J caught 31416 lbs of fish. The next day she caught 40% of what she caught the 1st day. If she caught exactly 40% of what she caught the prev. day, everyday for the next 20 yrs. how many lbs of fish, total, would she have caught?
2007-05-24 07:54:17 · 8 answers · asked by paul b 1 in Science & Mathematics ➔ Mathematics
This is essentially an infinite sum since the terms even after 1 year let alone 20 are negligible.
S = 31416 + 31416(.4) + 31416(.4)^2 + 31416(.4)^3 + ...
0.4S = 31416(.4) + 31416(.4)^2 + 31416(.4)^3 + ...
S - 0.4S = 0.6S = 31416
S = 31416/.6 = 52,360
2007-05-24 08:05:47 · answer #1 · answered by Astral Walker 7 · 0⤊ 0⤋
52360 lbs.
[ 31416 / (1 - 0.4) = 52360 lbs]
Dragonlord, your answer is not correct, you are adding 40% everyday, and on top of that you are exponentially increasing it. As Dr D. pretty much points out, it doesn't really matter whether you account for leap years or not because after about a month, you will be effectively adding about 0 lbs. of fish. (think about 40% of 0 lbs being 0 lbs)
2007-05-24 15:11:01 · answer #2 · answered by Megan L. 2 · 0⤊ 0⤋
Are you sure the question refers to 20 years? That seems like a long time.
If J catches only 40% of the fish she caught the day before, she will not be able to catch any fish after a while. In less than two weeks, according to my math, she will only catch 1 fish. After that it would basically be 0 fish. Pdouble-check the wording of the problem.
Easiest way to do this is use Excel.
2007-05-24 15:04:44 · answer #3 · answered by forestwizard1 6 · 0⤊ 3⤋
That essentially becomes the infinite sum of a geometric series, with a = 31416, r = 0.4
n = 365 * 20 is virtually infinity
Sum = 31416 / (1 - 0.4) = 52360 lbs
2007-05-24 14:58:21 · answer #4 · answered by Dr D 7 · 1⤊ 1⤋
On the nth day, she'll have caught 31416*1.4^(n-1) lbs. Assuming 5 leap years in the 20 year period, she be fishing for 7305 days. Therefore she'll have 31416*1.4^7304 lbs. of fish. Which is a ridiculously large number.
Edit: OK, so the series tails off, but I'll defend my statement by saying that even 31416 lbs of fish in one day is still rediculously large. If we accept that she can do that, we may as well be accept that she can catch 0.00086 lbs on the 20th day an so on. She is very mathematically precise in her job after all.
2007-05-24 15:02:53 · answer #5 · answered by dragonlord182 2 · 0⤊ 3⤋
This question was just posted and answered
on this forum!
The answer is 52360 lbs of fish.
When you solve it, don't forget to include
5 extra days for leap years!
Is this a homework problem for a class?
2007-05-24 15:07:48 · answer #6 · answered by steiner1745 7 · 0⤊ 0⤋
F = 31416(1+0.4+0.4^2++++++0.4^7299) taking that a year has 365 days
F=31416(1/(1-0.4)) = 52360 lb
I took the sum of the series to infinity as 0.4^7299 is too small
for any calculator
2007-05-24 15:05:49 · answer #7 · answered by santmann2002 7 · 0⤊ 0⤋
This question was already asked yesterday, verbatim. This is homework!!
2007-05-24 14:59:44 · answer #8 · answered by Anonymous · 1⤊ 1⤋