Sure! Let's walk through the process of deciding which model accurately represents the number of toys produced by the factory given an initial production and a yearly increase rate:
1. Initial Production:
- The factory produces 1,250,000 toys each year. This can be written as 1.25 million toys (since 1,000,000 is one million).
2. Yearly Increase Rate:
- The production of toys increases by 150% each year.
3. Understanding the Increase:
- A 150% increase implies that next year's production will be 2.5 times this year's production (because the original amount plus 150% of it results in 100% + 150% = 250% or 2.5 times the initial amount).
4. Choosing the Model:
- We need to model this as an exponential growth problem, where the new number of toys produced each year is a multiple (in this case, 2.5 times) of the previous year's production.
- The general formula for exponential growth is:
[tex]\[
n = \text{initial}_\text{value} \times (\text{growth}_\text{factor})^t
\][/tex]
- This translates to:
[tex]\[
n = 1.25 \times (2.5)^t
\][/tex]
where [tex]\( n \)[/tex] is the number of toys produced in millions, and [tex]\( t \)[/tex] is the number of years since the initial production.
5. Evaluating Given Options:
- The given models for [tex]\( n \)[/tex] are:
1. [tex]\( n = \frac{2.5(1.5)}{t}, \; t = 0 \)[/tex]
2. [tex]\( n = 15 t^2 + 1.25 \)[/tex]
3. [tex]\( n = 1.5 t + 1.25 \)[/tex]
4. [tex]\( n = 1.25 \cdot 2.5^t \)[/tex]
6. Matching Our Model:
- Looking at our derived model:
[tex]\[
n = 1.25 \cdot 2.5^t
\][/tex]
- This matches option 4 exactly.
Therefore, the correct model to find the number of toys [tex]\( n \)[/tex] (in millions) being produced in [tex]\( t \)[/tex] years is:
[tex]\[
n = 1.25 \cdot 2.5^t
\][/tex]
