Answer:
1864 R 2
Step-by-step explanation:
16,896 feet is the distance
Answer:p
pn gv=0753vc
Step-by-step explanation:
Answer:
The half-life of the radioactive substance is 135.9 hours.
Step-by-step explanation:
The rate of decay is proportional to the amount of the substance present at time t
This means that the amount of the substance can be modeled by the following differential equation:

Which has the following solution:

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.
After 6 hours the mass had decreased by 3%.
This means that
. We use this to find r.







So

Determine the half-life of the radioactive substance.
This is t for which Q(t) = 0.5Q(0). So







The half-life of the radioactive substance is 135.9 hours.