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.
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.
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.
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) = N0ABt 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 = N0AB x 10
Hence, AB 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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