1. 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
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

2. 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

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)