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$1,400 is invested in an account earning 4.3% interest (APR), compounded daily.Write a function showing the value of the account after t years, where the annualgrowth rate can be found from a constant in the function. Round all coefficients inthe function to four decimal places. Also, determine the percentage of growth peryear (APY), to the nearest hundredth of a percent.

Sagot :

Answer

[tex]A=1400(1.0001)^{365t}[/tex]

APY = 4.39 %

Step-by-step explanation

Compound interest formula

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

where

• A: final amount (value of the account after t years)

,

• P: principal (amount of money invested)

,

• r: interest rate (APR), as a decimal

,

• n: number of times interest applied per year

,

• t: time in years

Substituting, P = $1400, r = 0.043 (= 4.3/100), n = 365 (compounded daily means 365 times per year), we get:

[tex]\begin{gathered} A=1400(1+\frac{0.043}{365})^{365t} \\ A=1400(1.0001)^{365t} \end{gathered}[/tex]

APY formula

[tex]APY=\lbrack(1+\frac{r}{n})^n-1\rbrack\times100[/tex]

Substituting with the before mentioned values, we get:

[tex]\begin{gathered} APY=\lbrack(1+\frac{0.043}{365})^{365}-1\rbrack\times100 \\ APY=4.39\text{ \%} \end{gathered}[/tex]