Question
A certain type of bacteria in a lab experiment multiplies rapidly. Initially, there are 200 bacteria, and the population triples every 4 hours. The recursive formula for this growth is B(n)=B(nā4)ā
3
, where B(n)
is the number of bacteria after n hours, and B(0)=200
. If the experiment continues without interruption, how many hours will it take for the bacteria population to exceed 2,700,000? (1 point)
Responses
28 hours
28 hours
40 hours
40 hours
36 hours
36 hours
32 hours
, where B(n)
is the number of bacteria after n hours, and B(0)=200
. If the experiment continues without interruption, how many hours will it take for the bacteria population to exceed 2,700,000? (1 point)
Responses
28 hours
28 hours
40 hours
40 hours
36 hours
36 hours
32 hours
Answers
GPT-4o mini
To determine how many hours it will take for the bacteria population to exceed 2,700,000, we can use the recursive formula given for the bacteria population:
\[ B(n) = B(n-4) \cdot 3 \]
with the initial condition:
\[ B(0) = 200. \]
Since the population triples every 4 hours, we can derive a formula for \( B(n) \) when \( n \) is a multiple of 4.
If we express \( n \) in terms of multiples of 4, let \( n = 4k \) where \( k \) is a non-negative integer. Then,
\[ B(4k) = B(0) \cdot 3^k = 200 \cdot 3^k. \]
We need to find the smallest \( k \) such that \( B(4k) > 2,700,000 \):
\[
200 \cdot 3^k > 2,700,000.
\]
Dividing both sides by 200:
\[
3^k > \frac{2,700,000}{200} = 13,500.
\]
Now we need to find the smallest \( k \) such that \( 3^k > 13,500 \). We can calculate powers of 3:
- \( 3^1 = 3 \)
- \( 3^2 = 9 \)
- \( 3^3 = 27 \)
- \( 3^4 = 81 \)
- \( 3^5 = 243 \)
- \( 3^6 = 729 \)
- \( 3^7 = 2,187 \)
- \( 3^8 = 6,561 \)
- \( 3^9 = 19,683 \)
From this, we see that \( 3^8 = 6,561 \) is not sufficient, but \( 3^9 = 19,683 \) is greater than 13,500. Thus, \( k = 9 \) is the smallest integer satisfying the inequality.
Now, since \( n = 4k \):
\[
n = 4 \times 9 = 36 \text{ hours}.
\]
Therefore, it will take **36 hours** for the bacteria population to exceed 2,700,000.
The correct response is:
**36 hours**.
\[ B(n) = B(n-4) \cdot 3 \]
with the initial condition:
\[ B(0) = 200. \]
Since the population triples every 4 hours, we can derive a formula for \( B(n) \) when \( n \) is a multiple of 4.
If we express \( n \) in terms of multiples of 4, let \( n = 4k \) where \( k \) is a non-negative integer. Then,
\[ B(4k) = B(0) \cdot 3^k = 200 \cdot 3^k. \]
We need to find the smallest \( k \) such that \( B(4k) > 2,700,000 \):
\[
200 \cdot 3^k > 2,700,000.
\]
Dividing both sides by 200:
\[
3^k > \frac{2,700,000}{200} = 13,500.
\]
Now we need to find the smallest \( k \) such that \( 3^k > 13,500 \). We can calculate powers of 3:
- \( 3^1 = 3 \)
- \( 3^2 = 9 \)
- \( 3^3 = 27 \)
- \( 3^4 = 81 \)
- \( 3^5 = 243 \)
- \( 3^6 = 729 \)
- \( 3^7 = 2,187 \)
- \( 3^8 = 6,561 \)
- \( 3^9 = 19,683 \)
From this, we see that \( 3^8 = 6,561 \) is not sufficient, but \( 3^9 = 19,683 \) is greater than 13,500. Thus, \( k = 9 \) is the smallest integer satisfying the inequality.
Now, since \( n = 4k \):
\[
n = 4 \times 9 = 36 \text{ hours}.
\]
Therefore, it will take **36 hours** for the bacteria population to exceed 2,700,000.
The correct response is:
**36 hours**.