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Sagot :
To determine the inflation-adjusted cost of a [tex]$189,900 house in 5 years, we can use the formula for compound interest applied to inflation, which is:
\[ C(t) = C_0(1 + r)^t \]
Step-by-Step Solution:
1. Identify the given values:
- \( C_0 \) (the current cost of the house) = $[/tex]189,900
- [tex]\( r \)[/tex] (the annual inflation rate) = 3.5% = 0.035 (as a decimal)
- [tex]\( t \)[/tex] (the number of years) = 5 years
2. Plug the given values into the formula:
[tex]\[ C(5) = 189,900 \times (1 + 0.035)^5 \][/tex]
3. Calculate the inflation adjustment factor first:
[tex]\[ 1 + 0.035 = 1.035 \][/tex]
[tex]\[ 1.035^5 \][/tex]
4. Raise 1.035 to the power of 5:
[tex]\[ 1.035^5 \approx 1.187087 \][/tex]
5. Multiply this factor by the initial cost ([tex]\( C_0 \)[/tex]):
[tex]\[ C(5) = 189,900 \times 1.187087 \][/tex]
6. Perform the multiplication:
[tex]\[ 189,900 \times 1.187087 \approx 225,541.63 \][/tex]
Therefore, the inflation-adjusted cost of the [tex]$189,900 house in 5 years, when rounded to two decimal places, will be approximately $[/tex]225,541.63.
- [tex]\( r \)[/tex] (the annual inflation rate) = 3.5% = 0.035 (as a decimal)
- [tex]\( t \)[/tex] (the number of years) = 5 years
2. Plug the given values into the formula:
[tex]\[ C(5) = 189,900 \times (1 + 0.035)^5 \][/tex]
3. Calculate the inflation adjustment factor first:
[tex]\[ 1 + 0.035 = 1.035 \][/tex]
[tex]\[ 1.035^5 \][/tex]
4. Raise 1.035 to the power of 5:
[tex]\[ 1.035^5 \approx 1.187087 \][/tex]
5. Multiply this factor by the initial cost ([tex]\( C_0 \)[/tex]):
[tex]\[ C(5) = 189,900 \times 1.187087 \][/tex]
6. Perform the multiplication:
[tex]\[ 189,900 \times 1.187087 \approx 225,541.63 \][/tex]
Therefore, the inflation-adjusted cost of the [tex]$189,900 house in 5 years, when rounded to two decimal places, will be approximately $[/tex]225,541.63.
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