Greg buys a home for $$ \$ 328,500 $$. His home is predicted to increase in value $$ 4 \% $$ each year. What is the predicted value of his home in 30 years? $$ \$ 994,878 $$ $$ \$ 1,003,716 $$ $$ \$ 1,065,456 $$ $$ \$ 1,158,324 $$ Question 18 (4 points) Sarah bought her car for $$ \$ 17,440 $$. It is expected to depreciate an average of $$ 14 \% $$ each year during the first 6 years. What will the approximate value of her car be in 6 years? $$ \$ 7056 $$ $$ \$ 7148 $$ $$ \$ 7236 $$ $$ \$ 7424 $$

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**Greg's Home Value in 30 Years**

1. The home’s value increases by $$4\%$$ each year. Thus, after $$n$$ years, the predicted value is given by the compound interest formula:  
   $$
   V = P \cdot (1 + r)^n
   $$
   where  
   $$P = \$328,\!500$$ (initial value),  
   $$r = 0.04$$ (annual increase rate), and  
   $$n = 30$$.

2. Substitute the values into the formula:  
   $$
   V = 328,\!500 \cdot (1.04)^{30}
   $$

3. Calculating $$ (1.04)^{30} $$ approximately:  
   $$
   (1.04)^{30} \approx 3.243
   $$

4. Multiply to find the predicted value:  
   $$
   V \approx 328,\!500 \cdot 3.243 \approx 1,\!065,\!456
   $$

5. Therefore, the predicted value of the home after 30 years is approximately  
   $$
   \$1,\!065,\!456.
   $$

---

**Sarah's Car Value in 6 Years**

1. The car depreciates by $$14\%$$ each year, so each year its value is multiplied by $$(1 - 0.14) = 0.86$$. After $$n$$ years, its value $$V$$ is given by:  
   $$
   V = P \cdot (0.86)^n
   $$
   where  
   $$P = \$17,\!440$$ and  
   $$n = 6$$.

2. Substitute the values:  
   $$
   V = 17,\!440 \cdot (0.86)^6
   $$

3. Calculate $$ (0.86)^6 $$ approximately. One approach is to use logarithms or a calculator, which gives:  
   $$
   (0.86)^6 \approx 0.404
   $$

4. Multiply to find the approximate value of the car:  
   $$
   V \approx 17,\!440 \cdot 0.404 \approx 7,\!056
   $$

5. Therefore, the approximate value of the car after 6 years is  
   $$
   \$7,\!056.
   $$

---

**Final Answers:**

- Predicted value of Greg's home in 30 years: $$ \$1,\!065,\!456 $$.
- Approximate value of Sarah's car in 6 years: $$ \$7,\!056 $$.
- Greg's home will be worth approximately \$1,065,456 in 30 years. - Sarah's car will be worth approximately \$7,056 in 6 years.

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