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Mrs. Matthews wants to have $18,000 in the bank in 2 years. If she deposits $9000 today at 6% compounded quarterly for 2 years, how much additional money will she need to reach the desired $18,000?

Respuesta :

apologiabiology
depends if you want
1. find how much he will earn, find the differnce between that and 18000
2. see how much to invest till he will  get 18000


A=[tex]P(1+ \frac{r}{n})^{nt} [/tex]

A=futre amount
P=present amout
r=rate in decimal
n=number of times per year ccompounded
t=time in years


1.
A=?
P=9000
r=0.06
n=4 (quarter means 4 times per year)
t=2
?=[tex]9000(1+ \frac{0.06}{4})^{(4)(2)} [/tex]
?=[tex]9000(1+ 0.015)^{8} [/tex]
?=[tex]9000(1.015)^{8} [/tex]
?=10138.4 will be earned
18000-10138.4=7861.6 needed

2.
A=18000
P=9000+x
r=0.06
n=4 (quarter means 4 times per year)
t=2
18000=[tex](9000+x)(1+ \frac{0.06}{4})^{(4)(2)} [/tex]
18000=[tex](9000+x)(1+ 0.015)^{8} [/tex]
18000=[tex](9000+x)(1.015)^{8} [/tex]
divide both sides by 1.015^8
15978.8=9000+x
minus 9000 both sides
6978.8 needed




if he willnot be investing any more, he needs $7861.6 more
if he will invest more he will need to invest $6978.8 more

Answer:

She needs additionally 7,861.57 to reach the desired 18,000.

Step-by-step explanation:

Compounded interest formula is

[tex]A=P(1+\frac{r}{n} )^{nt}[/tex]

Where [tex]P[/tex] is the principal, [tex]r[/tex] is the interest rate in decimal number, [tex]n[/tex] is the number of compounded periods within a year and [tex]t[/tex] is time in years.

By given, we have

[tex]P=9,000\\t=2\\n=4\\r=0.06[/tex]

Replacing all values, we have

[tex]A=P(1+\frac{r}{n} )^{nt}\\A=9000(1+\frac{0.06}{4} )^{4(2)}= 9000(1.015)^{8}\\ A=10138.43[/tex]

After 2 years, Mrs. Matthews will have 10,138.43.

So, the difference she needs to reach 18,000 is

[tex]-10,138.43+18,000=7,861.57[/tex]

Therefore, she needs additionally 7,861.57 to reach the desired 18,000.

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