I flipped a coin and got heads 14 times and tails 11 times, if that helps
\left[x _{2}\right] = \left[ \frac{-1+i \,\sqrt{3}+2\,by+\left( -2\,i \right) \,\sqrt{3}\,by}{2^{\frac{2}{3}}\,\sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}+\frac{\frac{ - \sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}{24}+\left( \frac{-1}{24}\,i \right) \,\sqrt{3}\,\sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}{\sqrt[3]{2}}\right][x2]=⎣⎢⎢⎢⎢⎡2323√(432by+√(−6912+41472by+103680by2+55296by3))−1+i√3+2by+(−2i)√3by+3√224−3√(432by+√(−6912+41472by+103680by2+55296by3))+(24−1i)√33√(432by+√(−6912+41472by+103680by2+55296by3))⎦⎥⎥⎥⎥⎤
totally answer.
Compound interest formula = a=P(1+r/n)^nt
P= lump sum to deposit (solving for)
A= amount accumulated over the entire time (20000)
n= number of times interest is compounded annually (1)
r= rate of interest (0.82)
T= total number of years (15)
20000=P(1+0.082/1)^1*15
20000=P(1.082)^15
20000=P(3.26143638)
20000/3.26143638=P
P=$6132.2674
The submarine would be 900 below the ocean surface after 12 minutes assuming it started its descent at the surface of the ocean.
