Jowo Rinpoche (FSM!!)
#help-38
solid kilnBOT
zinc ginkgo
You don't have a logarithm in your problem
glossy gale
Sorry that was the entire question but mb for not showing you entirely what I wanted to do. I wanted to find the reference point
glossy gale
You don't have a logarithm in your problem
Oh yeah lemme show my work one sec
This was my work and how I got (-2, -3)
But I don't know exactly what "reference point" even means
Google didn't really tell me either
Did I do something wrong when converting?
glossy gale
Alr one sec lemme get my computer
zinc ginkgo
So if you follow their example, yours will be very similarly done
trim joltBOT
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glossy gale
Nvm I get it now! I had ot flip (-2, -3) since it was an inverse log function @zinc ginkgo
So it's (-3, -2)
thanks for ur help
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slate wagon
Hello I've got following problem: A company mass produces lamps, of which on average 10% are defective. An electrician now buys 50 packs of 10 lamps each. Calculate the probability that exactly 40 of the packs of 10 purchased contain at most one faulty lamp. Could anyone help me to solve it.
slate wagon
I tried using double Bernoulli chain
But since the probabilities that came out had crooked values and we only have a table book to look them up - I guess that is not be the solution but feel free to correct me if I'm wrong
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viscid flower
slate wagon
<@&286206848099549185>
tepid atlas
what do you mean by crooked values?
slate wagon
For instance 0.8156382...
tepid atlas
ok so im a little bit rusty on my probability, but how would you find the probability a pack of 10 lamps has at most one faulty lamp?
if you can calculate that number, then i know how to solve it
a double bernoulli chain, after reading about it, does seem like the right approach
tepid atlas
i mean i am not up to speed on probability
like the notation or the distributions
my intuition tells me that 8% could be slightly high
but also there is some at most stuff happening that could even it out
B is for binomial distribution, with p = .1?
slate wagon
Yeah
tepid atlas
yeah actually this looks right
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winged bough
are u people aloud to do work for me
trim joltBOT
are u people aloud to do work for me
delicate jasper
no
tepid hamlet
crisp ibex
I trying to find somebody to do problem for me to win an event
Thats probably the worst thing I have ever heard
In this server from a help channel
delicate jasper
lol
winged bough
Thats probably the worst thing I have ever heard
It is for nitro
winged bough
In server
winged bough
And they casually do these events it owned by YouTuber
crisp ibex
So you're trying to cheat in an event to win nitro?
winged bough
Yep
crisp ibex
We live in a society
winged bough
We do?
How about this I give u 2 of the problems worth 1 nitro each and we split 50 50
tepid hamlet
Yes normally you dont inform someone of the intention to cheat in our society you would just ask the question, duh
crisp ibex
How about this I give u 2 of the problems worth 1 nitro each and we split 50 50
Lol this is kinda funny
It's like I do both problems
And then you win half of it
(Tho from the looks of it if each problem is nitro I might not be able to do them)
winged bough
And then you win half of it
Yes it is the handling fee
crisp ibex
Yes it is the handling fee
Send me a problem, lets see how hard these are
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glacial aurora
My linear algebra book keeps stating that "infinite linear combinations" dont exist, does anyone have any clue what this is supposed to mean
glacial aurora
Im thinking he might mean that any element in a vector space V can be described as a linear combination of the basis. (since a basis is finite (i think? havent really gotten to basis yet))
marble wharf
if you want to take "infinite sums", you first have to talk about what that means
that is, you need to have a notion of convergence
linear algebra ignores all that and just doesn't allow infinite sums
glacial aurora
hmm okey, so i can see it as an "axiom" of linear algebra
marble wharf
if you do linear algebra "with" infinite sums you end up doing functional analysis
sort of I guess
infinite sums don't have meaning without more structure on the space you are considering
and for most of a linear algebra course you just don't look at stuff like that
glacial aurora
works for me, thanks alot!
a small question again, is it worth learning about sums and direct sums of subspaces if i havent gotten to the concept of basis yet
marble wharf
sure
glacial aurora
my book talks about sums and direct sums b4 basis but i dont really get its explanation. While everything on the internet seems to explain sums/direct sums using basis
glacial aurora
sure
Cool, if its do-able ill try rereading and understanding in a few
thanks for the help!
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mint peak
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mint peak
which of these graph are eulerian ?
shrewd ridge
it's easy, count the lines at each dot
shrewd ridge
it's supposed to be all even
uncut latch
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astral lily
Prove using cardinality definition that |AUB|=|A|+|B| if A and B are both disjoint and non empty finite sets
astral lily
If we suppose |A|=m and |B|=n then |AUB|=m+n so we need to prove that there is a bijection from N(m+n) to AUB
Not sure what to do after this. Probably need to define a new function but idk how
@marble wharf from what u weee saying earlier today
I’m guessing we’d define say h(i)=f(i) for 1<i<m where f is the bijection from Nm>A
marble wharf
1<=i<=m but yes
astral lily
But we need it to be a bijection from to Nn
So <=n
So i-m
Yeah
Ok and now proving it’s a bijection is just trivial
marble wharf
yeah
velvet raptor
How do i explain why 81^3/4 equals 27?
trim joltBOT
How do i explain why 81^3/4 equals 27?
astral lily
How do i explain why 81^3/4 equals 27?
Take the 4th root of 81 (I think it’s called 4th root lol not too sure)
Then cube it
velvet raptor
Thats 3 right?
astral lily
Yes sir
cerulean hemlock
$x^\frac{a}{b}= \sqrt[b]{x^{a}}$
solid kilnBOT
Nonna
velvet raptor
So i would just say something like “the 4th root of 81 cubed equals 27”?
I think i got it thanks
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wraith hinge
Hey guys
wraith hinge
In this question we have to find f(n) in each case, why does the solution in the first problem write the form as
an=(an+bn+c)2^n
And not an=(an+bn+c)K2^n
This table shows the forms
Is it because it already is getting multiplied by a, b and c?
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wraith hinge
Nope
wise dagger
A simple yes for your question.
For example
(an²+bn+c)K = Kan²+Kbn+Kc
Can be transformed into
a'n²+b'n+c'
By letting a'=Ka, b'=Kb and c'=Kc
For details, you can search for "dummy variables"
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wraith hinge
hello there,
first about me: iam a software developer near a deadline about a problem ive been working on with my team for amolst a week now we dont manage to solve (i would be surprised if someone knows the answer due to its large complexity)
problem: we all dont know much about math we all dropped out of school and just started our coding startup xD
mathematical problem:
A) example: you have number y of things number x of ppl. (short for people) and number z representing lets say a "rank" or a "role" for each x with the lowest z (z can be used for multiple x with same value) gets the equal amount of a (a is the output) then the x with the next highest z gets 5% more on value of a than the x compared to previous x and at the end every a must be equal or lower than the original y
B) restrictions: (doest really matter for you but for us as devs) that its a fast algorithm in terms of using this equation or algebra function or whatever its going to be to calcuate things we need in our production process
anyone here might have an idea?
what we tried alrdy was going with an neural network we train based on hand made random numbers who were matching our needed output but it gove very inaccurate numbers (only ints no floats as numbers)
in addition: please ping me if there are questions or solutions
trim joltBOT
hello there,
first about me: iam a software developer near a deadline about a pr...
@wraith hinge Has your question been resolved?
wraith hinge
<@&286206848099549185>
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@wraith hinge Has your question been resolved?
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@wraith hinge Has your question been resolved?
lean birch
Here I will look at yours while I wait for mine.
y Items
x People
z Ranks
Every Person gets a Rank
Each rank that is higher then others gets 5% more items then any previous ranks
You want a formula that divides up the items to fulfill the rank requirement correct?
Best way I can think of is Figure out all rank values
ranks = {1.05^0, 1.05^1, 1.05^2, ... 1.05^(z-1)}
then multiple each rank by the number of people in that Rank
Then sum those ranks
divide number of items by that sum
then take that value and mulitply it back into the ranks
you each how many item each rank should get
if you don't want to go over and whole numbers then you would floor them
I'll put example code in python
def main() -> None:
items = 10000
people = [5, 10, 5] # 5 rank 0, 10 rank 1, 5 rank 2
ranks = [1]
total = 0
# Auto generate ranks value
for i in range(len(people) - 1):
ranks.append(ranks[i] * 1.05)
# Figure out total we need to divide by to get items worth
for i in range(len(ranks)):
total += ranks[i] * people[i]
item_value = items / total
# reapply items worth to ranks
for i in range(len(ranks)):
ranks[i] = round(ranks[i] * item_value - 0.2) # Rounding Fix, could just floor
print("Per Person Items in Ranks: ", ranks)
print("Total Items in Rank: ", [a * b for a, b in zip(ranks, people)])
print("Total Items used: ", sum([a * b for a, b in zip(ranks, people)])
)
output:
Per Person Items in Ranks: [476, 500, 524]
Total Items in Rank: [2380, 5000, 2620]
Total Items used: 10000
flooring them gives you 9,985 items of 10k used
@wraith hinge
lean birch
feel free to dm me if you have more questions. I program for multiple companies
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wraith hinge
<@456226577798135808>
thanks alot iam going to try this out tomorrow till there i leave this open in case theres something wrong
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