Concept explainers
Loan problems The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for t ≥ 0, graph the solution, and determine the first month in which the loan balance is zero.
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- Q. At time k=0, an initial investment of $1200 is made into a savings account that pays % interest compounded monthly. (The monthly interest rate is 0.0025.) Each month, after the initial investment, an additional $200 is added to the account. Let Y be the amount in the atcount at the end of k months. a. Write the difference equation satisfied by Yk Yk+1= b. The amount of money that will be in the account after 3 years is $ The amount of money that will be in the account after 4 years is $ The amount of money that will be in the account after 6 years is $ The amount of interest in the 6-year total is $ Yo=arrow_forwardA certain manufacturer of Blu-ray players will make x thousand units available in the market per week when the unit price is p = 100(0.9x + 0.51/ 1 + x) dollars. What is the producers' surplus if the selling price is set at $370/unit? (Round your answer to the nearest dollar.) $_________arrow_forwardA universal life policy is issued to a life aged 45. Death benefit is 10,000 and the policyholder pays an annual premium of 200 at the beginning of each year. Expense charges are 30% of first year premium and 5% of renewal premiums. Interest credited is 6% per year and interest assumed in the cost of insurance is 4% per year. Cost of insurance is based on Makeham's mortality, H = 0.01+0.0001(1.05*). The account value at the beginning of the 7th year, before any premium is paid, is 1,500. Calculate, to the nearest integer, the account value at the end of the 7th year.arrow_forward
- The scrap value of a machine at the end of its useful life is given by S(n) = C(1-r)", where C is the original cost, n is the useful life of the machine in years, and r is the constant annual percentage of value lost. Find the scrap value of the following machine. Original cost, $49,000; life, 11 years; annual rate of value lost, 13% S=$ (Round to the nearest cent.)arrow_forwardAn investor deposits an initial amount of money (Yo) into a bank account that offers a simple (annually compounding) interest rate (r). The bank charges a fixed annual fee (b). The bank charge is debited after the interest is accumulated. Write down a difference equation that describes how the account balance (Yt) changes over time (t), then solve the equation analytically. If the interest rate is r = 0.1(10%), the bank charge is b = 100 and the investor projects that the account balance in year t = 3 is $2,464, then the %3D initial amount of money is: Select one: Select one: O yo = $2,000 O yo = $2, 100 O yo = $2, 200 O yo = $1,900arrow_forwardPurchasing Power If a retired couple has a fixedincome of $60,000 per year, the purchasing powerof their income (adjusted value of the money) aftert years of 5% inflation is given by the equationP = 60,000e-0.05t. In how many years will thepurchasing power of their income fall below half oftheir current income?arrow_forward
- The simple interest on an investment is directly proportional to the amount of the investment. By investing $5800 in a municipal bond, you obtained an interest payment of $221.25 after 1 year. Find a mathematical model that gives the interest I for this municipal bond after 1 year in terms of the amount invested P. (Round your answer to three decimal places.) I= 26.215P I=221.25P I= 0.038P I= 1,283,250P I=5800Parrow_forward6. Loan payments Let N(r) be the number of $300 monthly payments required to repay an $18,000 auto loan when the interest rate is r percent. What does the equation N(6.5) = 73 say in this context?arrow_forwardThe monthly sales S (in hundreds of units) of baseball equipment for an Internet sporting goods site are approximated by S = 60.1 - 43.9 cost where t is the time (in months), with t = 1 corresponding to January. Determine the months when sales exceed 7700 units at any time during the month. August through April May through September April through August March through August March through Septemberarrow_forward
- If $1 dollar is deposited in an account paying 27% per year compounded annually, then after t years the account will contain y = (1 + 0.27) = 1.27 dollars. (a) Use a calculator to complete the table. (b) Graph 1.27'. (a) Use a calculator to complete the table. 1 2 4 5 6 y 1.27 1.61 3.3 4.2 (Round to two decimal places as needed.) Help me solve this View an example Get more help - R D G Harrow_forwardPresent value is the amount of money that must be invested now at a given rate of interest to produce a given future value. For a 1-year investment, the present value can be calculated using Present value = Future value 1 + r , where r is the yearly interest rate expressed as a decimal. (Thus, if the yearly interest rate is 8%, then 1 + r = 1.08.) If an investment yielding a yearly interest rate of 13% is available, what is the present value of an investment that will be worth $4000 at the end of 1 year? That is, how much must be invested today at 13% in order for the investment to have a value of $4000 at the end of a year? (Round your answer to two decimal places.)arrow_forward5. Someone invests $3080 in an account at 6.3% interest compounded annually. Let f(t) be the value (in dollars) of the account at t years after she invested the $3080. a) Find the equation of f. b) What will the account's value be after 12 years? (Round to the nearest cent.)arrow_forward
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:Cengage