A doll sold for $$ \$ 296 $$ in 1976 and was sold again in 1986 for $$ \$ 411 $$. Assume that the growth in the value $$ V $$ of the collector's a) Find the value k of the exponential growth rate. Assume $$ \mathrm{V}_{0}=296 $$. $$ \mathrm{k}=0.033 $$ (Round to the nearest thousandth.) b) Find the exponential growth function in terms of t , where t is the number of years since 1976 . V(t) $$ =296 e^{0.033 t} $$ c) Estimate the value of the doll in 2009. $$ \$ \square $$ (Round to the nearest dollar.)

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To solve the problem, we will follow the steps outlined in the question.

### a) Finding the value $$ k $$ of the exponential growth rate

We know that the value of the doll in 1976 was $$ V_0 = 296 $$ and in 1986 (10 years later) it was $$ V(10) = 411 $$. The exponential growth model can be expressed as:

$$
V(t) = V_0 e^{kt}
$$

Substituting the known values into the equation for $$ t = 10 $$:

$$
411 = 296 e^{10k}
$$

To isolate $$ e^{10k} $$, we divide both sides by 296:

$$
\frac{411}{296} = e^{10k}
$$

Calculating the left side:

$$
\frac{411}{296} \approx 1.387
$$

Now, we take the natural logarithm of both sides:

$$
\ln(1.387) = 10k
$$

Calculating $$ \ln(1.387) $$:

$$
\ln(1.387) \approx 0.329
$$

Now, we can solve for $$ k $$:

$$
k = \frac{0.329}{10} \approx 0.0329
$$

Rounding to the nearest thousandth:

$$
k \approx 0.033
$$

### b) Finding the exponential growth function $$ V(t) $$

Now that we have $$ k $$, we can write the exponential growth function:

$$
V(t) = 296 e^{0.033t}
$$

### c) Estimating the value of the doll in 2009

To find the value of the doll in 2009, we need to determine $$ t $$ since 1976. The year 2009 is 33 years after 1976:

$$
t = 2009 - 1976 = 33
$$

Now we can substitute $$ t = 33 $$ into the exponential growth function:

$$
V(33) = 296 e^{0.033 \times 33}
$$

Calculating $$ 0.033 \times 33 $$:

$$
0.033 \times 33 = 1.089
$$

Now we calculate $$ e^{1.089} $$:

$$
e^{1.089} \approx 2.973
$$

Now substituting back into the equation:

$$
V(33) = 296 \times 2.973 \approx 879.768
$$

Rounding to the nearest dollar:

$$
V(33) \approx 880
$$

### Final Answers

a) $$ k \approx 0.033 $$

b) $$ V(t) = 296 e^{0.033t} $$

c) The estimated value of the doll in 2009 is $$ \$880 $$.
a) $$ k \approx 0.033 $$ b) $$ V(t) = 296 e^{0.033t} $$ c) The estimated value of the doll in 2009 is $$ \$880 $$.

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