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| rev | Admin | (Created page with "'''Solution: B''' We work with rate <math>i=.05</math> for 6 months. Let <math>\mathrm{X}</math> be the initial deposit. The Jan 1, 2005 deposit is <math>X(1.1025)^{10}</math>. Note that <math>1.05^2=1.1025</math>. The Jan 1 deposits accumulate, measuring from 2005 down, to <math display="block"> A 1=X(1.1025)^{10}(1.05)^2+\cdots+X(1.1025)^0(1.05)^{22}=11 X(1.1025)^{11} </math> The July 1 deposits accumulate, measuring from 2005 down, to <math display="block"> A 2=...") | Nov 26'23 at 21:48 | +965 |