Assignment 1(3)

# Assignment 1(3) - th year Your Answer No answer Correct...

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System Homepage » Class Homepage » Student Details » Student Assignment Details ActSci 231 W09 : Gradebook Welcome Po-Wei Chuang [ My Profile ] Assignment detail for Po-Wei Chuang in Assignment 1: Po-Wei Chuang Login: pchuang Email: Student ID: 20260696 Assignments completed: 9 Assignments active: 0 Question Grade 1 A principal amount of K grows to 1,850 after 2 years based on a simple interest rate of i, and after 4 years, it grows to 2,183.61. Determine the value K. Your Answer: No answer Correct Answer: 1516.39 Comment: K*(1+2*i)=1,850 and K*(1+4*i)=2,183.61. Solve for i and then find K=1,850/(1+i*2) Instructors Comment: 0.0 2 Given the amount function A(t) = 800 (1 + 0.18 t), find the equivalent annual effective rate of compound interest earned over the time period 2.5 to 5. Your Answer: No answer Correct Answer: .1142 Comment: i = (A(5)/A(2.5)) 1/2.5 - 1 = 0.1142 Instructors Comment: 0.0 3 Given the amount function A(t) = 1,200 (1.01) t , find the interest earned in the 8
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Unformatted text preview: th year. Your Answer: No answer Correct Answer: 12.87 Comment: I 8 = A(8) - A(7) = 12.87 Instructors Comment: 0.0 4 An accumulation function A K (t) is such that for all positive integers t, it is given by 1000*(1+i) t for a certain i, and linear interpolation is used to determine its value in between integers t. Given that A K (2) = 1,150, find A K (1.2). Your Answer: No answer Correct Answer: 1087.9 Comment: Find i=(1,150/1000)^0.5 and then A K (1.2)=1000*(1+i)*(1+(1.2-1)*i). Instructors Comment: 0.0 5 Find the rate of simple interest per year such that \$3,000 will accumulate to \$3,675 in 9 years. Your Answer: No answer Correct Answer: .25e-1 Comment: 3,000(1 + i(9)) = 3,675 =>i=[3,675-3,000]/3,000/9 = 0.025 Instructors Comment: 0.0 Gradebook Help Logout Maple T.A. - Gradebook http://maple-ta4.math.uwaterloo.ca:8080/mapleta/gradebook/gr. .. 1 of 1 4/6/09 9:04 AM...
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