Question

In a lab experiment, 80 bacteria are placed in a petri dish. The conditions are such that the number of bacteria is able to double every 16 hours. Write a function showing the number of bacteria after tt hours, where the hourly growth rate can be found from a constant in the function. Round all coefficients in the function to four decimal places. Also, determine the percentage of growth per hour, to the nearest hundredth of a percent.

Answer 1

Answer:

The hourly growth rate is of 4.43%.

The function showing the number of bacteria after t hours is $P(t) = 80(1.0443)^t$

Step-by-step explanation:

Equation of population growth:

The equation for the population 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 growth rate, as a decimal.

The conditions are such that the number of bacteria is able to double every 16 hours.

This means that $P(16) = 2P(0)$. We use this to find r.

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

$2P(0) = P(0)(1+r)^{16}$

$(1+r)^{16} = 2$

$\sqrt[16]{(1+r)^{16}} = \sqrt[16]{2}$

$1 + r = 2^{\frac{1}{16}}$

$1 + r = 1.0443$

$r = 1.0443 - 1$

$r = 0.0443$

The hourly growth rate is of 4.43%.

80 bacteria are placed in a petri dish.

This means that $P(0) = 80$.

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

$P(t) = 80(1+0.0443)^t$

$P(t) = 80(1.0443)^t$

The function showing the number of bacteria after t hours is $P(t) = 80(1.0443)^t$