I suck with questions like these never sure what forumla to use
#precalculus
maiden igloo
I tried a few last night but my final answer always comes out wrong
I think i got it
wow
im so dumb
fuuuuuu
fleet yew
Lol
viscid thistle
Hey
wary coral
Ive been stuck on this question on my practice exam and i cant figure it out
how would i do this?
maiden igloo
would this use the remainder theorem ?
wary coral
I tried to find the function of the points but i couldnt get it
lilac pier
@wary coral Dividing by x-2 and getting some remainder is same as plugging x=2 in f(x) and getting back some value for f(2).
You can see the point (2,4) given to you.
That means when x=2, f(2) = 4
maiden igloo
what is this called
i thought it was the same as square root 2 to the power of 9
but
i was wrong
wary coral
Isnt it ninth root of 2
fleet yew
2^(1/9)
wary coral
fleet yew
What calculayor
maiden igloo
i found it
fleet yew
Remember there are technically 9 solutions
maiden igloo
glossy yew
muted granite
Turn 2log(z^2 - 4) + (1/2)log(z - 2) - 3log(z + 2) into a single log. z is greater than 2. No clue what to do.
should i turn them into rational functions and then simplify?
upper kelp
It sounds like the only think you can do.
One thing that you may have to take into account in some problems is that using these manipulations could change the domain of the function.
For example, let's say that if you take the function:
f(x) = log (2x + 4) - log (x + 2)
and you simplify it and get:
f(x) = log (2),
then you still need to consider that the domain of the original function is x ≥ -2, and not all x.
Also, Subi, was your question answered?
torn iron
anybody have a good resource to just review precalc as a whole? I need to brush up on everything before I try to study any further.
proven marten
,rotate
obsidian monolithBOT
proven marten
Where is 1
proven marten
btw Daniel, Khan Academy?
torn iron
Kahn Academy is limited in what it covers I've found
torn iron
It barely touches on sin and cos graphs
torn iron
and no proofs either for the most part
proven marten
Honestly yeah, if you find a resource like that please share it with me
torn iron
I will let you know if something shows up for me
proven marten
thanks :)
proven marten
Is this what your pre calc curriculum covers
@viscid thistle Do you know the difference between soft bracket and hard bracket
torn iron
Thats pretty much what the curriculum covers except we don't cover limits, parametric functions, and area under a curve.
torn iron
@viscid thistle its asking for a range of numbers, it doesn't necessarily matter what the graph looks like as long as you take in the range of the y values. No matter what you'll probably be using interval notation
viscid thistle
Ok that makes sense
full garden
hey guys shouldn't these two equations have the same graph
serene heath
no?
upper kelp
Keep in mind that:
3(x - 40) = 3x - 120, not x - 120.
next kernel
Whaaaaaaaa
If x or y is standing on the right all by itself, a is positive or negative 1/4
maiden igloo
Never would take math online
viscid thistle
in general form
Ax + By + C
the x number always has to be positive
correct?
No
What do you mean the x number
A?
upper kelp
It doesn't even have to be non-zero, to be honest.
If you have a line of the form y = t for some real number t, which is a line perpendicular to the Y axis, then A "vanishes."
However, if A is non-zero, and you have an equation of the form:
Ax + By + C = 0
then you can multiply it by (-1) such that the coefficient of x will be A, if A isn't already positive.
maiden igloo
Is there a difference between foci and focus ?\
upper kelp
Foci is the plural form of focus, as far as i know.
next kernel
Man this problem you’ve got sucks @maiden igloo
next kernel
But x equals the square root of 18 and y is 2
maiden igloo
but mymath lab
is so picky
i put -5 for answer it marked it wrong
and said
x = -5
next kernel
You have to tell your teacher not to use it. It does that for no reason.
maiden igloo
-__-
next kernel
I actually rated my professor poorly because of his use of MML.
Then my next one never did.
maiden igloo
haha
My other teacher uses Aleks
which is nice
so nice
MML gives you 60 questions for HW on properties of Log
like why
steel badger
where did cos(theta) go
https://cdn.discordapp.com/attachments/540211747613704221/701595587241443370/unknown.png
upper kelp
It doesn't really "go" anywhere. It's more in the sense of that the inner product - the sum u1 v1 + u2 v2 defined this way, is used in order to define the angle between two vectors.
stuck lark
each side is really a separate definition of the dot product
steel badger
is something missing?
stuck lark
the left side is what i'd call a geometric definition which uses the angle theta between u & v
steel badger
wait yea I am missing something I re-read it
stuck lark
the right side is, i guess, the algebraic definition (or rather the definition of standard inner product on R^2)
steel badger
the cos(theta) didn't go anywhere because it wasn't like
expanding length of v and length of u times cos(theta)
it's a definition :S
god damn it lmao
stuck lark
what's relevant to YOU is that these are alternative definitions for the same thing
steel badger
I thought this is a definition of dot product I jut was tired and misread it as like
expanding
my bad for asking a really dumb question
stuck lark
it's not dumb, it's good to clarify
the left side is the geometric defn of u dot v, the right side is the algebraic one, that's why these were set equal to each other
steel badger
yea it is applying the cosine rule
trim carbon
I dont know if this is where im supposed to post for help but i have some questions of polar form and conversions
upper kelp
Are you referring to polar form:
z = r · (cos Θ + i sin Θ)?
vast steeple
Good luck y'all
muted steeple
Hi, I'm trying to check my answer, because the one online doesn't exactly match it.
My working out:
The unrolled rectangle represents $2\pi \cdot h$.
The $V=\pi R^2h-\pi r^2h$ when factored is $V = \pi h(R-r)(R+r)$. Where thickness is $(R-r)$ and average radius is $\frac{R+r}{2}$.
Then if I substitute them:
$V = 2 \pi \cdot \frac{(R+r)}{2} \cdot h \cdot (R-r)$
Yet the one online states $V = 2 \pi \cdot (R+r) \cdot h \cdot (R-r)$
obsidian monolithBOT
꧁༺Vocal༻꧂:
willow bear
$2\pi \cdot \frac{R+r}{2} \cdot h \cdot (R-r)$ and $\pi(R+r)h(R-r)$ are one and the same
uncut mulch
extra 2
muted steeple
I struggle to see it how that's true.
obsidian monolithBOT
Ann:
willow bear
$2 \cdot \frac{1}{2} = 1$
obsidian monolithBOT
Ann:
muted steeple
i get that part, but then pi is left on it's own, no?
muted steeple
I'm assuming you're saying that i should distribute 2pi
on the fraction
and cancel the 2's
i mean multiply*
willow bear
where did you get $\bd{2}\pi(R+r)h(R-r)$?
obsidian monolithBOT
Ann:
willow bear
can you link to the "online solution" which said that
muted steeple
I'm looking at the answers on the net.
i have no clue what he did with the average at the end.
native sequoia
they done oof
muted steeple
Is that a mistake there?
uncut mulch
the fked up at the very end
muted steeple
Looks like i'll need to buy an answer book...
uncut mulch
R+r isn't the average radius
muted steeple
Yeah
But he shows himself cancelling
I got no clue how, but i guess it's a mistake.
uncut mulch
he did the calculations for
2*(avg radius) but that doesn't really do much
muted steeple
Oh okay, so my answer is correct.
uncut mulch
the sort of went in a loop
and mistakenly wrote 2pi instead of pi
or wrote (R+r) instead of (R+r)/2 for avg radius on the last line
muted steeple
So I chose $x^4+8x^2+4$ and $4x^2+6x+3$
Which gives $x^4+12x^2+6x+3$
All i can see is that sum obtains the largest degree of the two polynomials.
The product is
$4x^6+6x^5+35x^4+48x^3+24x^2+12$
I'm not sure what I'm supposed to discover here besides the fact that the each term's degree is gradually decremented.
novel dirge
I need help with understanding probability problem
obsidian monolithBOT
꧁༺Vocal༻꧂:
muted steeple
Oh sorry, pancake.
novel dirge
np, sorry if I am interrupting
should I post later?
if you throw 2 dies, what is the probability of one being a number dividable with 3 or 4?
I don't understand the solution in the book
first it says the total number of options is 36
which is 6x6, and I understand that
then they want to find the probability of A-one of the dies landing on a number being dividable with 3, B-one of the dies landing on a number being dividable with 4 and AB one of the dies landing on a number being dividable with 3 and the other one being dividable with 4
I don't understand the last part
why look for AB
shouldn't just looking for the probability of A or B be enough
since both A and B can't happen on a single die anyway
viscid thistle
You are right
On a single dice throw you cant get a number that is dividable for 3 and 4 at the same time
@novel dirge still need help with the question?
(although this isnt precalculus, is probability)
novel dirge
@viscid thistle you stil here?
tardy ridge
if you throw 2 dies, what is the probability of one being a number dividable with 3 or 4?
viscid thistle
Bruh
tardy ridge
Find the probability a die number is divisible by 3 or 4. You can count that (3,6,4)
viscid thistle
I wanted him to answer to get the wording of the problem
tardy ridge
Then find the probability that neither dies roll a number divisible by 3 or 4
then 1- that.
viscid thistle
I wanted him to answer to get the wording of the problem
@viscid thistle
Uhhh
Kinda rude tbh
He just answered
@novel dirge yeah so answered without asking him
Rude but yeah there you go
viscid thistle
Np
tall gulch
Can anyone help me with 20 questions in geometry at around 12:50
tall gulch
its hw
tall gulch
its given everyday at 12:50
novel dirge
I still don't really understand the last part
viscid thistle
Yeah bc he just humped to the solution smh
novel dirge
why do i have to subtract the probability of one being dividable by 3 and the other one by 4
tall gulch
i need help going through every problem to understand it more
viscid thistle
In that way you cant understand it
novel dirge
can you explain it?
viscid thistle
why do i have to subtract the probability of one being dividable by 3 and the other one by 4
@novel dirge so learn it my way
What are they asking you for
novel dirge
what the probability is
if you throw 2 dies
for one to land in a way that it is dividable by 3 or 4
novel dirge
but I don't understand why I have to get the probability of the second die being 4 if the first one is 3
viscid thistle
Do you know the formulas?
$The
P(A\cup B)
P(A\cap B)?$
novel dirge
there are formulas?
obsidian monolithBOT
Al3dium:
Compile Error! Click the 
reaction for details. (You may edit your message)
novel dirge
ok
my teacher didn't explain anything, she just sent us the problems
some are solved and some are for homework
I am currently going through the solved ones
viscid thistle
$P(A\cup B) = P(A) + P(B) - P(A\cap B)$
obsidian monolithBOT
Al3dium:
viscid thistle
This is the formula for this problem
This means
P(A OR B)=P(A) + P(B) - P(A AND B)
And they asking you for
for one to land in a way that it is dividable by 3 or 4
@novel dirge
OR
So they are asking you for which of these variables above?
novel dirge
Not sure about the second question?
it is a chance that one of those 2 dies will be divisible by either a 3 or a 4
it can't be both I think
viscid thistle
Exactly
OR
it can't be both I think
@novel dirge it CAN be both
That its A or B means probability that both can ocurr
The tip is on the word OR
novel dirge
but there isn't a number between 1 and 6 that contains both 3 and 4
viscid thistle
P(A OR B)=P(A) + P(B) - P(A AND B)
@viscid thistle knowing this
novel dirge
but I guess it is just this example
viscid thistle
$P(A\cup B) = P(A) + P(B) - P(A\cap B)$
@viscid thistle that means this
obsidian monolithBOT
Al3dium:
viscid thistle
but I guess it is just this example
@novel dirge we'll go there later
Follow my questions
novel dirge
ok
viscid thistle
So
novel dirge
which one exactly?
viscid thistle
$P(A\cup B) = P(A) + P(B) - P(A\cap B)$
@viscid thistle if this equation
obsidian monolithBOT
Al3dium:
viscid thistle
P(A OR B)=P(A) + P(B) - P(A AND B)
@viscid thistle means this
What of the 4 variables they are asking you for?
If they said "OR" in the question
novel dirge
here, I understand that I want to add the probablities of either 3 or 4 being true
and I think that I also understand that formula
but not how to apply it to this problem
since it is already impossible to get both A and B to be true
on the same die
viscid thistle
P(A) = probability of being dividable by 3
And
p(B) = probability of being dividable by 4
novel dirge
but here they are on separate dies
viscid thistle
since it is already impossible to get both A and B to be true
@novel dirge we will get thereee
P(A) = probability of being dividable by 3
And
p(B) = probability of being dividable by 4
@viscid thistle got this?
novel dirge
yes
viscid thistle
And they are asking us for $P(A\cup B)$
obsidian monolithBOT
Al3dium:
novel dirge
I guess?
novel dirge
not sure how to know that it is AUB
viscid thistle
P(A OR B)=P(A) + P(B) - P(A AND B)
@viscid thistle
novel dirge
I think I understand now
obsidian monolithBOT
Al3dium:
novel dirge
yes
viscid thistle
What is $P(A\cap B)$
obsidian monolithBOT
Al3dium:
novel dirge
but in the example they said probability of them happening on separate dies
viscid thistle
Its impossible to happen at the same time as you said
Wait
but in the example they said probability of them happening on separate dies
@novel dirge what?
viscid thistle
Where?
novel dirge
(6,4),(3,5),(6,5)}, so mA = 20.
B={ (1,4),(2,4),(3,4),(4,4),(5,4),(6,4),(4,1),(4,2),(4,3),(4,5),(4,6)}, so mB = 11.
AB= { (3,4),(4,3),(4,6),(6,4)}, frhom which mAB = 4.
P(A B)=P(A)+P(B) - P(AB)=2036 +1136- 436 =2736 =34 .
You forgot the / sign on 2036 or 20/36.
viscid thistle
The probability of P(A B) = 34?!
novel dirge
this is what my teacher gave us
tardy ridge
3/4.
novel dirge
she didn't even explain anything
just sent the pdf
with these problems and homework
we have to do the homework which she will grade
and we will have an exam this week
without her explaning us anything
viscid thistle
Thats some bad teacher lmao
novel dirge
terrible
on the last exam she didn't eccept my answer because I didn't copy it correctly
viscid thistle
@novel dirge it'd be possible to explain this after 20 min? Im eating :p
novel dirge
one number was supposed to be ^-2, but I wrote ^2
viscid thistle
on the last exam she didn't eccept my answer because I didn't copy it correctly
@novel dirge lol
novel dirge
but solved that problem correctly
viscid thistle
Sucks
viscid thistle
Oof
@novel dirge it'd be possible to explain this after 20 min? Im eating :p
@viscid thistle
viscid thistle
Great
tardy ridge
the probability of A happening, or the probability that the first die rolls a number divisible by 3 or 4, is 3/6
the probability of B happening is probability that the second die rolls a number divisible by 3 or 4, which is as well, 3/6
So using the formula, P(A U B)=P(A)+P(B) - P(A N B)
=3/6+3/6 - (3/6)^2
=3/4
P(A N B)= the probability that both roll a number divisible by 3 or 4, which is 3/6 * 3/6.
since when they are independent, P(A N B)=P(A)*P(B) - the reason for this is because P(A N B) = P(A) * P(B|A) = P(A)*P(B). Since given first die rolls a number divisible by 3 or 4, it doesn't change the probability that the second die rolls a number divisible by 3 or 4.
viscid thistle
Basically this lmao
I was about to do this
@tardy ridge thx
Umm thats also a way of seeing it
Dividing into the rolls
novel dirge
I think I get it now
with the edit of the last message
but now I am stuck on another one
viscid thistle
Show it
novel dirge
if you take 2 cards from a deck of 32, what are the chances of those 2 cards having the same color or both being an ace
I will first write what I think
then you correct me
viscid thistle
Ok
novel dirge
when it comes to colors, I assume that it is 1/2 and 1/2
can't remember the last time I played with cards
novel dirge
if I had to pick one card, I would just write 8/32
no
16/32
since that is how many red or black cards there are compared to the total
but if I pick 2 cards at the same time
will it be 16/32+16/32 or 16/32 x 16/32
?
or maybe reduce th second one
16/32 +or x 16/31
novel dirge
ok
I have to go afk for maybe 5 or 10 min
but I will be back then
so just tag me
viscid thistle
Did you said 2
viscid thistle
It isnt written on the question?!
viscid thistle
Oh lol
viscid thistle
Oh
novel dirge
and she is basically trying to ruin our grades
I haven't even started preparing for collage entrance exams
I will probably be here a lot when I start preparing for those
novel dirge
I am just glad that I will leave soon
viscid thistle
ah okay
viscid thistle
so then you have P = 16/32x15/31+4/16+3/15 🙂
novel dirge
not sure about the last part
and don't i have to subtract anything?
to fit into that formula
for A U B
viscid thistle
My bad lmao
I was thinking of suites
Im rusty on probability
so then you have P = 16/32x15/31+4/16+3/15 🙂
@viscid thistle from where the hell you got that
I have never seen that logic in my life
....
don't you take the card out of the pile?
and then choose another?
novel dirge
you take 2 at the same time
novel dirge
I will be gone now for a bit, so sorry if I don't reply
viscid thistle
yes so once you have taken one, the probability of taking another one changes
novel dirge
Have to bring something to my aunt
viscid thistle
Like never seen that calculations
Its okay
Dividing
What Pascal did was number of wanted cards divided by all possible cards to draw
Dont know whats the actual problem but that probably is correct whats written above
,calc 16/32*15/31+4/16+3/16
obsidian monolithBOT
Result:
0.67943548387097
Result:
0.29193548387097
that's better
Do you know how it'd be done with the formulas
With P(A U B), P(A N B)
Being P(A) = probability of being red
And P(B) = probability of being ace
@viscid thistle
N means and?
Yeah
Or $\cap$
obsidian monolithBOT
Al3dium:
viscid thistle
The same
I wanted him to know that way, and i tried but im rusty
To show him*
Give me a second, helping someone else
Np
daring yarrow
uhhh i think they ommited a couple of steps
and i dont really get how they get 2
viscid thistle
divide it on two fractions
novel dirge
I am back
daring yarrow
lim(x/x) when x goes to infinity is 1
right?
if so i think ive just wasted about 20 minutes of my life
viscid thistle
It's been a while since I've used this notation, but I'll give it a go
We have P(A) = probability of being red?
viscid thistle
Okay
I'm sorry I don't know how to write it for you in that notation :/
Since it says OR we want to add the probability of having a red card twice and an ace twice
novel dirge
Here I am not sure if it is ok to have a red ace
there was a similar one
but it was just one card
viscid thistle
Since it says or I don't think it should matter
novel dirge
and it was the probabilty of it being either a spade or a queen
viscid thistle
if it were to say and, then you would have to take that into account
daring yarrow
novel dirge
so it was P(A)+ P(B) - P(AB)
daring yarrow
same thing as last time
viscid thistle
so it was P(A)+ P(B) - P(AB)
@novel dirge yes if you were doing probability of getting two reds AND two aces
novel dirge
if I understood correctly, it is either COLOR AND COLOR or ACE AND ACE
the last one was either red or ace
but not both together
viscid thistle
@daring yarrow this channel is occupied dude
novel dirge
so no red ace
daring yarrow
bruh
viscid thistle
<@&286206848099549185>
novel dirge
should i first get the chance of picking 2 cards of the same colors
would that be 1/2 + 1/2
or 1/2 ***** 1/2
daring yarrow
photomath cant solve the problem i sent apparently F
novel dirge
@daring yarrow it can
@viscid thistle the solution you wrote at the start makes sense if you pick the cards one after another
and under the condtion that the first one is indeed a red card
but if you pick them at the same time, should it also affect the outcome?
viscid thistle
Pick them at the same time? That is ambiguous, to me picking at the same time means you pick one and pick another so you don't return the card to the deck.
novel dirge
My guess is that you pick them simultaneosly
instead of taking one
and then the other
you take them both in one move
viscid thistle
how can u take 2 at the same time think about it
My guess is that you pick them simultaneosly
@novel dirge if you draw two cards simultaneously, they cannot be the same card. So the probability changes, it is the same as "one after another"
novel dirge
oh
so what if I want to get the probability just the 2 cards being the same color
what would the formula be?
or 1/2 * 1/2```or do I remove the cards
viscid thistle
The first one not at all
novel dirge
damn, I hate this
viscid thistle
are you returning them to the deck? or picking simultaneously ?
viscid thistle
Sums up to 1
Impossible in probability
Well
Let's say picking first red is A, P(A) = 16/32, because there are 32 cards in the deck and half are red
yes?
novel dirge
yes
viscid thistle
then the probability of getting a second red B, is P(B) = 15/31
You can try and do a tree diagram @viscid thistle
Maybe
viscid thistle
Then its black
novel dirge
then we still have 16 reds
viscid thistle
16/32
novel dirge
in the second one
novel dirge
so the second would be the probabiity of getting a second black?
viscid thistle
Yes, red or black doesn't really matter. We just found the probability of getting two of the same colour.
novel dirge
so can I just write this as (16/32 + 15/31) OR chance of both being an ace
viscid thistle
yes
novel dirge
which is 4/32
viscid thistle
not plus
novel dirge
or 30?
viscid thistle
so can I just write this as (16/32 + 15/31) OR chance of both being an ace
@novel dirge 16/32*15/31
viscid thistle
AND = multiply OR = add
chance of getting first ace = 4/32, chance of getting second ace = 3/31
viscid thistle
chance of getting first ace AND second ace = 4/32*3/31
novel dirge
I understand that one
and for the aces, themselves, it would be 4/32 * 3/31
but here I am not sure if I can leave it like that
or write it as 4/30 * 3/29
taking itno accont the first part
viscid thistle
no, the events are independent
viscid thistle
you add them
willow bear
i strongly disapprove of just saying
AND = multiply OR = add
without any of the details
novel dirge
but should I take into account what would happen if I pcik a black ace?
willow bear
P(A and B) = P(A)P(B) iff A and B are independent
P(A or B) = P(A) iff A and B are disjoint
if you don't say this you'll lead everyone hearing you say "AND = multiply OR = add" down an incorrect path
a lot
novel dirge
but Idk what the color is supposed to be
viscid thistle
Okay, Ann can explain it to you then.
novel dirge
@willow bear what is the difference between independant and disjoint?
willow bear
we say two events are disjoint if the probability of them happening simultaneously is zero
novel dirge
@viscid thistle thanks for the help so far
willow bear
one also hears the term "mutually exclusive" in the context of probability
it means the same thing
but is a bit bulky
novel dirge
I think I understand
willow bear
an example of two disjoint events would be "dice roll < 3" and "dice roll > 4"
novel dirge
and independant is what we did with the cards
willow bear
if i am to help you, then can i see the original problem
viscid thistle
No problem, sorry for not doing an adequate job in Ann's opinion, that's how I learnt it first. It seems she is going a bit too in-depth for the level of your question. Sorry if I confused you.
novel dirge
yes
np, don't worry
you didn't
I mean, I am still not 100% sure what everything is, but at least you didn't worsen it with the and vs or explanation
my teacher didn't even tell me that much
willow bear
i repeat, can i have the original problem
viscid thistle
if you take 2 cards from a deck of 32, what are the chances of those 2 cards having the same color or both being an ace
@willow bear 🙂
novel dirge
the problem is
if you simultaniosly pick 2 cards from a deck of 32, what are the chances of both being the same color or both being an ace?
yeah
willow bear
uh huh lemme see
viscid thistle
Have fun getting the exact same answer I did about half an hour ago.
willow bear
yeah the events "both cards are the same color" and "both cards are aces" are not disjoint
jsyk
you could draw the aces of spades and clubs, or the aces of hearts and diamonds
alright so personally i'd do this combinatorially
there are 32C2 ways to draw two cards if we don't care what they are
there are 2 * 16C2 ways to draw two cards of the same color
there are 4C2 ways to draw two aces
and finally there are 2 ways of drawing two aces of the same color
so if we were to write this down probabilitistically
i'll denote with A the event of drawing two aces
and i'll denote with C the event of drawing two cards of the same color
$P(C) = \frac{2 \times \binom{16}{2}}{\binom{32}{2}} \ P(A) = \frac{\binom{4}{2}}{\binom{32}{2}} \ P(A \cap C) = \frac{2}{\binom{32}{2}}$
and we're looking for $P(A \cup C)$
obsidian monolithBOT
Ann:
willow bear
does this make sense so far
novel dirge
I am still processing it
willow bear
whoops
obsidian monolithBOT
Ann:
willow bear
there we go, fixed a very embarrassing typo
novel dirge
didn't even get to that part
C is 2 cards of the same color?
why is therea 2 there?
novel dirge
for aces it basically means ther are 2 out of 4 ways to pick an ace when choosing 2 cards out of 32?
willow bear
...
novel dirge
sorry if I ask stupid questions but I really want to understand this as much as possible
I still have 2 other problems to solve and want to avoid having to ask for help for them, too
novel dirge
yes, I understand how we got that one
just asking if I wrothe there waas correct
if I understood it correctly
willow bear
your wording leaves a lot to be desired
novel dirge
we chose 2 because there is a total of 2 drawn cards, so it can't be more
English isn't my first language and I am especially bad with math terms
for the colors, if there were 3 colors in the deck would I write 3 instead of 2?
novel dirge
good, I tink I understad that part now
and because here are 2 black aces, it is 2/ 32C2 ?
still not reducing the number of cards?
we are leaving it to be 32 allthe time?
willow bear
the denominator is 32C2 because it's the number of possible draws without any restrictions
but no, the numerator in P(A and C) is not 2 because there are 2 black aces
novel dirge
but if we take one card out, won't the number be lower?
willow bear
we're not taking one card out, we're taking out both at once
novel dirge
ok, I understand
willow bear
anyway, it's 2 because there are only 2 ways to pick two cards that are both aces and the same color: either you get the A♡ and A♢, or you get the A♤ and A♧
it's impossible to have it any other way
novel dirge
but there are only 2 aces of the same color in that deck
willow bear
there are 2 aces of one color and 2 of the other
novel dirge
I think I understand
yeah
that is still 2 pairs
My problem is that I keep connecting everything
thinking that we can have both 2 cards of the same color and 2 aces
which ins't true
willow bear
thinking that we can have both 2 cards of the same color and 2 aces
which ins't true
so you're saying it's somehow impossible to draw the A♡ and A♢
novel dirge
no
I meant it as, In my head I keep thinking that there are separately 2 cards with the same color and 2 aces
which would be 4 cards
thus impossible
but it confused me
now I think I figure it out
can you tell me how I should approach this kind of problems, in general?
if it isn't to much for you to write
like what should I look for and what should I try doing
willow bear
formulating a general approach is kinda hard
novel dirge
not really a general approach
just a few tips and advice
like what do you generaly look for
willow bear
...
yeah that's still too hard to describe in general
combinatorics is a bit of a bag of tricks in this regard
viscid thistle
You okay @novel dirge 
?
hollow yacht
is this the chat for questions about hyperbolas
viscid thistle
Okay, please ask if you get stuck again. I'll try to help
novel dirge
yeah, Ann helped me with the formulas
viscid thistle
I dont see the result
Where is it
@willow bear is it possible with binomial probability?
viscid thistle
b(n,k) = (n)
(k) + p^n - q^n
@willow bear
I dont remember exactly the signs
That (n
k) is a nCk
willow bear
it sounds like you were going for $\binom{n}{k} p^k q^{n-k}$ but then messed up the formula horribly
obsidian monolithBOT
Ann:
willow bear
it isn't
viscid thistle
It isnt possible? Or it isnt, it is possible
willow bear
it isn't possible
viscid thistle
Okay
And being P(A) = being red and P(B) = being an ace
How'd it be
I think its impossible to be done with $P(A\cup B)$
obsidian monolithBOT
Al3dium:
viscid thistle
With those assumptions
willow bear
there is so much wrong with what you just said
viscid thistle
Yeah
viscid thistle
it sounds like you were going for $\binom{n}{k} p^k q^{n-k}$ but then messed up the formula horribly
@willow bear You could be a littttttle bit less judgmental, just saying. I personally would not want my teacher to tell me I messed up horribly..
obsidian monolithBOT
Nikky14:
novel dirge
viscid thistle
Is that the result from the previous problem
viscid thistle
It gets a different result than @viscid thistle did
novel dirge
A for colors and B for aces
novel dirge
What is the chance of throwing 2 dies so that their sum isn't dividable with 2 or 3?
I listed all the options
but am not sure if (3,2) and (2,3) are different
,rccw
obsidian monolithBOT
willow bear
yes they are
novel dirge
so i was planning to add all those up
and divide them with the number of combinations
willow bear
add them all up and divide by 6*6
viscid thistle
@viscid thistle why they get a different result than you on the previous problem?
Umm
I don't know, I am pretty confident my answer was correct. But Ann clearly knows better.
novel dirge
I checked with photomath
viscid thistle
Oh
novel dirge
so, is what I wrote good?
novel dirge
here is another one, which I did halfway
there are 6 bulbs
2 aren't working
what are the chances that by picking 4, 1 at most can be boken?
I figured that the total number of combinations is 4C6
which is 15
but I am not sure wha to do with the broken one
should I do it like 3C4* 1C2 + 4C4 ?
@viscid thistle
eager path
How do I find all possible parabolas given a focus and a containing point (also need to find directrix)
full gorge
Hey guys, if you are given a matrix, and it tells you to find 3A, do I multiply everything in the matrix by 3?
novel dirge
@willow bear I got confused again, what is the reason for writing 2* 16C2 in the problem you helped me with
with the colors
willow bear
16C2 ways to pick two red cards
16C2 ways to pick two black cards
obviously both can't happen at the same time
novel dirge
yes
so why multply with 2
to me it makes more sense without 2
since we want either red or black
viscid thistle
So you want to find the probability of getting no broken bulbs and then probability of getting one broken bulb
then add those probabilities
novel dirge
yes
viscid thistle
@novel dirge
novel dirge
I did 4C3*2C1 + 4C4
which is 3good ones and a bad one and all 4 good ones
but not sure if that is correct
?
how?
viscid thistle
You'd have.
viscid thistle
I'm sorry. I'm confused to, whether the bulbs are being picked simultaneously or not.
novel dirge
oh, right
the -1
I am guessing that they are
all the example problems are like that
we never did those where you reduce the number of choices
I do understand how they work, but just not sure when to use them
but I think here the number isn't reduced
viscid thistle
exactly
and then add the two
I got 1/3
And to the point about the numbers not being reduced, that's not quite the case. We can also calculate it like P(no broken)
and you'll find the answer is the same
viscid thistle
sorry the *'s got in the way
is it more clear now?
P(no broken) = 4/6 x 3/5 x 2/4 x 1/3
viscid thistle
this is only for no broken
it works out to the same as 4C4*2C0/6C4
then you can also find P(1 broken) = 4/6 x 3/5 x 2/4 *2/3
I need to refresh myself in probability lmao
Im rusty
Ill check all the problems of today again
novel dirge
makes sense
does either of you understand why @willow bear multiplied the options for each possible color?
there are 32C2 ways to draw two cards if we don't care what they are
there are 2 * 16C2 ways to draw two cards of the same color
there are 4C2 ways to draw two aces
and finally there are 2 ways of drawing two aces of the same color
so if we were to write this down probabilitistically
i'll denote with A the event of drawing two aces
and i'll denote with C the event of drawing two cards of the same color
$P(C) = \frac{2 \times \binom{16}{2}}{\binom{32}{2}} \ P(A) = \frac{\binom{4}{2}}{\binom{32}{2}} \ P(A \cap C) = \frac{2}{\binom{32}{2}}$
obsidian monolithBOT
Mr.Pancake:
novel dirge
here
viscid thistle
No idea what she did, sorry.
viscid thistle
I was starting to like probability until today lmao
novel dirge
I get the probability of around .25
which is close to what I get doing it your way
the difference is 0.05 or something like that
carmine elbow
Can someone help me with this?
viscid thistle
Go to one of the channels of #❓how-to-get-help
carmine elbow
I was able to get help from this channel before
I believe the other channels are taken
viscid thistle
i'd make use of trig identities, specifically double angle identity, did you learn those?
@carmine elbow
willow bear
@novel dirge i didn't multiply 16C2 * 16C2
the factor of 2 is only because i added 16C2 to itself
novel dirge
yes, but why did you do that?
viscid thistle
^
How would I use double angles for this problem?
@carmine elbow go look in internet
Or i think i have a pic
Wait
i'd do 1-2sin^2x and 2sinxcosx
Yeah thats the one
novel dirge
@willow bear I am really sorry for bothering you, but can you please explain why you did that?
willow bear
16C2 ways to pick two red cards
16C2 ways to pick two black cards
add them together
viscid thistle
@carmine elbow
novel dirge
either red or black
carmine elbow
I know that, but I don’t know how to use a double angle for my problem.
viscid thistle
sqrt3(sinxcosx) + 1 - 2sin^2x -1 =0
sqrt3 sinxcosx - 2sinx^2x = 0
sinx(sqrt3cos - 2sinx) = 0
Is the same
I know that, but I don’t know how to use a double angle for my problem.
@carmine elbow but with a sqrt 3
novel dirge
@willow bear can you explain the process of how you got that?
viscid thistle
Yeah
carmine elbow
Can you explain this in detail ?@viscid thistle
carmine elbow
Can you also verbally explain this? I’m not trying to be a pain, I just want to make sure that I fully understand
carmine elbow
I think so. I looks like you used sin^2theta + cos^2theta = 0
viscid thistle
@viscid thistle you could have made the cosines in terms of sines to operate them
Maybe
willow bear
@novel dirge because if you only count one you fail to account for the other
viscid thistle
I think so. I looks like you used sin^2theta + cos^2theta = 0
@carmine elbow Equal to 1, yes
willow bear
if you only count black-black pairs you fail to account for the red-red pairs and vice versa
novel dirge
but why would I even want to account for the other?
viscid thistle
@carmine elbow Equal to 1, yes
@viscid thistle Basically just manipulation of this equation
novel dirge
if I am trying to get the probability of 2 cards of the same colors being picked
willow bear
otherwise you go against the specification of "the same color" as in the problem
novel dirge
how?
viscid thistle
I did this to firstly get rid of the constant 1, because that makes it harder to factor
willow bear
wdym how
is it not obvious
there are 16C2 draws which consist of two blacks
these draws fit the "same color" description
and there are another 16C2 draws which consist of two reds
viscid thistle
@viscid thistle you could have made the cosines in terms of sines to operate them
@viscid thistle wouldnt you?
carmine elbow
@viscid thistle
willow bear
also, can the trig talk move to #geometry-and-trigonometry if our convo with pancake is to continue?
viscid thistle
@viscid thistle wouldnt you?
@viscid thistle There's many ways to solve it, not sure what you mean exactly but I'm sure it's possible. This is just the first thing that came to mind
Ok
willow bear
@novel dirge
viscid thistle
@carmine elbow I'll continue answering you in #geometry-and-trigonometry
novel dirge
and we are adding them together
carmine elbow
Ok @viscid thistle
viscid thistle
Yeah not sure why you're not moving your probability talk to #probability-statistics but okay.
novel dirge
there is a probability channel?
didn't know that
first time I posted about probability here, I was told that this is the one for it
viscid thistle
Yeah not sure why you're not moving your probability talk to #probability-statistics but okay.
@viscid thistle exactly
still wolf
is there a simpler way to write
obsidian monolithBOT
nokko:
viscid thistle
nope
viscid thistle
unless you want to do 2^1/2(2^1/2 + 1 )
still wolf
well actually that follows, if it's the simplest way
viscid thistle
but i don't think thats simpler
still wolf
and the sum of two irrational numbers can be rational, since
obsidian monolithBOT
nokko:
viscid thistle
yup!
tardy ridge
Math?
viscid thistle
haha, one proof at a time
tardy ridge
English is worse
still wolf
english is easy
still wolf
I learned it as a second language, and I'm really glad I'm able to speak fluently and get good marks
viscid thistle
do you need help with something? @tardy ridge
tardy ridge
No.
viscid thistle
:p
eager path
are there infinite amounts of parabolas that can be formed that goes through point(2,2) and focus (-2,-1)?
full jungle
Yes Einstein discovered that therum in 1780
viscid thistle
Lmao
upper kelp
If I'm not mistaken - yes.
A parabola is defined as the set of points whose distance from a focus F is the same as the distance to a guide p. (a line)
In this case, once you know what the point A(2, 2) on the parabola is and the focus F(-2, -1) is, then you already know what the distance between the point A (2, 2) and the line is. That distance will be equal to AF, which is:
√(2 - (-2))² + (2 - (-1))² = √4² + 3² = √(16 + 9) = √25 = 5.
However, you have the "freedom" to choose the "angle" of the line with respect to the point A.
Each such line will define a different parabola.
Here's one example of such parabola, where the guide is x = 7, and is 5 units away from the point A(2, 2).
Here's an additional example, where this time, the guide is:
p: 3x + 4y - 39 = 0.
hard hornet
what size width will the intervals be?
viscid thistle
this is all i got
viscid thistle
I’m a bit late, but if you know how to graph the parent function, translate it to find the new function or the function in question.
This is the standard form of one of these functions—
Sqrt(x-h) + k
(No dilations or stretches, just translations)
modern prairie
if a function is infinitesimal so its domain is:
[-∞,∞]
or
(-∞,∞)
?
viscid thistle
What does infinitesimal mean? @modern prairie
modern prairie
reaching infinty
willow bear
uh
viscid thistle
Isnt it
willow bear
[-∞, +∞] is normally not talked about at all, unless the words "extended real number line" have been brought up in your class
which it's very likely they haven't
the real number line is (-∞, +∞)
viscid thistle
So yeah the 2nd
upper kelp
By the way, if you're not sure, I suppose you can write ℝ, which is the symbol for the set of all real numbers.
willow bear
i don't know what they meant by that @viscid thistle