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

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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