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Price $P$. Residual value, or the amount for which you can successfully sell the car after $L$ months, $R$. Hence amount you are actually borrowing is effectively $B=P-R$. Now you are paying a total of $T = L\times M$ where $M$ is the monthly payment. Therefore the percentage difference in the amount you pay versus the amount you borrow is $$I=100\%\times( ... 1 When one does mathematics, it can be useful to go back to basic principles. With interest rate of 0.07, that is, 7\%, compounded annually, in n years A dollars grow to$$A(1.07)^n$$dollars. In our case, 1500 grew to 3750 in an unknown number n of years, so$$3750=1500(1.07)^n.$$It follows that$$(1.07)^n=\frac{3750}{1500}=2.5.$$Take the ... 1 This is good, subject to revising your typo. You have after n periods that your balance is$$1500* 1.07^n.$$Now set$$1500*1.07^n = 3750.$$Begin by dividing to get$$1.07^n = 2.5 $$so$$n = {\log(2.5)\over \log(1.07)} = 13.54.$$Your answer is off a bit because you transposed two digits. 1 If by chance you didn't want to use logarithm's you could make a table and write the formula out for the desired years: a(1+r)^n where a is the initial amount, r is the rate and n is the number of years. For year 12: 1500(1+0.07)^{12} = 3378.29 For year 13: 1500(1+0.07)^{13} = 3614.77 For year 14: 1500(1+0.07)^{14} = 3867.29 So if the ... 1 Here is an answer done "their" way- We have$$1300=500(1.06)^n\implies 13=5(1.06)^n\implies \frac{13}{5}=(1.06)^n\implies \log_{1.06}\frac{13}5 =n\implies n \approx 16.39830702\implies n \approx 16.4 Rounded to one decimal place. Please ask if you need any further clarification.

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The 1.102 comes from two places. First, "%" means hundredths. If you see "10.2%", that really means $\frac{10.2}{100}$, which is 0.102. Second, if your money is earning 10.2% interest annually, that means that at the end of the year, the bank takes your balance, multiplies it by 10.2%—that is, by 0.102—and hands you that much more in interest. Your new ...

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To get a solution, you'll need to specify an objective that you are trying to maximize. If you are just maximizing the expected profit, then your \$15,000 answer is correct. Typically, however, in finance we assume people are risk-averse. A standard objective function in that case is$E[profit] - \lambda * Var[profit]$where$\lambda\$ is the level of ...

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