Question
An animals total population is 4000 it is decreasing in numbers by 40 percent every year.
Create a sequence that describes your situation. Include at least 10 terms of
the sequence.
Create a sequence that describes your situation. Include at least 10 terms of
the sequence.
Answers
GPT-4o mini
To describe the situation mathematically, we can represent the total population of the animals as a geometric sequence. The initial population \( P_0 \) is 4000, and the population decreases by 40% each year. This means that each year, 60% of the population remains.
The first term (\( P_0 \)) of the sequence is:
\[
P_0 = 4000
\]
The common ratio (\( r \)) of the sequence, which represents the population remaining after each year, is:
\[
r = 1 - 0.40 = 0.60
\]
The sequence can be expressed as:
\[
P_n = P_0 \cdot r^n
\]
where:
- \( P_n \) is the population in year \( n \),
- \( n \) is the number of years.
Now, let's calculate the first 10 terms of the sequence:
1. \( P_0 = 4000 \)
2. \( P_1 = 4000 \cdot 0.60^1 = 4000 \cdot 0.60 = 2400 \)
3. \( P_2 = 4000 \cdot 0.60^2 = 4000 \cdot 0.36 = 1440 \)
4. \( P_3 = 4000 \cdot 0.60^3 = 4000 \cdot 0.216 = 864 \)
5. \( P_4 = 4000 \cdot 0.60^4 = 4000 \cdot 0.1296 = 518.4 \)
6. \( P_5 = 4000 \cdot 0.60^5 = 4000 \cdot 0.07776 = 310.32 \)
7. \( P_6 = 4000 \cdot 0.60^6 = 4000 \cdot 0.046656 = 186.624 \)
8. \( P_7 = 4000 \cdot 0.60^7 = 4000 \cdot 0.0279936 = 111.9744 \)
9. \( P_8 = 4000 \cdot 0.60^8 = 4000 \cdot 0.01679616 = 67.18464 \)
10. \( P_9 = 4000 \cdot 0.60^9 = 4000 \cdot 0.010077696 = 40.310784 \)
So the first 10 terms of the sequence representing the animal population over the years are approximately:
1. \( P_0 \approx 4000 \)
2. \( P_1 \approx 2400 \)
3. \( P_2 \approx 1440 \)
4. \( P_3 \approx 864 \)
5. \( P_4 \approx 518.4 \)
6. \( P_5 \approx 310.32 \)
7. \( P_6 \approx 186.624 \)
8. \( P_7 \approx 111.9744 \)
9. \( P_8 \approx 67.18464 \)
10. \( P_9 \approx 40.310784 \)
This sequence reflects the decreasing population of the animals over 10 years, showing how significantly the population declines as a result of the 40% annual decrease.
The first term (\( P_0 \)) of the sequence is:
\[
P_0 = 4000
\]
The common ratio (\( r \)) of the sequence, which represents the population remaining after each year, is:
\[
r = 1 - 0.40 = 0.60
\]
The sequence can be expressed as:
\[
P_n = P_0 \cdot r^n
\]
where:
- \( P_n \) is the population in year \( n \),
- \( n \) is the number of years.
Now, let's calculate the first 10 terms of the sequence:
1. \( P_0 = 4000 \)
2. \( P_1 = 4000 \cdot 0.60^1 = 4000 \cdot 0.60 = 2400 \)
3. \( P_2 = 4000 \cdot 0.60^2 = 4000 \cdot 0.36 = 1440 \)
4. \( P_3 = 4000 \cdot 0.60^3 = 4000 \cdot 0.216 = 864 \)
5. \( P_4 = 4000 \cdot 0.60^4 = 4000 \cdot 0.1296 = 518.4 \)
6. \( P_5 = 4000 \cdot 0.60^5 = 4000 \cdot 0.07776 = 310.32 \)
7. \( P_6 = 4000 \cdot 0.60^6 = 4000 \cdot 0.046656 = 186.624 \)
8. \( P_7 = 4000 \cdot 0.60^7 = 4000 \cdot 0.0279936 = 111.9744 \)
9. \( P_8 = 4000 \cdot 0.60^8 = 4000 \cdot 0.01679616 = 67.18464 \)
10. \( P_9 = 4000 \cdot 0.60^9 = 4000 \cdot 0.010077696 = 40.310784 \)
So the first 10 terms of the sequence representing the animal population over the years are approximately:
1. \( P_0 \approx 4000 \)
2. \( P_1 \approx 2400 \)
3. \( P_2 \approx 1440 \)
4. \( P_3 \approx 864 \)
5. \( P_4 \approx 518.4 \)
6. \( P_5 \approx 310.32 \)
7. \( P_6 \approx 186.624 \)
8. \( P_7 \approx 111.9744 \)
9. \( P_8 \approx 67.18464 \)
10. \( P_9 \approx 40.310784 \)
This sequence reflects the decreasing population of the animals over 10 years, showing how significantly the population declines as a result of the 40% annual decrease.
10 terms numbers only
GPT-4o mini
Sure! Here are the first 10 terms of the sequence (numbers only):
1. 4000
2. 2400
3. 1440
4. 864
5. 518.4
6. 310.32
7. 186.624
8. 111.9744
9. 67.18464
10. 40.310784
1. 4000
2. 2400
3. 1440
4. 864
5. 518.4
6. 310.32
7. 186.624
8. 111.9744
9. 67.18464
10. 40.310784