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any math wizzes in the house?
    #8944867 -

the owner of a candy store wants to sell, for $6 per pound, a mixture of choclate-covered raisins,which usually sells for $3 per pound, and choclate-covered macadamia nuts, which usually sells for $8 per pound. they have a 40 pound barrel of the raisins. how many pounds of the nuts should they mix with the barrel of raisins so that they hit their target value of $6 per pound for the mixture?.....it just makes me hungry


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It's not worth doing something unless you were doing something that someone, somewhere, would much rather you weren't doing.

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Re: any math wizzes in the house? [Re: kodyshowers]
    #8944883 -

nope


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Re: any math wizzes in the house? [Re: kodyshowers]
    #8944892 -

x*3+(1-x)*8=6
-5x+8=6
x=2/5

40% raisins, 60% macademia

So for a 40 lbs barrel of raisins, they need 60 lbs macademia

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Re: any math wizzes in the house? [Re: Entropymancer]
    #8944898 -

Entropymancer said:
x*3+(1-x)*8=6
-5x+8=6
x=2/5



*cough* *cough* show off *cough* *cough* :biggrin:


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Re: any math wizzes in the house? [Re: freeDOOM]
    #8944926 -

nice work Entropymancer
thanks


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It's not worth doing something unless you were doing something that someone, somewhere, would much rather you weren't doing.

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Re: any math wizzes in the house? [Re: kodyshowers]
    #8945049 -

Sorry, *cough* are you 18?


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Every Animal Is Someone 🌱 watchdominion.org

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Re: any math wizzes in the house? [Re: ivi]
    #8945077 -

ivi said:
Sorry, *cough* are you 18?



it was a joke, buddy.

calm down.


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Re: any math wizzes in the house? [Re: ivi]
    #8945090 -

Why don't they just through a random mixture in the barrel and sell it for $8 a pound?

Guaranteed profit.


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Word to your mom.

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Re: any math wizzes in the house? [Re: BrainChemistry]
    #8945096 -

Kinda like how the iPod shuffle used randomness as a marketing feature.

Yes, I'm paying extra to NOT have a screen and be able to see which song I'm selecting.  Because suspense just adds to that music-listening factor.


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We are all in the gutter, but some of us are looking at the stars.

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Re: any math wizzes in the house? [Re: deCypher]
    #8945112 -

Hmm...on that note

Has anyone noticed iPods and iTunes tendencies to pick songs from artists that have more play counts, or higher star ratings?

I've always wondered if that is something coded into the software or if its just dumb luck or me having a distorted view of things.


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Word to your mom.

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Re: any math wizzes in the house? [Re: BrainChemistry]
    #8945134 -

BrainChemistry said:
Hmm...on that note

Has anyone noticed iPods and iTunes tendencies to pick songs from artists that have more play counts, or higher star ratings?

I've always wondered if that is something coded into the software or if its just dumb luck or me having a distorted view of things.



it does that on purpose.

because typically the more popular the song is the one people buy/are looking for.


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Re: any math wizzes in the house? [Re: freeDOOM]
    #8945261 -

freeDOOM said:
ivi said:
Sorry, *cough* are you 18?



it was a joke, buddy.

calm down.



I think he was talking to the OP, considering the question was very basic high-school-level algebra.

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Re: any math wizzes in the house? [Re: kodyshowers]
    #8945389 -

Any math wizzes in the house?

I need to prove that all even numbers greater than two are the sum of two primes.

Oh and also that the function defined as z(s) = 1/1^s + 1/2^s + 1/3^s + ..., for complex s, has all its zeroes on the line Re = 1/2.

Plzzzz my teacher is going to kill me


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Formula: 0

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Re: any math wizzes in the house? [Re: isaacein]
    #8945399 -

.


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We are all in the gutter, but some of us are looking at the stars.

Edited by deCypher (01/21/09 08:31 PM)

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Re: any math wizzes in the house? [Re: deCypher]
    #8945446 -

The Cypher said:
Yeah, when I entered this thread I at least expected something in calculus or abstract algebra.  For shame.

And every even number greater than 4 is the sum of at most six primes, if that helps any.  :wink:



That must be as close as we've come to solving it. Who's it by?

I'm thinking it might just be one of those undecidable propositions :smile:.

I've got one for you, Cypher. This one I made up just for you. Can you prove that if n>10, pk is the greatest prime not exceeding n, and the product

p1*(n-p1)*p2*(n-p2)*...*pk*(n-pk)

does not divide n!, then n is the sum of two primes?


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Formula: 0

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Re: any math wizzes in the house? [Re: isaacein]
    #8945515 -

Olivier Ramare discovered the four primes fact.  Have you read Uncle Petros and Goldbach's Conjecture?  Quite an excellent book, IMO.

And God, I have no idea on where to start on that problem.  Presumably you'd start by assuming n - all prime numbers below n does not equal another prime and trying to find a contradiction, but I'm at a loss at what to do after this step.

How would you solve it?


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We are all in the gutter, but some of us are looking at the stars.

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Re: any math wizzes in the house? [Re: deCypher]
    #8945535 -

ive been on algebra level math for 3 years

pm me for assistance


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for all y’all niggaz out there that be puffin shit
when the music go on, y’all listen to this alright

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Re: any math wizzes in the house? [Re: deCypher]
    #8945536 -

Any math wizzes in the house?

if you have 28 apples, john has 16, john take 13 apples, how many apples do you have?

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Re: any math wizzes in the house? [Re: Prisoner#1]
    #8945537 -

0 appples and one testicle


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for all y’all niggaz out there that be puffin shit
when the music go on, y’all listen to this alright

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Re: any math wizzes in the house? [Re: deCypher]
    #8945543 -

It's only a matter of noticing that if n>10, less than half of the numbers <n are primes. Thus there are less than n terms in the product, and they are all <n. But obviously if they are all distinct, the product divides n!. Hence they are not all distinct, and there exist h and j such that

ph = n-pj

n = ph+pj

Q.E.D

:smile:


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Formula: 0

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