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.

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Answer:

The hourly growth rate is of 4.43%.

The function showing the number of bacteria after t hours is [tex]P(t) = 80(1.0443)^t[/tex]

Step-by-step explanation:

Equation of population growth:

The equation for the population after t hours is given by:

[tex]P(t) = P(0)(1+r)^t[/tex]

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 [tex]P(16) = 2P(0)[/tex]. We use this to find r.

[tex]P(t) = P(0)(1+r)^t[/tex]

[tex]2P(0) = P(0)(1+r)^{16}[/tex]

[tex](1+r)^{16} = 2[/tex]

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

[tex]1 + r = 2^{\frac{1}{16}}[/tex]

[tex]1 + r = 1.0443[/tex]

[tex]r = 1.0443 - 1[/tex]

[tex]r = 0.0443[/tex]

The hourly growth rate is of 4.43%.

80 bacteria are placed in a petri dish.

This means that [tex]P(0) = 80[/tex].

[tex]P(t) = P(0)(1+r)^t[/tex]

[tex]P(t) = 80(1+0.0443)^t[/tex]

[tex]P(t) = 80(1.0443)^t[/tex]

The function showing the number of bacteria after t hours is [tex]P(t) = 80(1.0443)^t[/tex]