo]

just do

5x-7a>0

all x and a that valid this equation

will give u solution for time waited

now u need a second equation in terms of x and a

to find x = a blabla

then u just a random a u find its x

for example

x=3

a=2

other way around

5x3-7x2= 15-14=1

1 is one of the waiting times

[timid silo]

.

so for x=3 the a is 2

for x = 5 the a is 3

5x5-7x3= 25-21= 4

4 is one of the waiting times

[uncut girder]

but then we have to try all the numbers

[timid silo]

for x = 6 a = 4

[timid silo]

unless u find a second equations that depends on x and a

then u can just chose a random a and get the x for it

[uncut girder]

can you say again what is x and a?

[timid silo]

x is number of trains

a is number of passengers

a = 2

we talk about second passenger

what train will he take ?

he will take 3d train

so x = 3

so lets part from the principle

that every passenger

will take a train

but not every train will take a passenger

train 1 takes no one

passenger 1 gets inside train 2

passenger 2 gets inside train 3

passenger 3 gets inside train 5

passenger 4 gets inside train 6

passenger 5 gets inside train 7

you need an equation

something like

7a = blabla

where a is the passenger

so passenger a will be taken in train x

[uncut girder]

yes

[timid silo]

what you need is

5x-7a>0

a is the passenger

x the train

for passenger 4

5x-7*4>0

5x-28>0

[uncut girder]

yep

but then x can be a lot of differnt numbers

[timid silo]

x>28/5

[timid silo]

we are looking for integer ones

and the first one

right after

so x>28/5

what is 28/5

[uncut girder]

yes that's what i solve in my way by using //

[timid silo]

5.6

next integer after 5.6 is 6

[uncut girder]

#help-10 message

[timid silo]

so passenger 4 takes train 6

nb-((nb//a)a)

with n an integer

[uncut girder]

but how do we mathematically express to continue calculation until a fuill circle is reached. And is a full circle always reached, even with numbers like every 4.399 minutes?

My end goal is to build a calculator where I can just plug in the two numbers

[timid silo]

what 2 numbers

?

[uncut girder]

and a program spits out average and standard deviation

[timid silo]

u will only be able to plug in 1

[uncut girder]

7 min 5 min

[timid silo]

no

thats a sequence then

thats ur calculator

a sequence

thats how i solved it

[uncut girder]

well if 1 of those two numbers changes our result will change

[uncut girder]

so I have to give both numbers

[timid silo]

u just need a calculator

that uses 1 number

the 7

min

plug in 3

7x3=21

25-21

4

7x5=35

35-35=0

passenger 5

waits 0 min

7x2=14

15-14

1

passenger 2 waits 1 min

7x1=7

10-7=3

passenger 1 waits 3 min

[uncut girder]

if I made a calculator that calculates waiting time based on passanger and train frequency, I need to plug in two numbers, that's what I meant

[timid silo]

oh ye

the thing remaining is how to plug in the train

if we talk in modulo 7

[uncut girder]

My formula above uses both, train and passanger frequency and it works I think

but how would I tell a program to repeat the process until coming full circle

[timid silo]

well full circle is just 5x-7a=0

x=7a/5

u need 7a/5 to be an integer

so a has to be a multiple of both 5 and 7

35

is 7 and 5

7 multiplied by what to get 35

5

a = 5

[uncut girder]

what about a train comes every 11.523465 minutes and passanger arrive every 23.4975 minutes

[timid silo]

for passenger 5

[uncut girder]

would it come full circle?

[timid silo]

eventually yes

everything can come in a full circle

periods will be different tho

[uncut girder]

but it will take so much time to calculate

[timid silo]

yes

thats why it wasnt the question

[uncut girder]

I picked a simple example to see if it can be solved in one step or fewer steps than my method

[timid silo]

oh

so ur simple example was

7 and 5

im pretty sure u can get around it with modulos

[uncut girder]

yes basically

[timid silo]

do u know what modulos are

?

6modulo5 is 1

[uncut girder]

no probably not

[timid silo]

7modulo5 is 2

8modulo5 is 3

9modulo5 is 4

[uncut girder]

is modulo just minus xD

[timid silo]

no

i mean yes

it means

u select any number

for example 5

and everytime i hit this number

i start counting back from 9

0

or multiple of 5

for example

16 is

1

because 5x3=15

16-15=1

and 15 is 0

10 is 0

5 is 0

20 is 0

its just how clocks work

60 min = 1 hour

you dont say 80 min

u say 1 hour and 20 min

u dont say 120 min

u say 2 hours

[uncut girder]

Ok I'm 50% following

[timid silo]

u dont say 61 seconds

u say 1 minute and 1 second

u dont say 60 seconds

u say 1 minute

so this is how u count cycles

with modulos

[uncut girder]

ok

[timid silo]

lets say one passenger every 6 min

and one train every 5 min

30

is where the cycle resets

passenger 1 gets in train 2 and waits 4 minutes

passenger 2 gets in train 3 and waits 3 minutes

passenger 3 gets in train 4 and waits 2 minutes

passenger 4 gets in train 5 and waits 1 minute

[uncut girder]

yeah eventually you reapeat the pattern

[timid silo]

passenger 5 gets in train 6 and waits 0 minutes

so to find when u wait 0 minutes

u just multiply them

5x6=30

in this case 30 is where the cycle repeats

in ur case

7x5=35

35 is where teh cycle repeats

35 is one cycle

35*a with a being an integer in R that is positive

will be the number of cycles

70 cycle repeats

105 cycle repeats

every 35 u have a repetition of a cycle

this is a good start

now from this we would need to make a formula

where we plug in minutes

of train

and minutes of passenger

and it gives us the average

[uncut girder]

what if the numbers are 0 point something

[timid silo]

same principle

multiply

[uncut girder]

0.3 minutes for passanger 0.6 for train

if u multiply

u get lower number

[timid silo]

u get

0.18

0.7 per passenger and 0.5 per train

is smarter choice

so we work on our example

0.35

same principle u see

[uncut girder]

no but my brain is a little fried now

[timid silo]

0.7

0.5x2= 1

1-0.7=0.3

0.3 minutes

it was 3 before

we just multiply everything by 10 if they are perfect decimals

make the problem

then divide the result

by 10

at 3.5

it will repeat

sec ill do it

[uncut girder]

at 3.5 ok

but not at 0.35

[timid silo]

yes

so u know

u know 3 is

0.3x10

and 7 is 0.7x10

u need a number that when u multiply it by 10

it gives 35

3.5

10x=35

x=35/10

x=3.5

3.5 is the new cycle

[uncut girder]

yeah but your first proposal was just to multiply the two numbers

[timid silo]

so if they are integers

yes

[uncut girder]

to get the cycle

[timid silo]

well u make exceptions

in programs

if numbers are

integers

multiply them

if they are not integers

solve

[uncut girder]

yeah i understand