Question 1096195: A population of bacteria has 1175 bacteria in its starting population. The half-life of this particular type of bacteria is 11 days. (This means that every 11 days the population decreases by half.) Write a function B(d) that represents the amount of bacteria B in the population after d days.
B(d)=
1175(1/2)
In this and every exponential problem, if you want to use decimals use 7 or 8 decimal places. If that doesn't work you will have to use fractions.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the formula to use with discrete compounding is f = p * (1+r)^n
when p = 2, the future value will be 1 when the population is halved.
you get 1 = 2 * (1+r)^n
r is the rate of growth per time period.
n is the number of time periods.
in this problem, our time period is days.
divide both sides of the equation by 2 to get 1/2 = (1+r)^n
when n = 11, this formula becomes 1/2 = (1+r)^11
take the 11th root of both sides of the equation to get (1/2)^(1/11) = 1+r
subtract 1 from both sides of the equation to get (1/2)^(1/11) - 1 = r
that's your interest rate per time period.
(1/2)^(1/11) - 1 is equal to -.0610690893
that's the growth rate per day.
test this out with your problem.
when p = 1175, the formula becomes f = 1175 * (1 - .0610690893) ^ 11.
this results in f = 587.5
587.5 * 2 = 1175, therefore 587.5 is exactly 1/2 of 1175.
if you want to know how much of the bacteria is left after d days, then replace 11 by d in the formula to get:
f = 1175 * (1 - .0610690893) ^ d
when d = 11, it's half.
when d = 22, it's half again.
when d = 33, it's half again.
1175 * (1 - .0610690893) ^ 11 = 587.5
1175 * (1 - .0610690893) ^ 22 = 293.75
1175 * (1 - .0610690893) ^ 33 = 146.875
587.5 * 2 = 1175
293.75 * 2 = 587.5 * 2 = 1175
146.875 * 2 = 293.75 * 2 = 587.5 * 2 = 1175
when d = 11, the result is 1/2 of the original.
when d = 22, the result 1/4 of the original.
when d = 33, the result is 1/8 of the original.
the formula you are looking at is:
f = p * (1 - .0610690893) ^ d
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