Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Get immediate and reliable solutions to your questions from a knowledgeable community of professionals on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

A model predicts that the population of Springfield was 15,000 in 2005. The model also predicts that each year for the next 5 years, the population [tex]\( p \)[/tex] increased by [tex]\( 4\% \)[/tex] of the previous year's population. Which equation best represents this model, where [tex]\( x \)[/tex] is the number of years after 2005, for [tex]\( x \leq 5 \)[/tex]?

A. [tex]\( p = 0.96(15,000)^x \)[/tex]
B. [tex]\( p = 1.04(15,000)^x \)[/tex]
C. [tex]\( p = 15,000(0.96)^x \)[/tex]
D. [tex]\( p = 15,000(1.04)^x \)[/tex]


Sagot :

To determine which equation best represents the model's prediction for the population of Springfield given the initial values and growth rate, let's evaluate each option one by one.

Given values:
- Initial population (in 2005): [tex]\( 15,000 \)[/tex]
- Annual growth rate: [tex]\( 4\% \)[/tex]
- Number of years after 2005 for evaluation: [tex]\( x = 5 \)[/tex]

The model predicts that the population increased by [tex]\( 4\% \)[/tex] of the previous year's population each year. This implies the population each year is multiplied by [tex]\( 1.04 \)[/tex] (which is [tex]\( 100\% + 4\% = 104\% \)[/tex]).

Let's analyze each equation:

Option (A): [tex]\( p = 0.96(15,000)^x \)[/tex]

We calculate this for [tex]\( x = 5 \)[/tex]:

[tex]\[ p_A = 0.96 \times (15,000)^5 \][/tex]
[tex]\[ p_A \approx 7.29 \times 10^{20} \][/tex]

Option (B): [tex]\( p = 1.04(15,000)^x \)[/tex]

We calculate this for [tex]\( x = 5 \)[/tex]:

[tex]\[ p_B = 1.04 \times (15,000)^5 \][/tex]
[tex]\[ p_B \approx 7.8975 \times 10^{20} \][/tex]

Option (C): [tex]\( p = 15,000(0.96)^x \)[/tex]

We calculate this for [tex]\( x = 5 \)[/tex]:

[tex]\[ p_C = 15,000 \times (0.96)^5 \][/tex]
[tex]\[ p_C \approx 12,230.59 \][/tex]

Option (D): [tex]\( p = 15,000(1.04)^x \)[/tex]

We calculate this for [tex]\( x = 5 \)[/tex]:

[tex]\[ p_D = 15,000 \times (1.04)^5 \][/tex]
[tex]\[ p_D \approx 18,249.79 \][/tex]

Now, let’s compare these results with the given population increase model's prediction:

- The population starting at [tex]\( 15,000 \)[/tex] should grow every year by [tex]\( 4\% \)[/tex]. This means if [tex]\( x = 5 \)[/tex], the population should be around [tex]\( 18,249.79 \)[/tex].

From these results, it is clear that option (D) [tex]\( p = 15,000(1.04)^x \)[/tex] best represents the model's prediction for the population growth in Springfield, as it accurately calculates the population increase by [tex]\( 4\% \)[/tex] each year for five years.
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.