John opens a savings account with $1000.00 that compounds daily interest. The Apr at the time is 3.5%. What is the apy to the nearest hundred of a percent after 1 year
+118723 I think this is what you are asking.
If the nominal interest rate is 3.5% pa compounded daily, what is the effective yearly rate.
$$(1+r)^1 = (1+\frac{0.035}{365})^{365}$$
effective annual rate = $${\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{0.035}}}{{\mathtt{365}}}}\right)}^{{\mathtt{365}}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{0.035\: \!617\: \!971\: \!057\: \!178\: \!2}}$$
So to the nearest 100th of a percent the effective annual interest rate is 3.56%
+118723 I think this is what you are asking.
If the nominal interest rate is 3.5% pa compounded daily, what is the effective yearly rate.
$$(1+r)^1 = (1+\frac{0.035}{365})^{365}$$
effective annual rate = $${\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{0.035}}}{{\mathtt{365}}}}\right)}^{{\mathtt{365}}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{0.035\: \!617\: \!971\: \!057\: \!178\: \!2}}$$
So to the nearest 100th of a percent the effective annual interest rate is 3.56%
