there are 3 amoebas in a jar. they double every minute. if the jar is completely filled in 20 minutes, when is the jar half full?
dont know hwo to set up the problem
a_n=a_1r^(n-1)
2007-01-03 15:03:58 · 8 answers · asked by rose64190 1 in Science & Mathematics ➔ Mathematics
more general formula is: a_n = a_m × r^(n-m).
a_20 = 3 × 2^(20-1) = 3 × 2^(19) = 1572864
The jar is full when there are 1572864 amoebas in it.
The jar is ½ full when there are 1572864 / 2 = 786432 amoebas in it.
a_n = 786432 = 3 × 2^(n-1), solve for n.
n = 19 minutes
alternative (easier) way:
since the amoebas double every minute, the one-minute-before amount of amoebas will constitute ½ jar.
so 20-1 = 19 minutes.
# of amoebas = 3 × 2^(19) = 786432.
2007-01-03 15:07:06 · answer #1 · answered by Esse Est Percipi 4 · 2⤊ 0⤋
19 min
This one was so simple you didn't need the formula. But the formula you cited is useful for figuring exponential grown and decay in general, so let's see how it would apply here.
Here,
a_n represents your amount at time n and a_1 represents your initial amount. Time is measured in units of n (years, minutes, seconds, it doesn't matter, as long as you're consistent), and r represents the rate of increase, in tihs case, 2, because your amoebas are doubling.
So here, a_n = 3x(2^(n-1))
at 20 minutes you have 3 x 2^19 bacteria which according to the info given, fills the jar. At 19 minutes you have half that amount which is 3x(2^18).
This formula gives you the amount at time n, it does not give you the time till the jar is some fraction of the way full. For that you'd have to work some algebra on it. Something like this...
(a_n)/(a_1) = r^(n-1)
ln((a_n)/(a_1)) = n - 1
1 + ln((a_n)/(a_1)) = n
So, for example, if you needed to know how long it takes for the quantity to increase by a factor of 5, you'd do
n = 1 + ln (5/3)
This form of the formula is useful for determining half life, etc.
2007-01-03 23:20:20 · answer #2 · answered by Joni DaNerd 6 · 0⤊ 0⤋
The previous poster is correct. Don't worry about fitting the question to a formula. Just realize that, if there are n amoeba in a full jar, then one minute earlier there had to be n/2 amoeba in the jar. You DON'T need to calculate n -- this is unnecessary to solve the problem, and in a test situation just wastes a lot of time.
2007-01-03 23:11:49 · answer #3 · answered by Anonymous · 1⤊ 0⤋
If the jar is full in 20 minutes, it must be half full in 19 minutes because they double every minute.
What you want is a formula for half-life, but in this problem the answer is too obvious for much math.
2007-01-03 23:11:31 · answer #4 · answered by xaviar_onasis 5 · 0⤊ 0⤋
the jar is half full in 19 minutes
2007-01-03 23:09:41 · answer #5 · answered by naynay1852 2 · 0⤊ 0⤋
it will be half full before it becomes double to completely fill the jar.. It will be half full 1 mt before it is completely full hence the jar will be half full at 19 mts...
2007-01-03 23:20:58 · answer #6 · answered by Eshwar 3 · 0⤊ 0⤋
19 minutes
2007-01-03 23:09:15 · answer #7 · answered by frank 5 · 1⤊ 0⤋
r= 6/3=2
Sn= (3-3^20)/-1
Sn= 3145725
half jar full =3145725/2
calculate n using
-1572862.5=(3-3*2^x)/
524288.5=2^x
solve for x
2007-01-03 23:23:22 · answer #8 · answered by Suhas 2 · 0⤊ 0⤋