Question 1068500
Annually,
{{{F=P(1+i)^t}}}
{{{120000=P(1+0.05)^14}}}
{{{P=120000/(1.05)^14}}}
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.
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Continuously,
{{{F=Pe^(it)}}}
{{{120000=Pe^(0.048(14))}}}
{{{P=120000/e^(0.672)}}}