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Fred has a savings account balance of $1,728.57. The interest rate on the account is 3.4% compounded daily. If he opened the account nine years ago, what was the value of his initial deposit? (2 points)

$1,272.91

$1,670.79

$508.40

$1,508.78

Respuesta :

we are given

amount is $1,728.57

so, [tex] A=1728.57 [/tex]

the account is 3.4% compounded daily

so,

[tex] r=0.034 [/tex]

[tex] n=365 [/tex]

If he opened the account nine years ago

so,

[tex] t=9 [/tex]

Let's assume

initial deposit is P

now, we can use formula

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

now, we can plug values

[tex] 1728.57=P(1+\frac{0.034}{365})^{365*9} [/tex]

now, we can solve for P

[tex] P\frac{365.034^{3285}}{365^{3285}}=1728.57 [/tex]

[tex] P=\frac{1728.57\cdot \:365^{3285}}{365.034^{3285}} [/tex]

[tex] P=1272.91 [/tex]

so,

initial deposit is $1272.91.........Answer

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