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<Safrole> thank you
<Safrole> my professor got stuck on this today and couldn't show it
<astrolabe_> <a^kx> is contained in <a^k>, you showed this, ok?
<Safrole> yep
<Safrole> I understand that part
<int80_h> does this channel keep logs anywhere?
<Knio1> hi.. could someone check this for me? if the average time for students writing a test is 100 minutes, normally distributed with a standard deviation of 10 minutes, what is the probability that a cl*** of 32 students will have an average time greater than 110 minutes?
<astrolabe_> <a^k> is contained in <a^ gcd(k,n)> because gcd(k,n) divides k.
<int80_h> if not , maybe I can make arrangements to hold public logs
<Safrole> gcd(k,n) divides k
<astrolabe_> Safrole: yes
<Safrole> let m = gcd(k,n) then k = qm where q is an integer
<astrolabe_> Safrole: yes. That is part of the defn of gcd.
<Safrole> explain why <a^k> is a subset of <a^gcd(k,n)> because gcd(k,n> divides k
<Safrole> explain why <a^k> is a subset of <a^gcd(k,n)> because gcd(k,n) divides k
<Safrole> can you I meant in front of explain
<Safrole> didn't mean to order :)
<astrolabe_> Safrole: If k = qm then (a^k)^p = (a^m)^(qp)
<astrolabe_> Safrole: No problem, I didn't take it like that.
<astrolabe_> The LHS is a typical element of <a^k>
<astrolabe_> The RHS is an element of <a^m>
<Safrole> let me write it down and think about it...
<MathKiD> hello dioid_
<Safrole> okay I think I understand astrolabe
<Safrole> thank you sir
<astrolabe_> Wow, I'm a sir!
<joblot> astrolabe_ : next level, excellency
<astrolabe_> :)
<astrolabe_> I have a question
<astrolabe_> I think I can guess how to find circles in the complex plane centered at the origin that contain as many gu***ian integers as possible for their size. What about if I relax the ***umption about their centres?
<f0ster> The primary coil of a transformer has 250 turns; the secondary coil has 1000 turns. An alternating current is sent through the primary coil. The EMF in the primary is 16V. What is the EMF in the secondary
<_paneb> do the vectors in a spanning set have to be independent? if so, what is the difference between a spanning set and a basis?
<RiskArb> interesting
<Mulder> no they dont
<_paneb> ok so they do not in the spanning set, but they do in the basis then?
<Mulder> a basis is a set of vectors that span a space
<Mulder> a minimum combo of vectors that is
<_paneb> ah
<RiskArb> anyone here good with geometric series
<RiskArb> or annuities
<Mulder> but there could be many basis that span a certain space
<_paneb> ok
<RiskArb> hello
<qed> just ask your question RiskArb
<RiskArb> oh ok
<_paneb> Mulder: thanks
<RiskArb> i have 35,000 and i need to pay this back over 3 years
<RiskArb> interest rate is .01 per month
<RiskArb> so 36 months = t r=.01 p =35k but instead of monthly payments i will pay it back every 6 months
<RiskArb> but interest is still compounded monthly
<RiskArb> im trying to find out what my payments would be
<RiskArb> i know the annuity formula is: a = (r(1+r)^n * P) / ( (1+r)^n - 1 )
<joblot> RiskArb : you adjust the interest rate, 12th root of i etc
<RiskArb> well the .01 is the monthly rate
<RiskArb> .12 is the annual rate
<RiskArb> of interest
<RiskArb> but im going to be making 6 equal payments
<RiskArb> 6 months apart
<joblot> RiskArb : the question is how many times a year it is compounded
<RiskArb> it is comppunded 12 times a year
<RiskArb> but paid twice a year
<joblot> RiskArb : then the annual rate is (i+1)^12-1, not .01->.12
<bah> How do you factor by grouping 4b^2-x^2+6xy-9y^2?
<RiskArb> joblot: what do you mean the annual rate
<RiskArb> the loan is compounded monthly at .01 per month
<RiskArb> it will be paid 6 times over the life of the loan
<RiskArb> only twice a year, yet it is compounded 12 times a year
<astrolabe_> bah: Factorise all but the first term to start.
<RiskArb> the principal is 35,000
<RiskArb> so i need to find 6 equal payments
<RiskArb> one payment every 6 months
<RiskArb> that will pay all interest and principal
<joblot> RiskArb : the basic idea is that if you charge 100% interest on a dollar, compounded yearly you owe $2, compounded twice per year you owe (1.5)(1.5)1 = 2.25
<RiskArb> i understand that
<RiskArb> this is compounded 12 times a year
<RiskArb> so 35000 * (1.01)^12
<RiskArb> would be the yearly interest
<RiskArb> or the principal + interest
<RiskArb> after one year
<astrolabe_> RiskArb: The interest over 6 months is (1.01)^6 -1
<astrolabe_> Think only in 6 month blocks
<RiskArb> yeah
<RiskArb> so after the first 6 months
<RiskArb> there will be 7153.21 in interest
<RiskArb> then there will be an amount that has to be paid
<RiskArb> then it is compounded again for another 6 months, monthly compounding
<RiskArb> and then another payment has to be made
<RiskArb> but the point is i need to find 6 equal payments
<RiskArb> that will pay off this loan
<RiskArb> it is easy when you pay the loan as often as it is compounded
<RiskArb> for example, its compounded monthly, you pay monthly
<RiskArb> i know how to do that
<RiskArb> now what if it is still compounded monthly, BUT i now only pay twice a year, semiannually
<astrolabe_> It is the same as if it was compounded 6 monthly
<astrolabe_> but with what I said as the interest over 6 months
<RiskArb> i guess you would then just manipulate one side of the equation
<RiskArb> the equation is
<RiskArb> a = (r(1+r)^n * P) / ( (1+r)^n - 1 )
<joblot> RiskArb : consider you owe $X, and you pay (an annuity) of $Y per interest period (and you must adjust the interest rate to the time period); then the cool thing is that the compounded account (interest on the principal) equals the value of the annuity
<RiskArb> so perhaps you make the first n= 6
<RiskArb> and the 2nd n is 36
<RiskArb> because the interest period is not equal to payment periods
<RiskArb> they are different
<RiskArb> there is 36 times that the pricipal is being compounded yet i'm only making 6 payments
<RiskArb> right?
<RiskArb> thats what im saying here
<RiskArb> i feel like no one is on the same page
<RiskArb> i feel like i ask this question
<RiskArb> and you are telling me the basics of what an annuity is
<astrolabe_> Sorry, just trying to understand why your equation is right.
<RiskArb> its ok
<RiskArb> maybe im explaining it poorly
<RiskArb> a = (r(1+r)^n * P) / ( (1+r)^n - 1 )
<RiskArb> For example.. this equation if I plug in the following: .01 = r 35000=P and n = 36
<astrolabe_> So (1+r)^n * P is what you'd pay if you payed the whole loan off at the end of the period.
<RiskArb> that would tell me how much my monthly payments would be
<RiskArb> yes
<RiskArb> so the top is $50,076.91
<RiskArb> thats how much it equals
<RiskArb> after 3 years of monthly compounding
<RiskArb> at .12 annual rate
<RiskArb> and the bottom divides it by the periods
<RiskArb> ((1+r)^n - 1 ) so i suppose i make n=6
<RiskArb> (1.01)^6 = .06152
<RiskArb> im not sure that makes any sense
<RiskArb> hmm
<astrolabe_> I get a different formula
<RiskArb> 1.06^6-1 perhaps
<astrolabe_> a = r(1+r)^n P (r-1)/(r^n-1)
<astrolabe_> Ooops, no I don't
<RiskArb> i dont have that
<RiskArb> im not sure how to manipulate my formula
<RiskArb> to achieve this answer
<astrolabe_> Ok, hang on a mo, I agree with your formula :)
<astrolabe_> Right, the first n is 36
<astrolabe_> I make the numerator $500.769
<astrolabe_> the second n is 6
<astrolabe_> So the denominator is 0.0615202
<astrolabe_> and the quotient is $8139.91
<astrolabe_> So, your total payments would be 6 lots of this: $48839.47
<RiskArb> i see
<RiskArb> wow
<RiskArb> seems like a lot of money
<RiskArb> if you take out a 35000 loan
<RiskArb> $35,000
<RiskArb> for 3 years
<RiskArb> 12% interest
<RiskArb> annually
<RiskArb> you have to pay almost $300,000
<RiskArb> wtf that doesnt seem right
<RiskArb> seems kinda crazy, no?
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