ntrog ChatGPT Math AI - Photo Math Solver wy.aleks.com/alekscgi/x/lsl.exe/1o_u-IgNsikr7j8P3jH-IB-ukBiXnxj_1SoekWMaP1As1fEhnCI79H4SSUelERg1SH4zrgehQQhBhWI4ArD9JmXW... Consumer Mathematics Fincing the periodic payment needed to meet an investment goal $$ 0 / 3 $$ Alan plans to set aside money for his young daughter's college tuition. He will deposit money in an ordinary annuity. that earns $$ 4.8 \% $$ interest, compounded quarterly. Deposits will be made at the end of each quarter. How much money does he need to deposit into the annuity each quarter for the annuity to have a total value of $$ \$ 71,000 $$ after 14 years? Do not round intermediate computations, and round your final answer to the nearest cent. If necessary, refer to the list of financial formulas. $$ \square $$

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UpStudy AI · Accepted Answer

We use the future value formula for an ordinary annuity:

$$
F = R\left(\frac{(1+i)^n - 1}{i}\right)
$$

where  
$$ F = 71000 $$ (the goal),  
$$ i = \frac{4.8\%}{4} = 0.012 $$ (quarterly interest rate), and  
$$ n = 14 \times 4 = 56 $$ (number of quarters).

Solving for $$ R $$ gives:

$$
R = F \cdot \frac{i}{(1+i)^n - 1}
$$

Substitute the known values:

$$
R = 71000 \cdot \frac{0.012}{(1.012)^{56} - 1}
$$

First, compute $$(1.012)^{56}$$. Using the exponential relation:

$$
(1.012)^{56} = e^{56\ln(1.012)}
$$

An approximate evaluation yields:

$$
\ln(1.012) \approx 0.011928, \quad 56 \ln(1.012) \approx 0.66797, \quad e^{0.66797} \approx 1.95
$$

Thus,

$$
(1.012)^{56} - 1 \approx 1.95 - 1 = 0.95
$$

Now, substitute back into the formula for $$ R $$:

$$
R \approx 71000 \cdot \frac{0.012}{0.95}
$$

Calculate the numerator:

$$
71000 \times 0.012 = 852
$$

And then:

$$
R \approx \frac{852}{0.95} \approx 896.00
$$

So, Alan needs to deposit approximately $$\$896.00$$ each quarter.
Alan needs to deposit approximately \$896.00 each quarter.

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