exercises\n1. vincent made a $2,000 deposit into an account on august 1 that yields 2% interest compounded…

exercises\n1. vincent made a $2,000 deposit into an account on august 1 that yields 2% interest compounded annually. how much money will be in that account at the end of 5 years?\nb = 2000(1 + \\frac{0.02}{1})^{5}=2,208.16\n2. on december 31, juan carlos made a $7,000 deposit in an account that pays 0.9% interest compounded semi - annually. how much will be in that account at the end of two years?\n3. liam was born on october 1, 2009. his grandparents put $20,000 into an account that yielded 3% interest compounded quarterly. when liam turns 18, his grandparents will give him the money for a college education. how much will liam get on his 18th birthday?\n4. colleen is 15 years from retiring. she opens an account at the savings bank. she plans to deposit $10,000 each year into the account, which pays 1.7% interest, compounded annually.\na. how much will be in the account in 15 years?\nb. how much interest would be earned?\n5. anton opened an account at bradley bank by depositing $1,250. the account pays 2.325% interest compounded monthly. he deposits $1,250 each month for the next two years.\na. how much will he have in the account at the end of the two - year period?\nb. write the future value function. let x represent each of the monthly interest periods.\nc. graph the future value function.\nd. using your graph, what will the approximate balance be after one year?\n6. sylvia wants to go on a cruise around the world in 5 years. if she puts $50 into an account each week that pays 2.25% interest compounded weekly, how much will she have at the end of the five - year period?

exercises\n1. vincent made a $2,000 deposit into an account on august 1 that yields 2% interest compounded annually. how much money will be in that account at the end of 5 years?\nb = 2000(1 + \\frac{0.02}{1})^{5}=2,208.16\n2. on december 31, juan carlos made a $7,000 deposit in an account that pays 0.9% interest compounded semi - annually. how much will be in that account at the end of two years?\n3. liam was born on october 1, 2009. his grandparents put $20,000 into an account that yielded 3% interest compounded quarterly. when liam turns 18, his grandparents will give him the money for a college education. how much will liam get on his 18th birthday?\n4. colleen is 15 years from retiring. she opens an account at the savings bank. she plans to deposit $10,000 each year into the account, which pays 1.7% interest, compounded annually.\na. how much will be in the account in 15 years?\nb. how much interest would be earned?\n5. anton opened an account at bradley bank by depositing $1,250. the account pays 2.325% interest compounded monthly. he deposits $1,250 each month for the next two years.\na. how much will he have in the account at the end of the two - year period?\nb. write the future value function. let x represent each of the monthly interest periods.\nc. graph the future value function.\nd. using your graph, what will the approximate balance be after one year?\n6. sylvia wants to go on a cruise around the world in 5 years. if she puts $50 into an account each week that pays 2.25% interest compounded weekly, how much will she have at the end of the five - year period?

Answer

2.

Explanation:

Step1: Identify compound - interest formula

The compound - interest formula for semi - annual compounding is $A = P(1+\frac{r}{n})^{nt}$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal), $n$ is the number of times compounded per year, and $t$ is the number of years. Here, $P=$7000$, $r = 0.009$, $n = 2$, and $t = 2$.

Step2: Substitute values into formula

$A=7000(1 +\frac{0.009}{2})^{2\times2}=7000(1 + 0.0045)^{4}$. First, calculate $(1 + 0.0045)^{4}=1.0045^{4}\approx1.01813$. Then, $A = 7000\times1.01813=$7126.91$.

Answer:

$7126.91$

3.

Explanation:

Step1: Determine values for formula

Liam was born on October 1, 2009, and will turn 18 on October 1, 2027. The principal $P=$20000$, the annual interest rate $r = 0.03$, the number of times compounded per year $n = 4$ (quarter - ly compounding), and the number of years $t=18$.

Step2: Use compound - interest formula

The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$. Substitute the values: $A = 20000(1+\frac{0.03}{4})^{4\times18}$. First, calculate $\frac{0.03}{4}=0.0075$ and $4\times18 = 72$. Then $(1 + 0.0075)^{72}$. Using a calculator, $(1 + 0.0075)^{72}\approx1.71257$. So, $A=20000\times1.71257=$34251.40$.

Answer:

$34251.40$

4a.

Explanation:

Step1: Identify the annuity formula

The future - value of an ordinary annuity formula is $FVA=A\frac{(1 + r)^{n}-1}{r}$, where $A$ is the annual payment, $r$ is the interest rate per period, and $n$ is the number of periods. Here, $A = 10000$, $r=0.017$, and $n = 15$.

Step2: Calculate the future value

First, calculate $(1 + 0.017)^{15}$. Using a calculator, $(1 + 0.017)^{15}\approx1.28324$. Then $(1 + 0.017)^{15}-1\approx0.28324$. $\frac{(1 + 0.017)^{15}-1}{0.017}=\frac{0.28324}{0.017}\approx16.6612$. $FVA=10000\times16.6612=$166612$.

Answer:

$166612$

4b.

Explanation:

Step1: Calculate total deposits

The total amount of deposits made over 15 years is $10000\times15=$150000$.

Step2: Find interest earned

The future value of the account is $166612$ (from part a). The interest earned $I$ is the future value minus the total deposits. So, $I=166612 - 150000=$16612$.

Answer:

$16612$

5a.

Explanation:

Step1: Identify relevant formula

We use the future - value of a compound - interest and annuity combination formula. The initial deposit is $P_0 = 1250$, the monthly deposit $A=1250$, the monthly interest rate $r=\frac{0.02325}{12}=0.0019375$, and the number of months $n = 24$. The future value of the initial deposit is $P_1=P_0(1 + r)^{n}=1250(1 + 0.0019375)^{24}$. The future value of the annuity is $A\frac{(1 + r)^{n}-1}{r}=1250\frac{(1 + 0.0019375)^{24}-1}{0.0019375}$.

Step2: Calculate future values

First, $(1 + 0.0019375)^{24}\approx1.0477$. $P_1=1250\times1.0477 = 1309.625$. For the annuity part, $(1 + 0.0019375)^{24}-1\approx0.0477$, $\frac{(1 + 0.0019375)^{24}-1}{0.0019375}=\frac{0.0477}{0.0019375}\approx24.61$. $A\frac{(1 + r)^{n}-1}{r}=1250\times24.61 = 30762.5$. The total future value $A_{total}=1309.625+30762.5=$32072.125\approx32072.13$.

Answer:

$32072.13$

5b.

Explanation:

Step1: Write the general form

The future - value function for the account with an initial deposit $P_0 = 1250$ and monthly deposit $A = 1250$ and monthly interest rate $r=0.0019375$ is $F(x)=1250(1 + 0.0019375)^{x}+1250\frac{(1 + 0.0019375)^{x}-1}{0.0019375}$, where $x$ is the number of monthly interest periods.

Answer:

$F(x)=1250(1 + 0.0019375)^{x}+1250\frac{(1 + 0.0019375)^{x}-1}{0.0019375}$

6.

Explanation:

Step1: Determine values for the formula

There are $n = 5\times52=260$ weeks in 5 years. The weekly deposit $A = 50$, the annual interest rate $r = 0.0225$, so the weekly interest rate $i=\frac{0.0225}{52}\approx0.0004327$. The initial deposit is assumed to be $0$. We use the future - value of an ordinary annuity formula $FVA=A\frac{(1 + i)^{n}-1}{i}$.

Step2: Calculate the future value

First, calculate $(1 + 0.0004327)^{260}$. Using a calculator, $(1 + 0.0004327)^{260}\approx1.1219$. Then $(1 + 0.0004327)^{260}-1\approx0.1219$. $\frac{(1 + 0.0004327)^{260}-1}{0.0004327}=\frac{0.1219}{0.0004327}\approx281.72$. $FVA=50\times281.72=$14086$.

Answer:

$14086$