Engage with knowledgeable experts and get accurate answers on IDNLearn.com. Our community provides accurate and timely answers to help you understand and solve any issue.

In 1990, Jamaica's population was 2,466,000 and was expected to grow exponentially by 1.1% each year. What would you expect the population to be in 2010?

A. [tex]\( 7.03 \times 10^{15} \)[/tex]
B. [tex]\( 8.75 \times 10^{15} \)[/tex]
C. 3,069,136
D. 16,590,015


Sagot :

To determine the population of Jamaica in 2010, given it was expected to grow exponentially by 1.1% each year since 1990, we follow these steps:

1. Initial Population and Growth Rate:
- Initial population in 1990: [tex]\(2,466,000\)[/tex]
- Annual growth rate: [tex]\(1.1\%\)[/tex]

2. Duration:
- Number of years between 1990 and 2010: [tex]\(2010 - 1990 = 20\)[/tex] years

3. Exponential Growth Formula:
- The exponential growth formula is [tex]\( P(t) = P_0 \times (1 + r)^t \)[/tex]
- [tex]\( P(t) \)[/tex] is the population at time [tex]\( t \)[/tex]
- [tex]\( P_0 \)[/tex] is the initial population
- [tex]\( r \)[/tex] is the growth rate (per time period)
- [tex]\( t \)[/tex] is the number of time periods

4. Substitute and Calculate:
- [tex]\( P_0 = 2,466,000 \)[/tex]
- [tex]\( r = 0.011 \)[/tex] (since [tex]\(1.1\%\)[/tex] as a decimal is [tex]\(0.011\)[/tex])
- [tex]\( t = 20 \)[/tex]

Now we plug these values into the formula:

[tex]\[ P(20) = 2,466,000 \times (1 + 0.011)^{20} \][/tex]

- First, we calculate [tex]\( (1 + 0.011) \)[/tex]:

[tex]\[ 1 + 0.011 = 1.011 \][/tex]

- Then, we raise [tex]\( 1.011 \)[/tex] to the power of 20:

[tex]\[ 1.011^{20} \approx 1.248092 \][/tex]

- Finally, multiply this result by the initial population:

[tex]\[ P(20) = 2,466,000 \times 1.248092 \approx 3,069,136 \][/tex]

So, the expected population of Jamaica in 2010 would be approximately [tex]\( 3,069,136 \)[/tex].

Therefore, the correct answer is:
[tex]\[ 3,069,136 \][/tex]