What rate compounded semiannually is equivalent to $11\mathrm{.}45\%$ compounded monthly?

Round your answer to $4$ decimal places.

$(4)$

$I_1=,I_2=0\mathrm{.}1145,m_1=2,m_2=12$

$(1+\frac{I_1}{m_1}{)}^{m_1}=(1+\frac{I_2}{m_2}{)}^{m_2}$

$(1+\frac{I_1}{2}{)}^2=(1+\frac{0\mathrm{.}1145}{12}{)}^{12}$

$(1+\frac{I_1}{2}{)}^2$

$(1+\frac{I_1}{2}{)}^2(\frac{1\mathrm{.}1145}{12}{)}^{12}$

$1+I_1=2\mathrm{.}117266313$

$I_1=1\mathrm{.}117266313$

$I_1=111\mathrm{.}7266313\%$

$I_1=111\mathrm{.}7266\%$

$2$

$f=(1+i{)}^{\text{'}\text{'}\text{'}}-1$

$S_{1\text{'}}=\frac{R((1+i{)}^n-1]}{i}$

$A_n=\frac{R(1-(1+i{)}^{-n}]}{i}$

Use the formula from the end of the page