In an experiment to study the growth of bacteria, a medical student measured 5000 bacteria at time 0 and 8000 at time 10 minutes. Assuming that the number of bacteria grows exponentially, how many bacteria will be present after 30 minutes? A. 14000 bacteria B. 20480 bacteria C. 17830 bacteria D. 24332 bacteria E. 29333 bacteria
In an experiment to study the growth of bacteria, a medical student measured 5000 bacteria at time 0 and 8000 at time 10 minutes. Assuming that the number of bacteria grows exponentially, how many bacteria will be present after 30 minutes? A. 14000 bacteria B. 20480 bacteria C. 17830 bacteria D. 24332 bacteria E. 29333 bacteria
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Understanding Exponential Growth in Bacterial Populations**
In an experiment to study the growth of bacteria, a medical student measured 5000 bacteria at time 0 and 8000 at time 10 minutes.
### Problem Statement
Assuming that the number of bacteria grows exponentially, how many bacteria will be present after 30 minutes?
#### Options:
- A. 14000 bacteria
- B. 20480 bacteria
- C. 17830 bacteria
- D. 24332 bacteria
- E. 29333 bacteria
### Explanation
To determine the number of bacteria present after 30 minutes, we need to understand exponential growth. Exponential growth can be modeled by the formula:
\[ N(t) = N_0 \times e^{kt} \]
Where:
- \( N(t) \) is the number of bacteria at time \( t \).
- \( N_0 \) is the initial number of bacteria.
- \( k \) is the growth rate constant.
- \( t \) is the time.
Given:
- \( N_0 = 5000 \) (initial number of bacteria)
- \( N(10) = 8000 \) (number of bacteria at time 10 minutes)
First, we use the given data to find the growth rate constant \( k \).
\[ 8000 = 5000 \times e^{10k} \]
\[ \frac{8000}{5000} = e^{10k} \]
\[ 1.6 = e^{10k} \]
Taking the natural logarithm (ln) on both sides:
\[ \ln(1.6) = 10k \]
\[ k = \frac{\ln(1.6)}{10} \]
Now, using the value of \( k \), we calculate the number of bacteria at 30 minutes:
\[ N(30) = 5000 \times e^{30k} \]
By solving the above equation, we can find the number of bacteria after 30 minutes.
In the context of the problem, students are required to calculate \( N(30) \) using the provided formula and select the correct answer from the options. The growth options provided are estimations and students need to use logarithmic calculations skills to solve for the precise number of bacteria present at the given time.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2bf6a656-c902-430b-8919-3920c6e4552b%2Ff375d498-c12e-4109-8d87-879bb8ba2e18%2Fg24cd5o_processed.png&w=3840&q=75)
Transcribed Image Text:**Understanding Exponential Growth in Bacterial Populations**
In an experiment to study the growth of bacteria, a medical student measured 5000 bacteria at time 0 and 8000 at time 10 minutes.
### Problem Statement
Assuming that the number of bacteria grows exponentially, how many bacteria will be present after 30 minutes?
#### Options:
- A. 14000 bacteria
- B. 20480 bacteria
- C. 17830 bacteria
- D. 24332 bacteria
- E. 29333 bacteria
### Explanation
To determine the number of bacteria present after 30 minutes, we need to understand exponential growth. Exponential growth can be modeled by the formula:
\[ N(t) = N_0 \times e^{kt} \]
Where:
- \( N(t) \) is the number of bacteria at time \( t \).
- \( N_0 \) is the initial number of bacteria.
- \( k \) is the growth rate constant.
- \( t \) is the time.
Given:
- \( N_0 = 5000 \) (initial number of bacteria)
- \( N(10) = 8000 \) (number of bacteria at time 10 minutes)
First, we use the given data to find the growth rate constant \( k \).
\[ 8000 = 5000 \times e^{10k} \]
\[ \frac{8000}{5000} = e^{10k} \]
\[ 1.6 = e^{10k} \]
Taking the natural logarithm (ln) on both sides:
\[ \ln(1.6) = 10k \]
\[ k = \frac{\ln(1.6)}{10} \]
Now, using the value of \( k \), we calculate the number of bacteria at 30 minutes:
\[ N(30) = 5000 \times e^{30k} \]
By solving the above equation, we can find the number of bacteria after 30 minutes.
In the context of the problem, students are required to calculate \( N(30) \) using the provided formula and select the correct answer from the options. The growth options provided are estimations and students need to use logarithmic calculations skills to solve for the precise number of bacteria present at the given time.
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