a. Developing an Exponential Model:
To model the rate of inflation, we can use the concept of exponential growth. The general form of the exponential growth equation is:
[tex]\[ y = P (1 + r)^t \][/tex]
where:
- [tex]\( y \)[/tex] is the final amount ([tex]$155 in 2009),
- \( P \) is the initial amount ($[/tex]130 in 2004),
- [tex]\( r \)[/tex] is the rate of inflation,
- [tex]\( t \)[/tex] is the number of years.
From the question, we know that the shoes cost [tex]$130 in 2004 and $[/tex]155 in 2009. Therefore, we have:
[tex]\[ 155 = 130 (1 + r)^{2009 - 2004} \][/tex]
[tex]\[ 155 = 130 (1 + r)^5 \][/tex]
First, solve for [tex]\( (1 + r) \)[/tex]:
[tex]\[ \frac{155}{130} = (1 + r)^5 \][/tex]
[tex]\[ 1.1923 = (1 + r)^5 \][/tex]
To solve for [tex]\( r \)[/tex], we take the fifth root of both sides:
[tex]\[ 1 + r = \left(1.1923\right)^{1/5} \approx 1.0358 \][/tex]
Thus, the rate of inflation [tex]\( r \)[/tex] is:
[tex]\[ r = 1.0358 - 1 \approx 0.0358 \][/tex]
So, the exponential model describing the inflation rate is:
[tex]\[ y = 130 (1.0358)^x \][/tex]
b. Estimating the Price in 2014:
Now, we need to estimate the price of the shoes in 2014 using the exponential model developed in part (a).
Using the model [tex]\( y = 130 (1.0358)^x \)[/tex], we need to find the price when [tex]\( x \)[/tex] is the number of years from 2004 to 2014:
[tex]\[ x = 2014 - 2004 = 10 \][/tex]
Substitute [tex]\( x = 10 \)[/tex] into the model:
[tex]\[ y = 130 (1.0358)^{10} \][/tex]
Using the calculation, we get:
[tex]\[ y \approx 130 \times 1.348 \][/tex]
[tex]\[ y \approx 175.32 \][/tex]
Therefore, the estimated price of the shoes in 2014, given the continuous rate of inflation, is approximately:
[tex]\[ \$ 184.81 \][/tex]
