Question

Researchers recorded that a certain bacteria population declined from 200,000 to 900 in 18 hours. At this rate of decay,

Answer 1

Answer:

9,942 bacteria were there at 10 hours.

Step-by-step explanation:

Equation for population decay:

The equation for population decay, after t hours, is given by:

$P(t) = P(0)(1-r)^t$

In which P(0) is the initial population and r is the decay rate, as a decimal.

Researchers recorded that a certain bacteria population declined from 200,000 to 900 in 18 hours.

This means that $P(0) = 200000$ and that when $t = 18, P(t) = 900$. So we use this to find r.

$P(t) = P(0)(1-r)^t$

$900 = 200000(1-r)^{18}$

$(1-r)^{18} = \frac{900}{200000}$

$\sqrt[18]{(1-r)^{18}} = \sqrt[18]{\frac{900}{200000}}$

$1 - r = (\frac{900}{200000})^{\frac{1}{18}}$

$1 - r = 0.7407$

So

$P(t) = P(0)(1-r)^t$

$P(t) = 200000(0.7407)^t$

At this rate of decay, how many bacteria was there at 10 hours?

This is P(10). So

$P(10) = 200000(0.7407)^{10} = 9941.5$

Rounding to the nearest whole number:

9,942 bacteria were there at 10 hours.