Mrs. stevens wants to have 18,000 in the bank in 3 years. if she deposits $9500 today at 4% compounded quarterly for 3 years, how much additional money will she need to add after 3 years to her investment to make her balance $18000
Mrs. stevens wants to have 18,000 in the bank in 3 years. if she deposits $9500 today at 4% compounded quarterly for 3 years, how much additional money will she need to add after 3 years to her investment to make her balance $18000
A=p(1+i/m)^mn
A=9,500×(1+0.04÷4)^(4×3)
A=10,704.84
The balance she needs
18,000−10,704.84
=7,295.16
$7259.16
Step-by-step explanation:
Hello
you can use the compound interest formula
[tex]A=P*(1+\frac{r}{n} )^{t} \\\\[/tex]
wherem A is the accumulated amount, P is the principal or initial amount, r is the interest rate and t is the number of periods
[tex]t=number of period = 3 years (\frac{4 quartely}{1 year} )=12 periods\\A=9500*(1+\frac{0.04}{4} )^{12} \\\\A=9500*(1.01^{12} )\\A= $10704.83\\\\she will need to add B\\\\B=18000-A\\B=18000-10704.83\\B=$7259.16[/tex]
Have a great day
[tex]\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$9500\\ r=rate\to 4\%\to \frac{4}{100}\to &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly means} \end{array}\to &4\\ t=years\to &3 \end{cases} \\\\\\ A=9500\left(1+\frac{0.04}{4}\right)^{4\cdot 3} \\\\\\ \textit{the difference willl then be}\qquad 18000 - A[/tex]
and 18000 - A is how much more she needs to make 18000
[tex]\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \qquad \begin{cases} A=\textit{current amount}\\ P=\textit{original amount deposited}\to &\$9500\\ r=rate\to 4\%\to \frac{4}{100}\to &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, 4 quarters a year} \end{array}\to &4\\ t=years\to &3 \end{cases}[/tex]
so.. if she deposits a principal of 9,500 today, compounding quarterly for 3 years, she'll have A amount
how much additional amount? well, 18,000 - A
[tex]\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$9500\\ r=rate\to 4\%\to \frac{4}{100}\to &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly means} \end{array}\to &4\\ t=years\to &3 \end{cases} \\\\\\ A=9500\left(1+\frac{0.04}{4}\right)^{4\cdot 3} \\\\\\ \textit{the difference willl then be}\qquad 18000 - A[/tex]
and 18000 - A is how much more she needs to make 18000
$7259.16
Step-by-step explanation:
Hello
you can use the compound interest formula
[tex]A=P*(1+\frac{r}{n} )^{t} \\\\[/tex]
wherem A is the accumulated amount, P is the principal or initial amount, r is the interest rate and t is the number of periods
[tex]t=number of period = 3 years (\frac{4 quartely}{1 year} )=12 periods\\A=9500*(1+\frac{0.04}{4} )^{12} \\\\A=9500*(1.01^{12} )\\A= $10704.83\\\\she will need to add B\\\\B=18000-A\\B=18000-10704.83\\B=$7259.16[/tex]
Have a great day
7,295.16
Step-by-step explanation:
A=p(1+i/m)^mn
A=9,500×(1+0.04÷4)^(4×3)
A=10,704.84
The balance she needs
18,000−10,704.84
=7,295.16