Thread: Exam tomorrow need to check answers on growth and decay

6) The radioactive isotope uranium 235 has a half-life of 8.8x10^8 years.
A) How much of a 1-gram sample will decay after 1000 years?
B) How long will it take for 90% of the mass to decay?

7) A bacteria culture is known to grow at a rate proportional to the amount present. After 1 hour, 1000 bacteria are present and after 4 hours, 3000 bacteria are present.
A) find an expression for the number of bacteria at any time T.
B) Find the number of bacteria that will be present after 5 hours.
C) When will the population reach 20,000?
D) How long does it take for the population to triple?

I did these out and i want to check my answers with anyone here
My answers:
6a) .99 or ~1
6b) 2.9x10^9 years

7a) Q(t)=1000e^(ln3/3)t
b) 6240 bacterias after 5 hours
c) 8.18 hours
d) 3 hours

thank you a lot

Originally Posted by tjnguyen

Ok i made a post about this on the pre-calc forum but I am in a hurry to get these checked because i have a exam tomorrow.



6) The radioactive isotope uranium 235 has a half-life of 8.8x10^8 years.

A) How much of a 1-gram sample will decay after 1000 years?

B) How long will it take for 90% of the mass to decay?



7) A bacteria culture is known to grow at a rate proportional to the amount present. After 1 hour, 1000 bacteria are present and after 4 hours, 3000 bacteria are present.

A) find an expression for the number of bacteria at any time T.

B) Find the number of bacteria that will be present after 5 hours.

C) When will the population reach 20,000?

D) How long does it take for the population to triple?



I did these out and i want to check my answers with anyone here

My answers:

6a) .99 or ~1

6b) 2.9x10^9 years



7a) Q(t)=1000e^(ln3/3)t

b) 6240 bacterias after 5 hours

c) 8.18 hours

d) 3 hours



thank you a lot

6(a) ... look carefully at what the question is asking.

6(b) is fine.


7. your equation is fine if the population is 1000 at t = 0 ... however, the problem statement says P = 1000 at t = 1. what can you easily do to correct your equation? (think horizontal shift)

would i have to convert 1gram to kg? so .001kg?

would it be .0001 for 6a?

and for 7 im still stumped, can you help me?

ok idk if this is right for 7a) but is the initial amnt bacteria would be 693? because i found K first its (ln3)/3 through and i found k through (((1000/(e^k)) * e^(4k)), and 1000/e^k = initial, plugged in 1000/e^((ln3)/3)) and got 693