help-0

[amber plover]

the other question though

this one

is this maybe just an error that my teacher did or somehing?

thanks for being so patient with me sorry for the trouble

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[median oar]

What sigma is this referring to

I’m up to here and I’m trying to write the lower bound of a k% CI interval

Would it be σ_p2?

Or is it (σ_p2 ^2/n)

[median oar]

In which case I’d have 1.645 * (σ_p2 ^2/n)/sqrt(n)

[lone heart]

@median oar Has your question been resolved?

[lone heart]

@median oar Has your question been resolved?

[lone heart]

@median oar Has your question been resolved?

[lone heart]

@median oar Has your question been resolved?

[alpine sable]

E

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[alpine sable]

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[alpine sable]

unsure how to do please help

[median oar]

?>?????

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[median oar]

kwla

[median oar]

eyo

mine

[median oar]

send help

[normal crater]

ill give you this

[median oar]

thank you

[normal crater]

its not over.

[median oar]

[untold forge]

kwla

[gusty gorge]

standard deviation of the iid random variables X_1, X_2, ..., X_n that get averaged to make X bar

[median oar]

so no the SD of the distribution itself (p2)

so i'd want since the µ_p2 hat is distributed normally to ~ N(µ_p2, σ²_p2/n)

then the σ/√n

i'd want √(σ²_p2/n)/√n

[keen plinth]

that doesn't look right

[median oar]

which one isn't right

[keen plinth]

this

you're dividing by sqrt(n) twice

[median oar]

well

the formula here wants σ/√n right?

im trying to figure out what σ i put in there

do i put the SD of µ_p2 hat?

do i put the SD of p2?

[keen plinth]

i would think p2

[median oar]

hmm

[keen plinth]

because /sqrt(n) gives you the sd of mu hat

and that would make sense

[median oar]

it does

hmmm

i suppose we are comparing to see if µ_p2 hat ± some interval contains µ_p2

so we should assume as our H₀ that it is so?

thus use √Var(p2) for our sigma here?

[keen plinth]

yeah

[median oar]

hmmmmm

🤔

it does make sense

[gusty gorge]

I feel like using only the sample mean won't result in a very powerful test

you're trying to test the null hypothesis that both the mean and the standard deviation are the given values

but just using the sample mean probably means you're giving up a lot of power

[keen plinth]

i have a feeling frosst isn't the one making up these tests

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[karmic spoke]

unsure

[chrome salmon]

Unsure of what?

[karmic spoke]

how to do this

i dont know where to start

😭😭

i tried subbing

but idk

[chrome salmon]

p = r^q , r = q^p so log_(q) p = q log_(q) r = q log_(q) q^p = qp

[lone heart]

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.

[chrome salmon]

I saw

[karmic spoke]

hmm

ohhhh

i reckon i understand

ty

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[waxen plinth]

Hi. If I take calc bc next year, I wouldn't need to take ab right? I wouldn't GO through another class of ab, I'd just go through ab concepts + bc concepts

[alpine sable]

Depends on how you do on the AP and how your university handles that sort of thing

It's pretty much impossible to give you an accurate answer that will take into account all possibilities

Calc ab is basically a full year course version of a 1 semester university course usually called calc 1

So basically it's at half the speed of the university course

Calc bc in theory covers both the university courses calc I and calc II

So in theory it goes at the same pace as the university courses

And if you do well on the AP and go to a university that accepts those credits it's possible that you won't even have to take calc I or even calc II when you go to university

This is the best way I can explain it

The most likely outcome though is that you will take bc and get to skip calc I at college

So you won't ever really have to "go over" ab again instead you'd go over the second half of bc in college

[lone heart]

@waxen plinth Has your question been resolved?

[median oar]

I’m just using it as an example to walk through all the ideas I’ve been taught thus far

[lone heart]

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[dull peak]

hi

[dull peak]

anyone that knows about powerpoint

its a quick thing

I cant find any servers related to pp

pls

[vale wigeon]

whats your issue

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@dull peak Has your question been resolved?

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[remote tangle]

Not sure where to start here. Previously we have been asked to find MLE's for continuous distributions (like gamma) where either alpha or beta is defined, so its solved through just plugging in values, but here I am not sure where to start

[last tendon]

Just write down the likelihood function for starters and drag a bunch of x_i's around the computation.

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[ionic sedge]

can someohe check if i did this correctly

[merry depot]

I don't see anything wrong so far.

[ionic sedge]

my work:

@alpine sable sorry haha

[slow sentinel]

Seems correct to me

[alpine sable]

Is this some form of epidemic or do teachers actually teach this horrible way of writing pi?

[slow sentinel]

The only thing wrong i see are those gaint holes in your book

[ionic sedge]

lol

[alpine sable]

How'd it taste?

[slow sentinel]

Jokes aside you could simply write it as $\frac{1}{10pi}$
Other than that it's all good

[ocean seal]

MeDumb

[ionic sedge]

oh really? even though the question is asking about 10 feet

it simplifies ahh

[slow sentinel]

Yeah

[ionic sedge]

thanks

[slow sentinel]

No problemo

[ionic sedge]

it would be 1/10pi feet per minute still right?

[slow sentinel]

Yess

[ionic sedge]

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[weak geyser]

please help me with this

[vale wigeon]

what is troubling you?

[weak geyser]

i need the answer

[vale wigeon]

we will not just give you the answer.

read #rules, #info, #❓how-to-get-help

[subtle birch]

!status

[lone heart]

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

[weak geyser]

1

[vale wigeon]

do you know how exponents work

[weak geyser]

yeah

[vale wigeon]

ok

[weak geyser]

please simplify the sum

[vale wigeon]

and do you know how fractions work and how to manipulate them

[weak geyser]

yeah

[vale wigeon]

again, no, we are NOT going to just give you the answer or the solution

this isn't the place for that.

[weak geyser]

okay help me up

[vale wigeon]

your fraction can be rewritten as $\frac{4 \cdot 6}{8} \cdot \frac{10^7 \cdot 10^{-5}}{10^4}$

[ocean seal]

Ann

[vale wigeon]

do you see how to proceed from here?

[weak geyser]

pls show me the steps

[vale wigeon]

there are shorter ways of saying "no"

are you able to simplify the fraction on the left, 4*6/8?

[weak geyser]

ok

i got it

next step

[vale wigeon]

... what did you get upon simplifying 4*6/8?

[weak geyser]

3

[vale wigeon]

okay

and... what about the other fraction?

[lone heart]

@weak geyser Has your question been resolved?

[weak geyser]

that's something tougher

[lone heart]

@weak geyser Has your question been resolved?

[weak geyser]

i mean the 2nd equation is tougher to solve

[lone heart]

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[rigid hound]

P = a - (b/c)

a= 15 to nearest 5
b = 8.5 to one decimal place
c = 5 to 1 significant figure

find lower bound of P

[limpid turret]

So $4.5\le c < 5.5$?

[ocean seal]

SWR

[rigid hound]

mhm?

for c = 4.5?

[chrome salmon]

4.5 for lower bound

[rigid hound]

right

what abt the other values?

[chrome salmon]

The question doesn't need it

[rigid hound]

?

whats the lower bound of p then

[limpid turret]

What have you tried, @rigid hound?

[rigid hound]

hm

i got

a = 12.5
b = 8.55
c = 4.5

wanted to check if its right

with these i got p = 10.6

[limpid turret]

That's what I got

[rigid hound]

u sure its right?

[limpid turret]

With the information you've given me, I'm very confident.

But I wouldn't say "p = "

It's the lower bound of p

[rigid hound]

ight thanks a lot

[rigid hound]

oh yeah sry abt that lol

thanks again

.close

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[frail grove]

Does the binomial theorem work with polynomials?

[frail grove]

Chat-GPT :

I thought it only works with 2 terms

[chrome salmon]

It's called multinomial then

[last tendon]

You can also just take a polynomial like 1 + p + p^2 and set x = 1 + p and y = p^2 or however you want to group them.

[frail grove]

Then apply binomial formula to (1+p)?

[last tendon]

Nothing is stopping you, but you should probably be asking yourself why are you doing something and if it will actually lead to a new form that is helpful in whatever you are trying to accomplish

[frail grove]

But it will take ages to develop the whole expression using ur method

I thought there's a direct formula just like the binomial theorem

[chrome salmon]

There is

Just told you about multinomial expansion

[frail grove]

Never heard of it

[ocean seal]

fäf

[frail grove]

it works with 3 terms only?

[chrome salmon]

This one

You get generalize this

[frail grove]

yeah i just found the general theorem

thnx man

.close

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[tall path]

Can this be broken down further or did I make a mistake along the way?

or would x not be there since there is a common factor of -3x?

[alpine sable]

you havent taken limits yet

[tall path]

Oh so do i need to plug in the limit first? 😅

[alpine sable]

well the limit is gone from the last line

so it is a fair assumption to have to evaluate the limit in that step

[tall path]

Oh okay gotcha. My apologies

So it would just be x

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[alpine sable]

hi

[alpine sable]

how to post image?

please help i been stuck with it for a month

I am stuck in trying to convert the a(r+1)^2 - 1+arar+2

<@&286206848099549185>

[lone heart]

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.

[keen sigil]

hello

[alpine sable]

hello

Hey can ayone help

Hello Helpers

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[quasi thicket]

how would you solve this partial fraction $\frac{3x-1}{(x^3+2)(3-x)}$

[ocean seal]

Dark_123

[quasi thicket]

i understand how you get $\frac{Ax^2+Bx+C}{(x^3+2)} + \frac{D}{(3-x)}$

[ocean seal]

Dark_123

[quasi thicket]

i have solved a = 8/29, b = 24/29, c = - 15/29 and d = 8/29, but i dont get how you substitute back in

[proven leaf]

$\frac{8x^2+24x-15}{29(x^3+2)}-\frac{8}{29(3-x)}$

put the 29s in the denom

[ocean seal]

XxMrFancyu2xX

[quasi thicket]

i get how you can put the 29s in the denominator if there werent other variables, but there are?

wouldnt it be this for the left fraction $\frac{(8/29)x^2 + (24/29)x - (15/29)}{(x^3 + 2)}$

[ocean seal]

Dark_123

[quasi thicket]

i understand the right one

[proven leaf]

each term has a common $\frac{1}{29}$ because $\frac{\frac{a}{b}}{c}=\frac{a}{bc}$ in your case $b=29$ and so I moved it to the denom, if it wasn't common you'd just have fractional coefficients which is fine

[ocean seal]

XxMrFancyu2xX

[quasi thicket]

oh ok

i was wondering how you simplify the common denominators from the coefficient to the denominator of the bigger fraction

i get it now

thanks

.close

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[elder crane]

im still new to vectors, do i just plot these points as is

arrow pointing in direction of terminal point

or do i need to do something first

<@&286206848099549185>

[lone heart]

@elder crane Has your question been resolved?

[elder crane]

<@&286206848099549185>

[slow sentinel]

Nope, you're correct.
Plot the points
Draw an arrow from intial point to terminal point
And you're done ✅

[elder crane]

ok sweet, and then what is meant by position vector

would that be <(3 - (-1)), (-7 - (-4))>

[slow sentinel]

A position vector just means that the intial point is the origin

[elder crane]

the question says express u as a position vector

[slow sentinel]

Could you show me the whole question

[elder crane]

heres my graph

[slow sentinel]

I think you're just supposed to draw it from the origin to the final point for position vector

[elder crane]

i dont understand, i dont think this question wants me to graph anything

that was just for q4

[slow sentinel]

https://youtu.be/MGEz57SVn_w
This should help

[elder crane]

.close

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[paper stratus]

how

[paper stratus]

what did they do in the circled step?

i dont get that step

[alpine sable]

polynomial long division

[paper stratus]

ohh

but A + Bx+C would also be fine right?

.close

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[peak haven]

helpo

[peak haven]

find the length of the curve defined by the parametric equations x=t^3,y=3t^2/2 with 0<=t<=rad3

[pine kettle]

arc length

[peak haven]

The length

[pine kettle]

arc length

,w arc length integral

[pine kettle]

yes

😊

[peak haven]

can you please solve

i dont know how

[pine kettle]

no you can 😊

[peak haven]

i really cant

i tried

[pine kettle]

is it calculator

?

[peak haven]

?

i didnt understand

[pine kettle]

can you use a calculator

[peak haven]

yea

[pine kettle]

ok

find dx/dt and dy/dt

then plug in it

it in

[peak haven]

how

i need steps

[pine kettle]

using differentiation rules

with respect to t

[peak haven]

i dont know how

[pine kettle]

what

you are in calculuis

[peak haven]

yea

calc 2

[pine kettle]

you guys didnt do power rule

?

[tacit arch]

did you just like forget everything in calc 1?

,tex .diff rules

[ocean seal]

riemann

[peak haven]

no

ik them

[tacit arch]

use power rule to find dx/dt

[peak haven]

but arent i supposed to use the trig sub

[tacit arch]

and constant multiple + power rules to find dy/dt

[peak haven]

ik how to find dx/dt

[tacit arch]

[peak haven]

the step after idk

[pine kettle]

ok

so what is dx/dt

[peak haven]

3t^2

[pine kettle]

ok

find dy/dt

[peak haven]

3t

[pine kettle]

plug in

to

arclength integral

and dont forget to use the correct limits

[peak haven]

What do I do after this

I'm stuck here

[pine kettle]

just plug it into your calculator

?

[peak haven]

i have never do tha t

hoiw do i do it

[warped lintel]

you can take the $9t^2$ from under the square root to make the integrand $3t\sqrt{9t^2+1}$ and then use $u$-substitution

[ocean seal]

jacob

[peak haven]

tyy

[warped lintel]

ofc

[peak haven]

wait

what dfo we use as u

?

[tacit arch]

try anything

[peak haven]

not working

[tacit arch]

what'd you try

[peak haven]

i used 9t^2 +1 as u

[tacit arch]

that's right

[peak haven]

i get 18 du

[tacit arch]

.

[peak haven]

and?

[tacit arch]

what is this supposed to be

18 du = ?

[peak haven]

du/18=dt

[tacit arch]

nope

use the table

[peak haven]

can u guy just give me the answer its 2 am and i want to sleep

[tacit arch]

then go to sleep

and no that's not what this server is for

[peak haven]

if i take 9t^2+1 as u

the du should be du=18dt

right

?

[tacit arch]

that line specifically

n = 2

[peak haven]

18t

du=18tdt

then?

there is no place for du to be in the equation

[tacit arch]

yes there is

,tex .int rules

[ocean seal]

riemann

[tacit arch]

18 = 3 times what?

[peak haven]

6

will it be du/18 ,then we put it in the equation and get 1/6

right?

[tacit arch]

what is "it"

[peak haven]

1/18

[tacit arch]

1/18 will be du / 18?

[peak haven]

yea

[tacit arch]

put "it" in the equation

what is "it"

[peak haven]

du/18=t dt

[tacit arch]

correct

[peak haven]

so the t will disappear and the remaining wil be rad u du

and 1/6 as a constant

1/6 inegral = rad u du

?

will the answer be 3.11

[tacit arch]

Just show your full work

[peak haven]

i go t it right

thank you

sorry for bothering you

[royal ocean]

@peak haven are you done with this? Can you close it please using .close

[peak haven]

.close

[lone heart]

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[left shore]

Find the area of triangle

[left shore]

Can anyone explain why (4, 1) is = (0, 0) and not

(3, 6)

[alpine sable]

those are cords

[tacit arch]

If you're not gonna help, don't bother sending messages

[alpine sable]

?

[left shore]

Wym cords?

Sorry the question is

Find the area of the triangle who's vertices are:

[alpine sable]

those numbers you gave. Are they the co-ordinates of the different points of the triange?

[tacit arch]

Show parts a and b too

[left shore]

If that's what you mean yes

Only part C

[tacit arch]

Why do you think it's saying (4,1) is (0,0)

[left shore]

Determines by which one is the highest in the X and Y graphs

Maybe am wrong

[tacit arch]

Well 4>3

But 1<6

So there's preference on the second coordinate it looks like

[left shore]

Hmm or I could use the distance formula

[tacit arch]

Maybe try drawing it

[left shore]

Have you heard of the Heron’s formula?

Trying to use these into the heroin formula

[tacit arch]

That should work

[left shore]

Does the S represents 1?

[tacit arch]

S is the 1/2 the perimeter

[tacit arch]

Or 1/2 the sum of these 3

[left shore]

So it would be 1/2(1/2 - A) etc

[lone heart]

@left shore Has your question been resolved?

[tacit arch]

What is "it"

[left shore]

The formula

[tacit arch]

What is A

[tacit arch]

Either show your steps or explain what you did

[tacit arch]

There's no A here

[lone heart]

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[shut pewter]

I already made a line segment where p is the midpoint of a and b. I'm just confused on the P is a moving point on the number axis, "and the corresponding number is x".
Is x like point p at the number line? or is it like the minutes passed.

[trim pike]

What is pi? is it 3.14159265359 or 3.13159265369?

[lone heart]

Please read #❓how-to-get-help

[trim pike]

ok bot

[serene junco]

This seems like a typo?

It seems like they're saying points A, B and P have coordinates 1, 5, and x respectively

but then they said if P is the midpoint of AB, x=2, which is not right

but 2 is half the distance of AB, maybe that's what they meant

[shut pewter]

Ah its -1

[serene junco]

oh

[shut pewter]

A coordinates is -1

[serene junco]

in that case yeah x=2 is right

[shut pewter]

Yes

I think at x =2 , what it's saying that when p is at 2, it's the middle point of a and b

[serene junco]

yeah, kinda awkward wording

[shut pewter]

Yeah it's from a chinese test that I'm teaching

[serene junco]

oh okay, I suspect when they called P a "moving" point, they meant its position is variable

[shut pewter]

I think applying the velocities first at -1 and 5 first would do the trick? and then just go with it whenever if it was divided by 2 would become 2? or something like that?

[serene junco]

I think you can make a linear equation for each point

for example point A starts at x=-1 and has velocity -1

[shut pewter]

Ah yes

[serene junco]

so after t minutes, A = -1 - t

[shut pewter]

Then see where they intersect?

[serene junco]

Well we can make expressions for PA and PB in terms of t

and then just substitute them in the given equation

and solve for t

Like

A = -1 - t
P = 2 - 3t

PA = |P-A|

[shut pewter]

Ah then (A-B)/2 = 2?

Since we want to find t when P = 2

[serene junco]

I believe P = 2 at t=0

unless I'm misunderstanding the problem

I interpreted 2 as the starting position for P

And since the points are moving with different velocities, P will not be the midpoint of AB anymore

Besides, if P were still the midpoint, the equation PA = 1/3 PB would not have any solutions

[shut pewter]

I think that P at 2 isn't the starting position for me. It's when A at -1 and B = 5

Then I just thought that maybe doing your suggested equation earlier. Since when P =2, it would become the midpoint of AB then A - B all over 2 = 2.

But we already have the equation for A and b, the equation you suggested

But I'm not entirely sure

[serene junco]

Well we need to know the initial position of P

or if not, we need some other information

what is condition (1) ?

[shut pewter]

I think it's the statements above

[serene junco]

Well

hm

if we're not expected to take 2 as the starting coordinate of P, then it seems like there's not enough information

I don't think they meant that P is always just the midpoint

because then the problem is trivial

In that problem, there would be no solutions

because PA = 1/3 PB implies that P is not the midpoint

[shut pewter]

Hey what about this

3 = PA/PB

Since we have the equation for that

we have 3 = |-1-t|/|5-3t|

[serene junco]

well -1-t is just A

PA is the distance between P and A

but we can't know that without knowing the coordinate of P

[shut pewter]

Ey wait

Is PA Point P - point A?

or is it Point A?

[serene junco]

It usually means the distance between P and A

|P - A|

or |A - P|

which are equivalent

[shut pewter]

How about we just make this

2-2t

[shut pewter]

Then use this but with 2-2t?

like -1-t - (2-2t)

then also at the bottom

[serene junco]

2-3t

but yeah

[shut pewter]

Wait why 2-3t?

I thought that the velocity for 2 is 2

and the velocity for 5 is 3

[serene junco]

"A, B, P move left with velocities 1, 2, and 3"

seems to imply the velocity for P is 3

I suppose we're working with something that might not be a perfect translation though

[shut pewter]

Ah YOu're right

[serene junco]

I would put those in absolute values too

like |(-1-t) - (2-3t)|

[shut pewter]

Wait I already have a class

[lone heart]

@shut pewter Has your question been resolved?

[lone heart]

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[slender quiver]

!status

[lone heart]

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

[edgy sequoia]

1

i don’t know where to begin

!status

[lone heart]

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

[limpid aurora]

so in PSR PS/PR is equal to the long leg over the hypotenuse

in what triangle is PR the long leg?

[edgy sequoia]

prs

?

[limpid aurora]

uh PR is the hypotenuse in triangle PRS

[edgy sequoia]

what’s a hypotenuse

[limpid aurora]

the longest side of a right triangle

[edgy sequoia]

oh

so hypotenuse can’t be long leg?

is it PS then?

[lone heart]

@edgy sequoia Has your question been resolved?

[edgy sequoia]

got this too

[slender quiver]

a^2 + b^2 = c^2

[edgy sequoia]

<@&286206848099549185>

[stark path]

Try to draw pictures of the triangles side by side to get the similarity among those

btw PQR & PRS
PQR & RSQ . these are similar

Do u know the rules that makes two triangles simillar?

[edgy sequoia]

equal angles

[lone heart]

@edgy sequoia Has your question been resolved?

[lone heart]

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[alpine sable]

can some one help me with geometry pls 😭

[lone heart]

@alpine sable Has your question been resolved?

[hushed locust]

when you want to ask a question just post it

[lone heart]

@alpine sable Has your question been resolved?

[tacit arch]

Come back when you're ready to post the question

.close

[lone heart]

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[rain flicker]

Does anyone know the fastest way to multiply accurately any 4 digit number

[rain flicker]

Is that a doubt

[drowsy meadow]

.close

[rain flicker]

<@&286206848099549185>, please solve this

[lone heart]

@rain flicker Has your question been resolved?

[lone heart]

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[vestal obsidian]

The x^6 is giving me trouble

<@&286206848099549185>

[lone heart]

@vestal obsidian Has your question been resolved?

[vestal obsidian]

nvm its fine I figured it out

[lone heart]

@vestal obsidian Has your question been resolved?

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[barren portal]

When we solve for z, we get both + and - right?

how do we know which one to choose?

[slender marten]

We do. The positive square root will be the part of the sphere above the xy-plane and the negative square root will be the part of the sphere below the xy-plane.

[barren portal]

right

This is what he did after

I was wondering, if -sqrt was taken shouldn't it become the bottom function all of a sudden?

[barren portal]

later to find the volume as, volume of top function - volume of below function

he choose the +sqrt part but wondering what If i choose -sqrt

[slender marten]

We are between the sphere and the cone z = r. Remember this cone is above the xy-plane so the region bounded is bounded above the sphere and below by the cone.

Hence r <= z <= sqrt(2 - r^2).

[barren portal]

oh so tho algebraically we get +- sqrt

only +sqrt count for the cone

[slender marten]

Well z = sqrt(x^2 + y^2) could only possibly be the positive one.

z^2 = x^2 + y^2 would represent both z = -sqrt(x^2 + y^2) and z = +sqrt(x^2 + y^2) but since you were explicitly given z = +sqrt(x^2 + y^2) you only have the part above the xy-plane.

[barren portal]

so the cone intersects the sphere only above xy plane

[slender marten]

Yes.

[barren portal]

my point of confusion was I thought the surface intersects at two different points

[slender marten]

This specific cone anyway.

The cone and the sphere intersect at infinitely many points.

[barren portal]

level curves i mean

[slender marten]

Not too sure what that means.

[barren portal]

the equation that we obtain when we equate both functions

[slender marten]

Oh.

Well if you use z^2 = r^2 = 2 - r^2, you obtain r^2 = 1. This represents the cylinder x^2 + y^2 = 1 which does intersect the sphere in two places. You just have to use a bit of thought since we used z^2 = r^2, this includes both cones when in reality we only have one.

[barren portal]

I get it now

I just saw the region that we are finding the volume

it's the region trapped inside both cone and sphere

[lone heart]

@barren portal Has your question been resolved?

[lone heart]

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[vale wigeon]

no

we in fact don't know at all how likely you are to get hired if you score under 80 on the test

it might be that you get rejected outright (meaning this probability would be 0) or that there's simply a lower pass rate for those for any other reason (it might be something like 0.5 compared to the 0.65 for those who do better on the test)

the probability that an applicant who scores 80+ DOESN'T get hired

no

P(not hired | score ≥ 80)

why did you send the same message twice?

oh, no, you replaced "and" with "who"

that didn't really make the sentence make any more sense.

this might be a language-barrier issue

0.35 is the probability that a good-scoring applicant doesn't get hired.

[lone heart]

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[stray stirrup]

What will be the equation for this ?

[pseudo ice]

Note that OD is also automatically the same as OC and CE (why?)

[stray stirrup]

Idk

Odc Is isosceles right

[subtle birch]

Finding ODC (in terms of x) will solve the problem

[subtle birch]

OD=OC because they are radii so we have two isosceles triangles DOC and COE

[subtle birch]

Try to find DCO in terms of x
(using exterior angle property)

[stray stirrup]

Ok

Ohhh i get it now

Ty

It’s 23 right

Cuz 111= 180-3x

.close

[lone heart]

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[slim thunder]

Coucou there is someone who studies Mathematics in French!

[lone heart]

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[alpine sable]

i dont understand last question

[rocky grove]

Hmm

So they gave your the ratios

[subtle birch]

find b in terms of a using 1st proportionality

[rocky grove]

a:b = 5:6

And b:c = 3:8

Can you find the factors for b in both ratios?

[alpine sable]

i dont understand it although i do see that they have given ratios a:b and b:c

and then they ask is to find a : b : c

[covert ferry]

They want you to find the ratio for all three but it has to match

I have 5 a's for every 6 b's, and 3 b's for every 8 c's. Put it altogether: for 5 a's, how many c's do I have?

[alpine sable]

a = 5

b = 3

okay i need a minute to understand

[covert ferry]

ok

[alpine sable]

can you like tell me

they ask to find a : b : c

now i know that a = 5

and its b = 3

[alpine sable]

is it 6/3

[rocky grove]

Hmm

What are the factors of b in the first ratio?

[alpine sable]

in the first ratio it is a : b and that is 5:6

b is 6

[rocky grove]

Okay

What are the factors of 6?

[alpine sable]

they are 1,2,3 and 6

[rocky grove]

Alright

Now what is b in the second ratio?

[alpine sable]

its three

is it needed to know factors of six

i had no idea about it before

its three in the second ratio

for b = 3 - my bad

[solid turtle]

How do you make a romb with d1-6cm and d2-4cm ( i dont know what d1 or d2 is)

[rocky grove]

b = 3 yes

Now what do you need to multiply 3 with to make it equal 6?

[alpine sable]

3 * 2

do you have to know factors before you do these type of questions?

[rocky grove]

Yeah. Ratios is all about factors

:)

[alpine sable]

this sounds very very new to mw

ohh

.close

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[flat turtle]

How do I find d?

[flat turtle]

Y is a random variable, and the density function is given

[vale wigeon]

integral of f(y) over the entire real line should be 1

[flat turtle]

sorry so do I integrate dy^2, and d(6-y^)^2

and then add it to equal 1?

[median oar]

For its domains

Yeah

[flat turtle]

so definite integral with 0,1 for dy^2

[flat turtle]

and 1,2 for d(6-y)^2 ?

[median oar]

You also need it the cdf to be monotonic I believe

[flat turtle]

oh sry could you explain what that means?

[vale wigeon]

eh?

[median oar]

Plus the definite integral of the second part on its domain

[vale wigeon]

$\int_{\bR} f(y) \dd{y} = \int_0^1 d \cdot y^2 \dd{y} + \int_1^2 d \cdot (6-y)^2 \dd{y}$

[ocean seal]

Ann

[median oar]

As you move from -inf to inf your cdf must increase

Your cdf can’t decrease in that sense

As in, you can’t have negative probabilities for any part of your pdf

[flat turtle]

that should = 1 right, then solve for d

[median oar]

The pdf curve is never negative

[vale wigeon]

yes

[flat turtle]

uh so what should I do afterwords

[vale wigeon]

yeah but we are given the pdf not the cdf

[median oar]

Which means it’s integral can’t have negative slope anywhere

[vale wigeon]

so moot point

[vale wigeon]

once you find d you are done, no?

unless youre asked for sth else

[median oar]

Couldn’t you find a d that violates that?

[flat turtle]

oh sorry I didnt write it in the question, to find the distribution function\

[median oar]

You could probably solve for 1 region to be 1 more than the other

[vale wigeon]

can you maybe like

show the problem exactly as it was stated

[median oar]

Then your “total” would still be 1 but scuff the cdf

[vale wigeon]

just to be clear as to what you're given and wahat you're asked to find

[vale wigeon]

that violates WHAT

i think youre making shit up that doesnt exist

[median oar]

That your cdf can’t decrease

Ok whatever

[flat turtle]

.close

[lone heart]

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[pliant estuary]

Could somebody please check if my solution of this simultaneous equation of diferential equations is correct?

[pliant estuary]

,rotate 270

[pliant estuary]

.close

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[vital whale]

this is a really simple question but why does d/dx of 1/4-x =
1/(4-x)^2?

[vital whale]

I thought it was -1/(4-x)^2 because:
d/dx (1 /4-x)
= d/dx ((4-x)^-1)
=-(4-x)^-2
= -1/(4-x)^2

[golden canyon]

Chain rule

The derivative of 4 - x is -1

[chrome salmon]

Yes jelle

I agree with you for sure

That seems to be right

[vital whale]

thank you

.close

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[lone heart]

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[visual wolf]

hello

[visual wolf]

okay so basically I got 75/100 on a Uni test, that's worth 20% of the module

how many marks is that out of the entire 100%?

would it be 20% of 75?

so I've got 15 marks out of the available 100 currently?

[vale wigeon]

presumably yes

[visual wolf]

ty

[limpid turret]

Or 75% of 20

[wind cloak]

Metamath

[limpid turret]

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[obsidian crane]

I , need help

[obsidian crane]

sec

So I have this

I know the conditions. The first condition is true (0, 0, 0, 0) is in H

The second condition is to prove (or check?) whether it is closed wrt addition

So, for-example I got two vectors

a = (x, y, z, u) belonging to R4
b = (e, f, g, h) belonging to R4

Need a + b

so it will be come

a + b = (x+e, y+f, z+g, u+h)

Now we need to evaluate and check if this new a + b is inside the vector space, that is, its >= 0

so we put it in xyzu, it gives:

Question
(x+e)(y+f)(z+g)(u+h) >=0

is it true? if yes , how to prove it

[vale wigeon]

do you want intuition or do you want formalism

if both, which one first

[obsidian crane]

formalism, i need to prove it

in exam

both i guess

[vale wigeon]

(2, -1, -1, 1) and (0, 2, 0, 0) both belong to H but their sum does not

[obsidian crane]

oh

right

but how do i prove it using mathematic notation etc that stuff

[vale wigeon]

(2, -1, -1, 1) ∈ H because 2 * (-1) * (-1) * 1 = 2 ≥ 0
(0, 2, 0, 0) ∈ H because 0 * 2 * 0 * 0 = 0 ≥ 0
(2, -1, -1, 1) + (0, 2, 0, 0) = (2, 1, -1, 1)
(2, 1, -1, 1) ∉ H because 2 * 1 * (-1) * 1 = -2 < 0

[obsidian crane]

hmm right

thx

[vale wigeon]

do you still want the intuition behind this

[obsidian crane]

Its okay this was sufficient, thanks.

🙂

.close

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[wispy wave]

could someone explain how this became 1/cos(x) - 2/cos^3(x)

[gray isle]

apply
tan^2(x) + 1 = sec^2(x)

[proven quarry]

Not yet

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@wispy wave Has your question been resolved?

[wispy wave]

does this need to be simplified?

on mathways it shows the non simplified version. Is this good enough?

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[violet flare]

yoooo

[violet flare]

i have $\int_{0}^{\infty}e^{-x}sin^{4}xdx$

[ocean seal]

Zouni

[violet flare]

don't need to solve it just need to check for convergence

what method should I use?

[ruby current]

comparison?

[violet flare]

so i'd compare $e^{-x} and sin^{4}x$

[ocean seal]

Zouni

[violet flare]

?

[ruby current]

no i mean comparison with e^(-x)

since sin^4(x) <= 1

[violet flare]

ah i see

could you elaborate further

[ruby current]

this implies e^(-x)sin^4(x) <= e^(-x)

[violet flare]

so pretty much squeeze theorem with integrals

[ruby current]

yes

[violet flare]

nice

thanks

[ruby current]

https://tutorial.math.lamar.edu/classes/calcII/improperintegralscomptest.aspx

[violet flare]

.close

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[violet flare]

.reopen

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✅

[violet flare]

@ruby current

so i'd get $\lim_{x->\inft}sin^{4}x<=e^{-x}$

[ruby current]

no

[violet flare]

with a limit

[ocean seal]

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

[ruby current]

$\sin^4(x) \leq 1 \implies e^{-x}\sin^4(x) \leq e^{-x}$

[ocean seal]

tushar

[ruby current]

just multiplying both sides of the first inequality by e^(-x), which is always nonnegative

[violet flare]

and then i just take the limit?

[ruby current]

no

nothing to do with limits

see the link i posted above

[ocean seal]

tushar

tushar

tushar

[ruby current]

by using part 1 of the image

[ocean seal]

tushar

[ruby current]

you should evaluate this expression and verify that it converges

[violet flare]

$\int_{0}^{\infty}\frac{1}{e^{x}} = 1$

[ocean seal]

Zouni

[ruby current]

that's right

[violet flare]

so the integral converges

ok very cool, thank you

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[forest sail]

Let there be a uniform circular cake of diameter x. How would you remove a slice that is one fifth of the cake without making a pentagon? You can only use a scale and a compass.

[stark yew]

Hi Meta, what have you tried?

[molten pivot]

what?

removing a fifth of a circle doesn't make a pentagon

it just makes 4/5ths of a circle

[analog falcon]

they probably meant a sector that is one fifth of the circle

and if so, its a good question ig

[forest sail]

I do hope it's not impossible

[molten pivot]

still doesn't make any sense

[forest sail]

Is there more than one way to cut one fifth of a cake using a ruler and a scale?

[stark yew]

A ruler and a scale refer to the same thing. I believe you meant a ruler and compass. The answer is: yes.

[forest sail]

I meant that if you connect all the chords it would essentially make a pentagon

[forest sail]

Can you please explain all of them

[molten pivot]

i guess you could cut two tenths if you'd like.

[stark yew]

I can think of two ways off the top of my head.

1. Construct five 72 degree angles inside the circle and remove one of them.
2. Adjust your compass to half the radius and draw another circle. Remove the portion between the inner radius and the outer radius (why adjusting to half the radius works requires further elaboration).

[forest sail]

I see

I'll try that thanks

[stark yew]

Please close the channel with .close if your question has been resolved. @forest sail

[forest sail]

.close

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[supple knot]

.claim

i need help with this

[supple knot]

its in arabic

tthey want the number that should be in "?"

[heady pollen]

what are the numbers?

also what is the rule here? a triangle with some numbers around it? do you add them up or what

[supple knot]

the first on the left

5,3,10

[supple knot]

the second one at the left: 6 up, right down: 9 , middle:3 , left down:2

[supple knot]

no rule stated

critical thinking

[supple knot]

3rd one above b{

3 at the top, middle is 4, right down: 6 , left down:2
}

[supple knot]

its an exam for teachers university

[heady pollen]

so there is 4 triangles where the middle is unknown and the 3 around it are known? so weird

[supple knot]

yes

they want to know which number should be in "?"

[heady pollen]

i cant see an obvious rule for this. strange task

[supple knot]

mmm idk really thats why am asking

think and see what u get

https://tenor.com/view/panic-irene-irenepanic-gif-20856890

i got the first one

lucky

6

[supple knot]

3

[supple knot]

🤔

.close

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[rotund crater]

hekp

[rotund crater]

how i get

[ruby current]

https://tutorial.math.lamar.edu/classes/calci/volumewithrings.aspx

[heady pollen]

for a circle we get the area pi r^2
consider a big circle with Radius R and we cut out a smaller circle of readius r
then the area will be piR^2-pir^2=pi(R^2-r^2)
here if we draw a diagram, we see that the big circle has radius 1 and the small circle has radius x=f(y)
therefore we get
Area=pi(1-x^2)

then to get the volume of a disk, we multiply the area of it times it's height, which is dy

[rotund crater]

show me the step i cant see

[heady pollen]

which one?

[rotund crater]

.

[rotund crater]

.

[heady pollen]

we have this

which we can "turn over" to get this

now if we look at the 3d figure, we can look at a slice of this figure

it will look like this