help-28

[raw ruin]

2³ = 8?

[velvet sedge]

once you have all the cases of 2 tails you can just mirror it to get the cases of 2 heads

in total, yes, 8

because you have 2 possiblities for each of the 3 trials

so 2 * 2 * 2=8

[raw ruin]

So 6/8?

3/4?

[velvet sedge]

yep :)

[raw ruin]

So basically

2³ = 8

2 - heads or tails of a coin
³ - cuz of the toss counts
6 - the total number of heads or tails combined?

[velvet sedge]

6 comes from the number of cases out of the 3 coin flip trials that have 2 tails or 2 heads

[raw ruin]

So in simple terms, 6 comes from the possibility of getting heads or tails in the 3 coin flip trials?

[velvet sedge]

6 represents the total possible ways to get 2 heads when flipping a coin 3 times + the total possible ways to get 2 tails when flipping a coin 3 times

that's 3+3=6

[raw ruin]

Ohhhhhhhh

[raw ruin]

So is this correct? Except for 6?

@velvet sedge I still don't get the idea, how about for number 2?

Atleast a tail or 3 heads

So P(1 tail, 3 heads)

[velvet sedge]

think of them as two separate events

not occuring together

[velvet sedge]

think of or as 'adding' the probabilties together if you can't consider them together

$P(A \cup B)=P(A)+P(B)-P(A \cup B)$

[glossy valve]

MSC2020 55N31 (Moosey)

[velvet sedge]

$P(A \text{ or } B)=P(A)+P(B)-P(A \text{ and } B)$

[glossy valve]

MSC2020 55N31 (Moosey)

[velvet sedge]

but P(A and B) in this case doesn't make sense, because you can't have 3 heads and 1 tail in 3 coin flips

so we say its 0

[raw ruin]

So P(0, 0)?

[velvet sedge]

no...

let A=atleast 1 tail in 3 coin flips and B=3 heads in 3 coin flips

whats the P(A)

whats the P(B)

[raw ruin]

I really don't get it

P(A) + P(B) - P(A and B)

P(A) = 1/3
P(B) = 1

[velvet sedge]

no..

P(event)=(how many ways out of the 8 total ways does this event occur)/8

so out of the 8 total possiblities of flipping a coin 3 times, how many of those have 3 heads

1 right?

[raw ruin]

Yeah

[velvet sedge]

so P(B)=1/8

P(A) is also wrong, can you see what it should be

[raw ruin]

Idk

[velvet sedge]

it's also easier to think of this as 1-P(not A)

what's the opposite of atleast 1 tail

[raw ruin]

Atleast 1 heads?

[velvet sedge]

no...

atleast 1 tail means there is always atleast 1 tail, there's always a tail.

what is the opposite of having a tail in your 3 coin flips

[raw ruin]

Getting all heads.???

[velvet sedge]

yes!

[raw ruin]

But why???

[velvet sedge]

well think about what atleast 1 tail means

THH has atleast 1 tails, TTT has atleast 1 tail

[raw ruin]

Ohhhhhh

[velvet sedge]

the only case which doesnt have atleast 1 tail is

HHH

[raw ruin]

I see

[velvet sedge]

so you can see that P(A)+P(B) is...

7/8+1/8 right?

[raw ruin]

Where did u get 7?

[velvet sedge]

1-1/8

or you can think about just getting rid of the HHH case out of the 8 total cases

[raw ruin]

Oh so basically
P( atleast 1 tail) = 7/8
P(3 heads) = 1/8?

7/8 + 1/8 = 8/8
= 1

@velvet sedge

[velvet sedge]

yes

or you can think about it more simply like P(atleast 1 tail)=1-P(3 heads)

then 1-P(3 heads)+P(3 heads)=1

[raw ruin]

The reason why P(B) = 1/8 is because you can only get 1 HHH out of the coin flips?

[velvet sedge]

yes

[raw ruin]

3.)

P(A) = 7/8 + P(B) = 2/8?

I'm not sure about B

@velvet sedge

[velvet sedge]

well this one you HAVE to account for P(A and B)

[raw ruin]

Why is that?

[velvet sedge]

P(A or B)=P(A)+P(B)-P(A and B)

because both can occur at the same time

you can have a head and 2 tails

[raw ruin]

Is P(B) = 2/8 correct?

[velvet sedge]

what is B? the probability of having 2 tails in 3 coin flips?

[raw ruin]

Yeahh

[velvet sedge]

well we already calculated that earlier didn't w

e

in part 1

[raw ruin]

Yeah but we calculated both heads and tails

So P(B) = 3/8?

@velvet sedge

[velvet sedge]

yes

[raw ruin]

Is this correct?

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[raw girder]

not sure how the hint helps

[full marsh]

Well what's one of of computing it?

[raw girder]

wdym?

[full marsh]

holy shit I was so confused, like "why is that thing raised to the fourth power"

[raw girder]

yeah that confused me too

[full marsh]

now I see it's just a reference to somewhere else

Well

<v, Av> = ..?

What can you simplify here

[raw girder]

<A*v, v>

[full marsh]

Yes that's one way

[raw girder]

= <-Av, v>

[full marsh]

Yes

Go on

[raw girder]

what now

[full marsh]

-Av = ?

[raw girder]

hm

[full marsh]

Remember what v is?

An eigenvector of eigenvalue lambda

[raw girder]

ohh

yeah

so 0

[full marsh]

0??

[raw girder]

wait

Av = lambda * v

[full marsh]

Yes

[raw girder]

so we get <-lambda * v, v>?

[full marsh]

Yes

Continue...

[raw girder]

= (-lambda)* <v, v>

[full marsh]

Yes

And now do the same but without switching to <A*v, v> at the start

brb

[raw girder]

ohh

<v, Av> = <v, lambda*v> = lambda * <v, v>

[full marsh]

Yes

And now compare the two results

[raw girder]

so lambda<v, v> = (-lambda)* <v, v>

[full marsh]

Wait

You forgot the conjugate

[raw girder]

which part?

[full marsh]

Here

Like

[raw girder]

lambda<v, v>

my lecture notes

[full marsh]

$\inner{v}{Av} = \inner{v}{\lambda v} \neq \lambda\inner{v}{v}$

[glossy valve]

RedstonePlayz09

[full marsh]

At least not over C

[raw girder]

i think my lecturer defined it in the opposite way

[full marsh]

Oh

Then you'll need conjugate on the first argument

[full marsh]

That's weird why would they define it this way lmao

Yeah ok sure

$$\inner{v}{Av} = \inner{v}{\lambda v} = \lambda\inner{v}{v}$$

\medskip
And also:

\medskip
$$\inner{v}{Av} = \inner{A^*v}{v} = \inner{-Av}{v} = \inner{-\lambda v}{v} = \conj{-\lambda}\inner{v}{v} = -\conj{\lambda}\inner{v}{v}$$

[glossy valve]

RedstonePlayz09

[full marsh]

well my latex isnt very good

but its ok ig

[raw girder]

yep

makes sense

tysm

[full marsh]

np

Also note that

<v, v> is nonzero

since v is nonzero (it's an eigenvector)

[raw girder]

this part makes sense i think

[raw girder]

yeah

[full marsh]

So you got anymore questions?

[full forum]

@raw girder Has your question been resolved?

[raw girder]

nope, thanks for ur help 🙏

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[surreal lynx]

I can help! 😁

What do you need help with?

I can't either lol

mb

[vernal lava]

give an example

rectangles

90degrees

what is the question here?

here, every intersection is perpendicular

theer are 3 right angles in every corner

in 3 different faces of the cuboid

[surreal lynx]

they

they're parallelograms

[vernal lava]

so total of 24 right angles (8 corners of 3 right angles each)

done?

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@torn jolt Has your question been resolved?

[vernal lava]

what is the actual question in the book?(or the source)

make a grid of the 3d shape

like what you get by cutting open a paper cube box

then combine the results

deal with 1 face at one time

like?

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✅

[vernal lava]

you would need to use trigonometry

think of the faces as rectangles, so the corner has a right angle

even if you rotate along the edge, the angle will remain the same, as long as the adjacent edges are moved

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[sinful totem]

hey can anyone help me with this

[sinful totem]

for the first one i did 22190/2 so 11,095 is the expected number of males

for the second one i did 2-1 so 1 degree of freedom

[timid heart]

hi

[sinful totem]

hey

[timid heart]

whats your prob?

[sinful totem]

i posted the picture

[timid heart]

ok

[sinful totem]

would it be 0.05>p>0.01?

[timid heart]

i think its.

[timid heart]

ye prob

[sinful totem]

huh

ah okay

what about the next one

[timid heart]

you need to calculate

do it with a calculator

[sinful totem]

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[lean mortar]

If I have a ratio of 1:74, and the numerator is 17,952, what will the denominator be?

[sudden summit]

I believe you start by setting up a proportion 1/74 = 17952/x and solve for x

[lean mortar]

Okay thanks

But like I know the equation for a ratio

I'm asking for the answer

[sudden summit]

I don't think they allow giving the answers directly ill check the rules

[lean mortar]

Oh for real? its fine mb

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[royal kiln]

Is the answear for i) wrong or hiw is it 0?

[full marsh]

It's definitely not 0

[royal kiln]

Thx

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[warm cargo]

did i get these right?

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Please don't occupy multiple help channels.

@warm cargo Has your question been resolved?

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@warm cargo Has your question been resolved?

[warm cargo]

<@&286206848099549185>

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@warm cargo Has your question been resolved?

[severe coral]

I agree with your answers for 1 and 4, but for 2, you should check the roots of that again. If you plug x=8 back in, it doesn't work out

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[ruby wolf]

Uhh

10.51

:D

[ruby wolf]

oh srry uh

i actually dont even know how to set up the equation

😭

<@&286206848099549185>

[warm perch]

20 cherry 30 grape

what’s the ratio

[ruby wolf]

2/3

breh

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[warm perch]

💀

would’ve been 20 - c - x / 30 - g = 20 - c / 30 + 5x

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[meager garnet]

circle b and circle c is 1/8 of original circle

so combined 1/4

the overall area of the triangle is 18

@craggy tapir

are you gonna help me?

[craggy tapir]

im looking at the problem

[craggy tapir]

can you calculate the radius of the big circle?

hint: draw AH perpendicular to BC

the radius will be AH

[meager garnet]

what is AH?

theres no point h

[craggy tapir]

a point you construct it yourself

[meager garnet]

so what is the length of AH

3*sqrt2?

@craggy tapir

bruh

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@meager garnet Has your question been resolved?

[graceful yoke]

alright, the radii are easy to find, are you done calculating that?

angle B and angle C is 45 degrees

area of a sector of a circle is given by $\frac{\theta}{2}r^2$

[glossy valve]

zelphanine

[graceful yoke]

where r is the radius

find the areas of all the sectors

then find the area of the full triangle

the difference should give you your answer

@meager garnet

[meager garnet]

can you confirm that this is the radius of circle A?

@graceful yoke

[graceful yoke]

no it will be sqrt(3)

draw a line from point A to the midpoint of BC

call the new point D

triangle ADC will be a right angled isoceles triangle

apply pythogoreas theorem on it

ah sorry my bad i did a calculation mistake

yes it is 3*sqrt(2)

sorry about that i forgot to square the 6 while applying pythagoreas

now to find the radius of the smaller circle, observe that the sum of the radius of the big circle and the radius of the small circle is just 6

[meager garnet]

oh

ok

is the answer 1.44?

[graceful yoke]

yes

i got 1.4379

big circle area is 14.1371

and the two smaller circles make up 2.4255

and the whole triangle is 18

[meager garnet]

so rounded to the nearest hundreth is 1.44

alr thanks

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[cunning radish]

can someone help me with composite functions

[cunning radish]

f(x)=4x+1 and g(x)=x^3+1 and h(x)=2x+1/3

find the value of gf(0)

help plz

[delicate veldt]

g(f(x) = (4x+1)^3 +1

[cunning radish]

how thoo

i dont understand

it

the whole lesson

[delicate veldt]

to find g(f(0) first evaluate the value of f(0)

[cunning radish]

0 is x right

[delicate veldt]

yeah

f(0) = 4(0)+1 [ replace x with 0]

therefore f(0) = 1

now replace f(0) with 1 in g(f(0) = g(1)

[cunning radish]

oh so 4x0+1

[delicate veldt]

On replacing x with 1 in g(f(0) = g(1) = (1)^3+1 = 2

therefore g(f(0)) = 2

[cunning radish]

i didnt get the last part

[delicate veldt]

from which part exactly ?

[cunning radish]

after when we found out that g(1)

after that

i didnt

get it

..

[delicate veldt]

substitute 1 in place of x of g(x) function

[cunning radish]

yes so g(1)

[delicate veldt]

g(x) = x^3+1 , put 1 in place of x

= 1^3+1 = 2

[cunning radish]

where do u get x^3 from

[delicate veldt]

g(x) = x^3+1

[cunning radish]

then why did we use 4x+1

[nova topaz]

When you see g.f(0)

You start with the last function listed and work towards the first one

[cunning radish]

is it bc g is in the first one and f is in ther 2nd one

but what if it was gh(0) then would we use the one that contains h in it like this one

[nova topaz]

Exacty

[cunning radish]

OHH

[nova topaz]

You start with H first because it’s the second one listed

Then when you find the result you use that in g

[cunning radish]

ok ok so wait the vlaue of g is g(1)

now we go to f

and there it says x^3

so should we replace

that

1

?

[nova topaz]

For g(f(0))?

[cunning radish]

yes

f(x)=4x+1 and g(x)=x^3+1

[nova topaz]

You substitute 0 into f first

Then you substitute your result from that into g

[cunning radish]

ok one sec lemme try to solve it

can we say like since we found f now we will go to g and for g there is x^3 so we replace this with 1 and it would become 1+1=2

can we solve liek thid

[nova topaz]

Yes that’s correct

[cunning radish]

ok i will ssolve the 2nd one its

find fg(x) simplfy ur answer

wait i dont know how to solve this

[nova topaz]

Ok so this time notice you don’t have a value to substitute into g(x)

[cunning radish]

ya

[nova topaz]

So instead you replace x in the second function ( which is f(x) ) with the entire function g(x)

[cunning radish]

so f=x^3+1?

[nova topaz]

f(x^3+1)

[cunning radish]

oh

[nova topaz]

So replace every x in f(x) with (x^3+1)

[cunning radish]

so 4(x^3+1)+1

[nova topaz]

Dont forget the +1 at the end

[cunning radish]

yep i forghet

[nova topaz]

Yeah that’s right

[cunning radish]

ok so it would be

4x^3+4+1

final ans 4x^3+5

[nova topaz]

Yep

[cunning radish]

oh

ok left the last one

whihc is

find h^-1(x)

[nova topaz]

Have you learnt to find inverse functions?

[cunning radish]

h(x)=2x+1/3

[cunning radish]

kindaa i just know at the last step it must be f^-1 something

and that

we

need

to

rearrange

it

[nova topaz]

Yeah so replace x in the function with h^-1(x)

And replace h(x) with x

[cunning radish]

ok so 2(x)+1/3

like this?

[nova topaz]

x=2*h^-1(x)+1/3

Then rearrange to get h^-1(x) as the subject

[cunning radish]

alr one sec

why is it -1?

[nova topaz]

-1 means the inverse function

$x=2h^{-1}(x)+\frac{1}{3}$

[glossy valve]

Luca M

[nova topaz]

You could also, if it makes it easier, let y = h^-1(x)

Then it becomes

$x=2y+\frac{1}{3}$

[glossy valve]

Luca M

[nova topaz]

Now solve for y

Your simply swapping x and y, which is the definition of an inverse function

[nova topaz]

Yeah

[cunning radish]

can u help me

with

a

diffrent

lesson

[nova topaz]

Do you know the formula for the volume of a cylinder?

[cunning radish]

ya

pi r^2H

[nova topaz]

Right. What is the radius in this case?

[cunning radish]

uhhh

[nova topaz]

Oh it tells you in the question

[cunning radish]

oh yaa

6cm

[nova topaz]

Length 11cm is the same as height 11cm

[cunning radish]

oh

[nova topaz]

Now all you have to do is substitute those values in and divide your result by 2

[cunning radish]

pi is 2.14?

[nova topaz]

3.14159

[cunning radish]

oh ya

why divide by 2

[nova topaz]

Notice it is the volume of half a cylinder

pi*r^2h is the volume of a full cylinder

So we must divide by 2 to get the volume of one half

[cunning radish]

so 3.14x6^2x11

[nova topaz]

Yes, then divide that by 2

[cunning radish]

7,460.64

dived by 2

[nova topaz]

And make sure to write down cm

[cunning radish]

3,730.32 i got this

i think its wrong

[nova topaz]

I got 621.72 cm

[cunning radish]

oh

let me check again

[nova topaz]

Did you enter it right in the calculator

[cunning radish]

oh ya i got 621.72

[nova topaz]

Nice

[cunning radish]

i put something wrong

ok i have a hard question

this

is

thw

lastt

this one

[nova topaz]

Do you know where to start for a) or b)

[cunning radish]

nope not rlly..

[nova topaz]

Do you know formula for volume of the prism

[cunning radish]

yep

half bh

(1/2 bh)xL

[nova topaz]

Yeah

You have h and L. What are they?

[cunning radish]

arents hight and lenghth

same

thing

[nova topaz]

Not in this case

Base and height refer to the cross section of the prism

Which is the right angle triangle

Length is how far in the 3d world the triangle goes for

[cunning radish]

oh

hey i gtg but thanks for helping me !!!

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[spring ivy]

can anyone help me with this, not sure how to do it

[glossy lintel]

ill try

[spring ivy]

ty

@glossy lintel

got anything?

[glossy lintel]

wait

i might

5

its 5

[spring ivy]

how?

[glossy lintel]

ok

so

pipe A takes 6 hours to fill it fully

pipe B takes 4 hours

[spring ivy]

ya

i don't get the alternate part

[glossy lintel]

meaning A takes 3 hours to fill it halfway

[spring ivy]

yep

[glossy lintel]

and B takes 2 hours

so

turn on A for an hour

then B

then A

Then B

then A

the 2 hours

B was on

it filled up halfway

and

the 3 hours

A was on

it filled up half way

now the tank if full

its all about thinking how much can they fill up in an hour

i couldnt do that

so i just thought

how many hours to fill it halfway

do you get it?

[spring ivy]

wait

sorry

i dont get it

[glossy lintel]

np bro

ok

[spring ivy]

can you explain from the halfway thing

[glossy lintel]

it said in the question

that pipe A takes

6 hours to fill

the tank

Fully

[spring ivy]

yes

[glossy lintel]

so

if we half the time

the tank will also be filled halfway

[spring ivy]

why half it though?

[spring ivy]

yes

true

[glossy lintel]

because we want to shorten the time interval

to calculate more precisely

[spring ivy]

oh ok

[glossy lintel]

the same applie to pipe B

half the time

half the tank

so 2 hours fill up half way

now we just

play it our

turn on pipe A

for an hour

then pipe B

then pipe A

then pipe B

then pipe A

the 2 hours B was on

it filled up halfway

and the 3 hours A was on it filled up halfway

[spring ivy]

how do we know how many times they have to be turned on and off till the tank is full

[glossy lintel]

the options

try the options

A option

is 4 hours

and i can see on hour 4

its imposible for the tank to be full

so naturally i turn it back on for more time

ok

and when i tried 5

i saw it fit perfectly

[cunning radish]

1st hour: Pipe A fills 2/12.
2nd hour: Pipe B fills 3/12 of remaining 10/12.
3rd hour: Pipe A fills 2/12 of remaining 7/12.
4th hour: Pipe B fills 3/12 of remaining 5/12.
5th hour: Pipe A fills 2/12 of remaining 2/12.
6th hour: Pipe B fills the remaining 2/12.

So, the tank shall be full after 6 hours.

[spring ivy]

that ain't even in the options

[glossy lintel]

bro

do you understand now?

if you have a doubt on something let me know

[spring ivy]

can't we solve using maths

algebraic expressions?

i understand it better like that

[glossy lintel]

probably can but i cant think of it

[spring ivy]

its ok

[glossy lintel]

like i solve using more logic

[cunning radish]

1st hour: Pipe A fills 2/12.
2nd hour: Pipe B fills 3/12 of remaining 10/12.
3rd hour: Pipe A fills 2/12 of remaining 7/12.
4th hour: Pipe B fills 3/12 of remaining 5/12.
5th hour: Pipe A fills 2/12 of remaining 2/12.

If we add up the fractions filled by each pipe, we get:

2/12 + 3/12 + 2/12 + 3/12 + 2/12 = 12/12
yep sorry ans is 5h

[spring ivy]

i solve using algebra

bro where did you get 12

[glossy lintel]

i think this is algebra

[spring ivy]

ya but i don't get it

what is 2/12

[cunning radish]

2+3+2+3+2 is what?

[spring ivy]

12

but where did you get 2 and 3 and 2 and 3 and 2 from

[cunning radish]

1st hour: Pipe A fills 2/12.
2nd hour: Pipe B fills 3/12 of remaining 10/12.
3rd hour: Pipe A fills 2/12 of remaining 7/12.
4th hour: Pipe B fills 3/12 of remaining 5/12.
5th hour: Pipe A fills 2/12 of remaining 2/12.

Starting with Pipe A: Since pipe A is opened first, it fills the tank for the first hour.
Second Hour with Pipe B: Then, pipe B fills the tank for the second hour.
Alternating Hours: This pattern continues, with pipe A filling the tank during odd-numbered hours and pipe B filling it during even-numbered hours.
Now, let's see how much of the tank each pipe fills in one hour:

Pipe A fills 1/6 of the tank in one hour (because it takes 6 hours to fill the tank completely).
Pipe B fills 1/4 of the tank in one hour (because it takes 4 hours to fill the tank completely).
Now, if we alternate between these pipes every hour, we can calculate the total time it takes to fill the tank.

1st hour: Pipe A fills 1/6 of the tank.
2nd hour: Pipe B fills 1/4 of the tank.
3rd hour: Pipe A fills another 1/6 of the remaining tank.
4th hour: Pipe B fills another 1/4 of the remaining tank.
5th hour: Pipe A fills the remaining 1/6 of the tank.

Adding up these hours, we see that it takes 5 hours in total to fill the tank.

So, the tank shall be full after 5 hours.

did u get it now?

[spring ivy]

yep thanks a lot bro

[glossy lintel]

do you want us to close this

i am gona close it

so others can use i

it

.close

whoever started this

close it

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[spring ivy]

yap

yup

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[torn jolt]

ok so

for each roll, the possible outcomes are numbers from 1 to 6

right?

[versed flicker]

Yes

[torn jolt]

now We want to find the number of sequences of rolls where the largest number appearing is not 4

consider cases where the largest number is not 4

so the largest number appearing can be 1,2,3,5 or 6

for each throw

there are 5 possible outcomes to meet that solution

or condition

[versed flicker]

Yes

[torn jolt]

so

5 choices for each 4 throws

[versed flicker]

Mhm

[torn jolt]

how do you write that as a number

[versed flicker]

5^4

But

[torn jolt]

good

?

[versed flicker]

There's a catch

[torn jolt]

what

[versed flicker]

Suppose I took the dice 4 times and I get 1,1,2,2
I do it again, I get 1,2,1,2

In the list they will be treated the same

But 5^4 counts it 6 times to be precise

Similarly 1112, 1211,1121,2111

Same thing with 4 ways

[torn jolt]

ok then the lowest one

[versed flicker]

The number of time each case gets repeated

Can't divide by just 1 number

It's complicated ig

Like aaaa will occur once time
aaab will occur 4 times
aabb 6 times
aabc 12 times
abcd 24 times

I hope I'm making sense

Ig

[torn jolt]

yeah i know what you are saying

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[white delta]

Bit stuck with this

[white smelt]

multiply the two things

use vieta's relations to evaluate the extra terms that come up

[white delta]

I was just wondering if there was a faster way

but Ill do that now

I just have a big expression and I can't really see how to apply it

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@white delta Has your question been resolved?

[white delta]

<@&286206848099549185>

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@white delta Has your question been resolved?

[granite torrent]

faster way, see:

https://artofproblemsolving.com/wiki/index.php/Newton's_Sums

[white delta]

yeah never seen this before, I was jsut doing papers from other schools and thought I'd learnt what i'd needed to do it

thanks though

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[winged fog]

A train arrives at a station 3 times an hour. It stops 2 minutes each time.
What are the chances Helena has to wait 10 minutes or less?
This is the question but I'm struggling to understand the explanation for how to solve it.
"In one hour, there are 60 minutes, and the train stops three times per hour.

If the train is already at the station, she can board it directly. Since the train stops for two minutes each time, it spends a total of 3⋅2=6 minutes at the station in one hour.

She needs to wait less than ten minutes if she arrives 10 minutes or less before the train arrives. Therefore, we can add 3⋅10=30 minutes to the 6 minutes the train spends at the station 36/60 = 60%"
I don't really understand the adding 30 minutes part?

[full forum]

@winged fog Has your question been resolved?

[fast peak]

imagine you have all 60 minutes of the hour and you call them either a good minute if you have to wait less than 10 minutes for the train, a neutral minute if the train is already there or a bad minute otherwise

then there are 30 good minutes, 6 neutral minutes and 24 bad minutes

so 36 minutes out of 60 are what you want

(this is assuming that the train leaves regularly and not something like 3 times within 15 minutes and then never in the remaining 45 minutes)

[winged fog]

Okay thank you!

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[next basalt]

I'm supposed to find least squares solution.

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[next basalt]

The matrix $A_{8\times 6}$ has 6 linearly independent column vectors from R^8.\
Can one claim that any of the matrices Are $A^T * A$ and $AA^T$ surely invertible, or, alternatively, singular (ie, lacking an inverse)?
Hint: $B^{−1}$ does not exist if $Bx = 0$ for some x ̸= 0. In that case x is also $x^T Bx = 0$. Interpret then $B = AT A or AAT$

[craggy tapir]

use \times

[next basalt]

where

[craggy tapir]

8 \times 6

[next basalt]

🤷🏼 w/e

[craggy tapir]

The matrix $A_{8\times 6}$ has 6 linearly independent column vectors from $R^8$. Can one claim that any of the matrices Are $A^T * A$ and $AA^T$ surely invertible, or, alternatively, singular (ie, lacking an inverse)?
Hint: $B^{−1}$ does not exist if $Bx = 0$ for some $x\neq0$. In that case x is also $x^T Bx = 0$. Interpret then $B = AT A $or $AAT$

[glossy valve]

FungusDesu MSC2020 34A05
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[craggy tapir]

eh good enough

[craggy tapir]

for helpers, this is the question

[acoustic bobcat]

For the $A^T A$ case, try showing $A$ and $A^T A$ have the same kernel

[glossy valve]

AlphaNull

[next basalt]

kernel?

[acoustic bobcat]

null space

set of all vectors that A sends to 0

[next basalt]

Ideally i'd like to understand the question first

[acoustic bobcat]

what part do you not understand?

[next basalt]

I'm given a matrix which has 6 colon vectors that are R^8? Right?

[acoustic bobcat]

yes

[next basalt]

"Can it be said that $A^T * A and AA^T$ are inversible or are missing invers"

[glossy valve]

Merineth

[next basalt]

What does inversible mean?

Is that essentially the same thing?

Inversible and missing an inverse

[acoustic bobcat]

no, they are opposites

invertible means there exists some matrix $A^{-1}$ such that $A \cdot A^{-1} = A^{-1} \cdot A = I$

[glossy valve]

AlphaNull

[next basalt]

Okay so it's asking me if the matrix is inversible or missing an inverse?

[acoustic bobcat]

yes

for A^T A and AA^T

[next basalt]

Is there a specific condition that has to be fulfilled for it to be inversible / missing an inverse? Doesn't it have something to do with determinants ?

If the determinant is not 0 then there exists an inverse?

[acoustic bobcat]

that is true but you can't use that

have you heard of the invertible matrix theorem?

[next basalt]

No

[acoustic bobcat]

you should probably look it up

it has a lot of equivalent statements for invertibility

[next basalt]

all of this?!

[acoustic bobcat]

you dont need all of it, just a few

null space of A being {0} only is a key one

[next basalt]

Seems crazy if they expect us to learn all?

Is that what the hint says?

[acoustic bobcat]

by the end of a linear algebra class most of them will seem quite obvious

yeah the hint says use the null space property

[next basalt]

We don't have a teacher so kind of hard to learn this myself

[acoustic bobcat]

if you are using a good book you should be fine

[next basalt]

I am not

[acoustic bobcat]

then use a better one?

[next basalt]

I don't know what null space of A being 0 is

[acoustic bobcat]

why are you doing this question if you havent learnt null space yet?

[next basalt]

Because it's an assignment due soon

[acoustic bobcat]

well i suggest learning null space then

[next basalt]

No time

Has to be done today

[acoustic bobcat]

im not allowed to give you answers sorry

[next basalt]

fantastic

👍

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[next basalt]

tysm 👍 :)

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[burnt sequoia]

i dont understand c

[burnt sequoia]

they need to find a formual

the worked example explained it: but idk where the formula came friom

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[void nova]

Where f is above or exactly equal to g

Yep

[tired blade]

so is that [-1.5, 1] U [4.5, infinity)?

[void nova]

I can't tell you the precise values with the only info present in this graph

[tired blade]

do I have to make it clearer?

[void nova]

No no

It was not the image quality the problem, but the fact that you're not given the expression of those two functions

Not only there

[tired blade]

where else?

[void nova]

Where it wasn't here

[tired blade]

hm not sure

[void nova]

If it's not ≤ it will be >

There are no other cases

[tired blade]

do you mind showing me where on the graph, I'm still confused a bit

[void nova]

Easier than you might think, actually

Green: f < g
Orange f > g

[tired blade]

ight got it now, last question, where are they equal?

[void nova]

Where they touch

Yes

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[torn jolt]

Wanting to know wether this process is right or wrong

[plain fox]

u can take derivative of the final answer , if its equal to the question then its right

[hallow walrus]

It’s wrong

The formula you’re using is only valid when it’s just x

[flat mesa]

it looks like you made a u-sub error

[hallow walrus]

Not x^{3/2}

[torn jolt]

Oooh i see

[flat mesa]

this integral is a terrible one to do by hand

[torn jolt]

Haha I know how to solve , was just curious about this style😂

Nvm thnx for the help

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[lilac parcel]

If I have a non-square m x n matrix whose columns are linearly independent, does that mean if I multiply this matrix by its own transpose, the columns are still linearly independent?

[torn jolt]

is m<=n?

[lilac parcel]

m>n

[torn jolt]

umm then it's definitely false right

[torn jolt]

Yes, mostly

A is of size mxn

A^t is nxm

so A A^t is mxm

the rank of A and hence the rank of AA^t is n (because it has n independent columns)

so there's no way an mxm matrix has all independ. rows but rank only n < m

[verbal musk]

n is the number of columns so its the number of vectors

if n > m then the columns are def not independent

[torn jolt]

yup that's also true

so i guess the only case this can work is m=n

in whcih case the answer is indeed yes

otherwise its always no

[lilac parcel]

hm alright

but let's say i multiply A^t A instead of A * A^t then, is it still false?

this time it's nxm * mxn, so the matrix is now nxn if i'm not schizo

[torn jolt]

sure

[verbal musk]

no no you can have lin independent vectors if m>n

[next basalt]

[torn jolt]

A can have lin indep. columns if m>n yes. But AA^t cannot

[torn jolt]

yes in this case the answer is yes

because it's rank will be n

and the matrix is of size nxn

[next basalt]

If $A_{6x8}$ then $AA^T a$ results in a 8x8 matrice

[glossy valve]

Merineth

[next basalt]

Shouldn't it be inversible then?

$AA^T = 8x8 \
A^TA = 6x6$

[glossy valve]

Merineth

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[buoyant saffron]

could someone help with this problem

I have

some work if helpers would like me to send

essentially I plugged in each f(x) into the given formula to set up a system of equation

[velvet sedge]

yes send your work

[buoyant saffron]

i also graphed it just for me to understand but i know its not necessary

[velvet sedge]

it shouldn't have an 'x' in it

w1f(x1)+w2f(x2)

[buoyant saffron]

oh shit

i misread

I simplified

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[astral relic]

X1245 generates $100 in free cash flow every year, but it has a 50% chance of breaking down every year. What is the value of X1245?

How would I go about solving this?

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@astral relic Has your question been resolved?

[astral relic]

<@&286206848099549185>

[tribal urchin]

This is a somewhat open ended question and I don’t know if there’s some kind of business algorithm or whatever for it, but I’d say just divide its value by 2 every year and sum the infinite series

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@astral relic Has your question been resolved?

[astral relic]

Haven’t taken a math class in a bit, any advice on how to sum an infinite series?

It would be this right?

.50(100+.5(100+.5(100+…)))

[tribal urchin]

Yes

Although you should also have a 1(100) term

To sum a geometric series you take a/(1-r) where a is the starting value ($100) and r is the value you’re compounding exponentially (in this case .5)

When you write it as $100 + $50 + $25 + … you can intuitively get the sum as well (and it nicely corroborates the formula)

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@astral relic Has your question been resolved?

[astral relic]

holdup Bot

let me think

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[pine rune]

Topic: Calculus, Convergence

I get that from 0<x<2pi sinx will always be between -1 and 1, but
can someone explain the underlined part?

[glacial pasture]

if x is an odd multiple of pi/2 then sin(x)=1 or -1

[pine rune]

ooh right cause -1 and 1 are not included omg

[glacial pasture]

yeah

|sin(x)|<1

[pine rune]

aight thx

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[torn jolt]

<@&286206848099549185> what does recursive formula means and what do we mean also by closed formula please

[gritty rose]

!15m

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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.

[torn jolt]

It already passed 😬😳

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@torn jolt Has your question been resolved?

[torn jolt]

Not uet

Yet*

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@torn jolt Has your question been resolved?

[austere cove]

@torn jolt a recursive formula is a formula that is defined by previous values in the sequence.

For instance the Fibonacci sequence is defined recursively.

F_0 = 0
F_1 = 1
F_n = F_{n-1} + F_{n-2}

A closed form solution to a recursive formula is when you solve the recursion and remove it

For the Fibonacci sequence this is

F_n = ((φ)^n + (-φ)^(-n))/√5

Where φ is the golden ratio: (1 + √5)/2

For linear recurrences it is generally possible to find a closed form solution. There are a handful of techniques. For non-linear ones, though, it's pretty much hopeless.

[torn jolt]

So for example we have got p(n)=p(n-1)×2+1 with p(1)=1 , that's the recursive formula, when i want to write the closed formula, it will be p(x)=2powern -1 , so what i did here is that i I generalized the expression for the sequence to include all terms ? Is that correct?

Sorry but i am just new to this and it's confusing me😭

[austere cove]

You removed the dependence on p(n-1) from p(n)

I don't think "generalized the expression for the sequence to include all terms" is accurate

The sequence already is defined for every positive integer.

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[cerulean lynx]

can anyone help me with these probability questions? all help appreciated!!

[marble cedar]

What question?

[cerulean lynx]

can i start with 1B?

I don’t know what the bars above the numbers mean

letters sorry

[marble cedar]

Are they replaced or not?

[cerulean lynx]

replaced?

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@cerulean lynx Has your question been resolved?

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[broken geode]

For the joint probability 𝑃(𝐶,𝐷, 𝑅, 𝑆) state its factorised form in terms of its constituent conditional probabilities for all events. What does this question mean?

[broken geode]

So this is the question:• C for the coin
• D for the dice
• R for the event that coin is heads OR dice is six
• S for the event that coin is heads AND dice is six.
and the answer I got is -> P(C) X P(D) X P(R given C OR D) X P(S given C AND D AND R)

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@broken geode Has your question been resolved?

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@broken geode Has your question been resolved?

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@broken geode Has your question been resolved?

[red matrix]

still wanting the answer?

[broken geode]

Yes

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@broken geode Has your question been resolved?

[fossil crest]

what is that

[torn jolt]

logarithm

[sacred yarrow]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[austere cove]

Based on what you have said, I don't see anything wrong with your work. However, it would probably be more useful to everyone if you posted the original question in full context.

[broken geode]

Okay

This is the question

And im looking at bi

And i am not sure how to represent R

R | c or d

But how would i write that? As p(R|c,d) or p(R|c or d)

[austere cove]

The second.

I read P(R|C, D) as R given C and D

[broken geode]

Okay

So its alright to express a joint probability as p(R| c or d)

[austere cove]

Yes, but I would not consider that to be a fully factorized version

[broken geode]

So how can i fully factorise it?

[austere cove]

Using the definition of conditional probability

[broken geode]

so like the equation on the left side?

[austere cove]

yup

well, you'll want to put it in terms of the right side

[broken geode]

Oh so this will need to be expanded more?

I see

Hmm

[austere cove]

You should find that a lot of things cancel.

You'll probably need to also use the independent of C and D to break apart P(C and D) and P(C or D) into P(C) P(D) and (P(C) + P(D) - P(C) P(D)) respectively.

[broken geode]

oh my

so

is there any resources u can reccommend ?

[austere cove]

I think you have all of the information that you need to tackle this problem. But your textbook should have any relevant formulae if I missed something

[broken geode]

hmm

ill have a look

thank you

i think u can close this now

[austere cove]

.close

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[sinful wasp]

yo

[sinful wasp]

2(5x+21)

can i get help with this

i got this far but idk what to do now that there is x on both sides

[dense fractal]

take away 9x from both sides

54 = x + 42

then take away 42 from both sides

12 = x

[sinful wasp]

for this one would i make them equal to eachother?

nvm i would

can someone check this i got 12

is the answer 12 @dense fractal

[hard veldt]

x = 12, but the question is asking for $m\angle 3$

[glossy valve]

Dork9399

[sinful wasp]

ohh

yea

21?

[hard veldt]

gj

[sinful wasp]

alr

[sinful wasp]

for this i got 162

but i dont think thats right

[hard veldt]

no ur right

[sinful wasp]

alr

65?

[hard veldt]

yea

whoops

[sinful wasp]

alr

96?

[hard veldt]

yea

[sinful wasp]

i dont even know where to start

it looks hella confusing

wait