A sample of bacteria covers an area of 1 cm² in a petri dish and grows by 50% each day. In another petri dish a sample of mold covers an area of 8 cm² and shrinks by 10% each day. Which sample covers more area after 8 days and by how much more?
A sample of bacteria covers an area of 1 cm² in a petri dish and grows by 50% each day. In another petri dish a sample of mold covers an area of 8 cm² and shrinks by 10% each day. Which sample covers more area after 8 days and by how much more?
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![### Problem Statement:
**Context**: A study is conducted on the growth and shrinkage of bacteria and mold samples in petri dishes over time.
**Description**:
- A sample of **bacteria** starts by covering an area of \( 1 \, \text{cm}^2 \) and grows at a rate of **50% per day**.
- In another petri dish, a sample of **mold** starts with an area of \( 8 \, \text{cm}^2 \) and shrinks at a rate of **10% per day**.
### Task:
Calculate the area covered by each sample after **8 days**. Determine which sample covers a larger area and the difference in area between the two.
### Options:
- \(\bigcirc \) mold, \(3.7 \, \text{cm}^2 \)
- \(\bigcirc \) bacteria, \(3.7 \, \text{cm}^2 \)
- \(\bigcirc \) mold, \(40.4 \, \text{cm}^2 \)
- \(\bigcirc \) bacteria, \(40.4 \, \text{cm}^2 \)
---
### Detailed Solution:
To solve this problem, we need to calculate the area for both mold and bacteria after 8 days.
**For Bacteria:**
The bacteria’s area grows by 50% each day. This forms a geometric progression.
The area on the \( n \)th day is calculated by:
\[ A_{\text{bacteria}}(n) = A_{\text{initial}} \times (1 + \text{growth rate})^n \]
\[ A_{\text{bacteria}}(8) = 1 \, \text{cm}^2 \times (1 + 0.5)^8 \]
\[ A_{\text{bacteria}}(8) = 1 \times 1.5^8 \]
\[ A_{\text{bacteria}}(8) \approx 25.63 \, \text{cm}^2 \]
**For Mold:**
The mold’s area shrinks by 10% each day. This forms another geometric progression.
The area on the \( n \)th day is calculated by:
\[ A_{\text{mold}}(n) = A_{\text{initial}} \times (1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4fcdecb9-56df-4c9c-aece-60df7fbe5b7a%2F2366563c-da39-4e6d-afbe-b868c26fd09b%2Fiqyf3u_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
**Context**: A study is conducted on the growth and shrinkage of bacteria and mold samples in petri dishes over time.
**Description**:
- A sample of **bacteria** starts by covering an area of \( 1 \, \text{cm}^2 \) and grows at a rate of **50% per day**.
- In another petri dish, a sample of **mold** starts with an area of \( 8 \, \text{cm}^2 \) and shrinks at a rate of **10% per day**.
### Task:
Calculate the area covered by each sample after **8 days**. Determine which sample covers a larger area and the difference in area between the two.
### Options:
- \(\bigcirc \) mold, \(3.7 \, \text{cm}^2 \)
- \(\bigcirc \) bacteria, \(3.7 \, \text{cm}^2 \)
- \(\bigcirc \) mold, \(40.4 \, \text{cm}^2 \)
- \(\bigcirc \) bacteria, \(40.4 \, \text{cm}^2 \)
---
### Detailed Solution:
To solve this problem, we need to calculate the area for both mold and bacteria after 8 days.
**For Bacteria:**
The bacteria’s area grows by 50% each day. This forms a geometric progression.
The area on the \( n \)th day is calculated by:
\[ A_{\text{bacteria}}(n) = A_{\text{initial}} \times (1 + \text{growth rate})^n \]
\[ A_{\text{bacteria}}(8) = 1 \, \text{cm}^2 \times (1 + 0.5)^8 \]
\[ A_{\text{bacteria}}(8) = 1 \times 1.5^8 \]
\[ A_{\text{bacteria}}(8) \approx 25.63 \, \text{cm}^2 \]
**For Mold:**
The mold’s area shrinks by 10% each day. This forms another geometric progression.
The area on the \( n \)th day is calculated by:
\[ A_{\text{mold}}(n) = A_{\text{initial}} \times (1
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