Each female of a certain species of fish lays 100 eggs each fall and then dies. If only 30% of these eggs mature to become egg-producing females the next year , then how many egg-producing females will descend from a single egg-producing female 2 years after she first lays eggs ?

To find the number of egg-producing females that will descend from a single egg-producing female 2 years after she first lays eggs, we need to calculate the number of eggs that mature into egg-producing females each year.

Year 1:
- The initial egg-producing female lays 100 eggs.
- 30% of these eggs mature into egg-producing females.
- Therefore, in Year 1, there will be 100 * 0.30 = 30 egg-producing females.

Year 2:
- Each of the 30 egg-producing females from Year 1 will lay 100 eggs.
- Again, 30% of these eggs will mature into egg-producing females.
- Therefore, in Year 2, there will be 30 * 100 * 0.30 = 900 egg-producing females.

So, two years after the initial egg-producing female lays eggs, there will be 900 egg-producing females that descend from her.

30% of the 100 eggs from the 1st female will mature, or 30 eggs.

Start: 1 female lays 100 eggs, and 30 mature for the next year

1 year later: 30 females lay 3000 eggs, and 900 mature for the next year

2 years later: 900 females lay 9000 eggs, and 2700 will survive

1800

amount=P(1+i)^t

P=100
i=.3
t=2