Question 1186153
Here's how to solve this problem:

**1. Set up the differential equation:**

Let *B(t)* be the number of bacteria at time *t*. The problem states that the rate of growth is directly proportional to the number present and the difference between A and the number present.  This can be written as:

dB/dt = k * B * (A - B)

where *k* is the proportionality constant.

**2. Use the given information to find *k*:**

We know that A = 1,000,000, and when B = 1000, dB/dt = 60.  Plugging these values into the equation:

60 = k * 1000 * (1,000,000 - 1000)
60 = k * 1000 * 999,000
k = 60 / (1000 * 999,000)
k = 6 / 999,000
k ≈ 6.006 x 10⁻⁸

**3. Find the rate of growth when B = 100,000:**

Now we want to find dB/dt when B = 100,000.  Using the same equation and the calculated value of *k*:

dB/dt = (6.006 x 10⁻⁸) * 100,000 * (1,000,000 - 100,000)
dB/dt = (6.006 x 10⁻⁸) * 100,000 * 900,000
dB/dt = 5405.4 bacteria per minute

**Answer:**

The rate of growth when there are 100,000 bacteria present is approximately 5405.4 bacteria per minute.