A factory produces [tex]$1,250,000$[/tex] toys each year. The number of toys is expected to increase by about [tex]$150\%$[/tex] per year. Which model can be used to find the number of toys, [tex]n[/tex] (in millions), being produced in [tex]t[/tex] years?

A. [tex]n = \frac{2.5(1.5)}{t}, t = 0[/tex]
B. [tex]n = 1.5 t^2 + 1.25[/tex]
C. [tex]n = 1.5 t + 1.25[/tex]
D. [tex]n = 1.25 \cdot 2.5^t[/tex]



Answer :

To determine the correct model for predicting the number of toys produced over time given the initial production and the rate of increase, we'll break down the problem step-by-step.

1. Understand the Initial Values and Growth Rate:
- Initial production is [tex]\(1,250,000\)[/tex] toys per year.
- This can be expressed as [tex]\(1.25\)[/tex] million toys, for simplicity in our model.
- The production increases by [tex]\(150\%\)[/tex] each year. A [tex]\(150\%\)[/tex] increase means the production grows to [tex]\(250\%\)[/tex] of the initial value each year because [tex]\(100\% + 150\% = 250\%\)[/tex].

2. Convert the Increase Rate to a Multiplier:
- A [tex]\(250\%\)[/tex] growth means that each year the production is multiplied by [tex]\(2.5\)[/tex].

3. Formulate the Exponential Growth Model:
- The general formula for exponential growth is [tex]\( n(t) = n_0 \cdot a^t \)[/tex] where:
[tex]\( n(t) \)[/tex] is the amount after [tex]\( t \)[/tex] years.
[tex]\( n_0 \)[/tex] is the initial amount.
[tex]\( a \)[/tex] is the growth factor per time period.
[tex]\( t \)[/tex] is the number of time periods (years in this case).

- For this problem:
[tex]\( n_0 = 1.25 \)[/tex] million (the initial production in millions).
The growth factor [tex]\( a = 2.5 \)[/tex].

4. Substitute the Given Values into the Model:
- Replacing [tex]\( n_0 \)[/tex] and [tex]\( a \)[/tex] in the model, we get [tex]\( n(t) = 1.25 \cdot 2.5^t \)[/tex].

5. Check the Options Provided:
- The model [tex]\( n = 1.25 \cdot 2.5^t \)[/tex] matches our derived formula.

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]