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| rev | Admin | (Created page with "'''Solution: A''' We first need to calculate Jeffery’s accumulated interest at the end of each 12 years. His interest each year is: 4000(.07) = 280, so: <math display = "block"> \begin{aligned} 280s_{\overline{12}|i} = 7500 \\ s_{\overline{12}|i} = 26.786 \\ i = .137 \end{aligned} </math> Jane’s interest would be: 1000(.1) = 100 at the end of 2nd year 2000(.1) = 200 or 2(100) at the end of 3rd year 3000(.1) = 300 or 3(100) at the end of 4th year . . . 24000...") | Nov 22'23 at 0:31 | +971 |