8. A certain strain of virus grows in numbers at the rate of 50% per hour. If its present population is 611,000 what will be its population count in 4 hours?

Question

8.	A certain strain of virus grows in numbers at the rate of 50% per hour. If its present population is 611,000 what will be its population count in 4 hours?

anonymous · Answer

Nt=No*e^(r*t)


Nt = Future Population 
No = Starting population 
r = growth rate 
t= number of years.

anonymous · Answer

"t" in this case i in hours not in years

anonymous · Answer

Can you please explain a little better. I am so confused.

dumbcow · Answer

the rate of growth is 50% per hour not continuously, i dont think we use "e" here it will grow too quickly
$$P _{t}=611*1.5^{t}$$
each hour it will be 150% or 1.5X the prev population

anonymous · Answer

ok. Let's do with V variables for Virus:$$V(t) = (Vo) e^{rt}$$
V(t) = Future Population of virus
Vo = Present/initial population of virus
r = growth rate (grows in numbers at the rate of )
t = how much time (in this case hours)

From the question we know that:
V(t) = we are trying to find this
Vo = 611,000
r = 50% = 0.50
t = 4 hours

hence,
$$V(4) = 611,000*e^{0.5*4}$$

anonymous · Answer

hmm... i might be wrong thou

dumbcow · Answer

your math is good, but i dont think it applies to this problem
e^rt is usually used when modeling a rate compounded continuously

anonymous · Answer

It would be something like P*(1+r/100)^t where,
P -> current population
r -> rate of growth
t -> time period

anonymous · Answer

so it would come up to 3093187.5

anonymous · Answer

ooh... i see.. you are right, it's just a basic form P = Po*(1+r)^t

anonymous · Answer

thanks everyone. this is so confusing I think I am may be seeing the light with these problems.