Math Question Please Need Help????

Question

The doubling period of a baterial population is 10 minutes. At time t = 90 minutes, the baterial population was 90000. What was the initial population at time t=0 ?

Answer 1

The population can be modelled using the following equation.
P(t) = P(initial) * 2^(t/10)

You know P(90) = 90000:
So plug this in:
P(90) = 90000 = P(initial) * 2^(90/10)

90000 = P(initial) * 2^9

Solve for P(initial):
P(initial) = 90000 / 512
P(initial) = 175.78125

Here are the details:
P(0) = 175.78125
P(10) = 351.5625
P(20) = 703.125
P(30) = 1,406.25
P(40) = 2,812.5
P(50) = 5,625
P(60) = 11,250
P(70) = 22,500
P(80) = 45,000
P(90) = 90,000

You can also do this working backwards, just dividing by 2. So at time t = 80, you have half the population (45,000) at t = 70 you have half of that (22,500), etc. Until you get to t = 0.

So the answer is:
Initial population = 175.78125 ≈ 176 bacteria

Answer 2

I got 176 as the initial population.

t90 = 90,000
t80 = 45,000
t70 = 22,500
t60 = 11,250
t50 = 5,625
t40 = 2,813
t30 = 1407
t20 = 704
t10 = 352
t0 = 176

I followed appropriate rounding rules too generate whole numbers. I suspect that the method you are supposed too use to answer this question is a half-life aquation.

Good luck with that!

Answer 3

you take 2 and do 2^9. then take 90000 and divide it by that

(90000/512) =175.78125

Answer 4

90000/2^9

= 176.781

But you cant have 0.781 of a bacteria so 177 bacteria is your answer.

Answer 5

approx. 88

err 176 i mean. accidentally used the 10th power the first time.

Answer 6

P= population
t=time

P=10t
P=0.............actually....i'll ask sum1 cuz i dont think i did that right. Im gonna ask my Bio teacher. brb