Answered: The doubling period of a baterial…

Trigonometry (11th Edition)

11th Edition

ISBN:9780134217437

Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Chapter1: Trigonometric Functions

Section: Chapter Questions

Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.

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Question

The doubling period of a baterial population is 1010 minutes. At time t=80t=80 minutes, the baterial population was 70000.

What was the initial population at time t=0t=0?

Find the size of the baterial population after 4 hours.

Expert Solution

Step 1

Let the population equation will be:

N(t) = N₀A^Bt where N(t) is the population at any time t and N₀ is the population at t = 0 (i.e. the initial population)

Step 2

The bacteria population doubles every 10 minutes.

Hence, when t = 10; N(t) = 2N₀

Hence, N(t) = 2N₀ = N₀A^{B x 10}

Hence, A^{B x 10} = 2 = 2¹

Hence, A = 2 and B x 10 = 1

Hence, A = 2 and B = 1/10

Hence, the population equation takes the form:

N(t) = N₀(2)^t/10

Step 3

Now when t = 80; N(t) = 70,000

Hence, 70,000 = N₀(2)^80/10 = N₀(2)⁸ = 256N₀

Hence, N₀ = 70,000 / 256 = 273

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