Mth126_Week_1_Assignment_1

Real Life Application Problems Denise Guilbeault MAT 126 Instructor Stepp May 21, 2012 Week 1 Assignment 1 This was an A+ paper...got 100 Real Life Application Problems The assignment at hand requires me to solve two word problems that pertain to real world applications. In mathematics “two types of reasoning can be used; they are inductive and deductive reasoning” (Bluman, 2011). Inductive reasoning helps you arrive at a general conclusion based on what you are observing with the equation. Deductive reasoning allows you to come to a conclusion based on previously accepted statements. The most direct approach to an answer for an equation will be to these two types of reasoning. As we look at an arithmetic sequence, it is important to use inductive and deductive reasoning to solve the equation in the most logical way. The mathematical approach to these two problems will be arithmetic or geometric sequencing. The first problem, # 35 asks: “A person hired a firm to build a CB radio tower. The firm charges $100 for labor for the first 10 feet. After that, the cost of the labor for each succeeding 10 feet is $25 more than the preceding 10 feet. That is, the next 10 feet will cost $125; the next 10 feet will cost $150, etc. How much will it cost to build a 90 foot tower'” I used deductive reasoning and the arithmetic sequence to come up with my solution to this equation. It will cost $100.00 for the first 10 feet, $25.00 for each additional 10 feet, plus the original $100.00. So, it is fairly simple to figure this problem out. 100+125+150+175+200+225+250+275+300 = $1,800.00. The common difference in this arithmetic sequence is (25), the first term of this sequence (100) and the nth is equal to (9) however, we only add 25 8 more times because the 100 is already there and that is our first term. The second question, # 37 asks: “A person deposited $500 in a savings account that pays 5% annual interest that is compounded yearly. At the end of 10 years, how much money will be in the savings account'” For this answer I used the geometric sequence. So, it should look like this: a(n) = (a1)r^(n-1) (I cannot make it look like it should because I do not how to use the keys). Or I could write it out: a small 1 = a small one r ^n-1 over the right of the r. A(10) = 500(1.05)^10-1, the next step is = 500(1.05)^9, then the final step is the total which is; =$814.45 When I did the math, I took 500 and I multiplied it by 1.05, I got that answer and multiplied again by 1.05. I did this 9 times to arrive at the answer of 814.4473134. I rounded the numbers to the right of the decimal point to 45. Hence, I got $814.45. (500(1+0.05)10=814.45) A is = to the $500.00 deposited, the R is = to the annual rate of 5% and the N is = to 10 years. The money accrued for this account in ten years was an additional $314.45. So, this person made $314.45 for leaving his money alone. In conclusion, I found that using both inductive and deductive reasoning at the same time was important to help me distinguish which sequence was more useful. I had an easier time using the arithmetic method as opposed to the geometric method. Both sequences took me a long time to decipher. I have a good handle on the arithmetic sequence and I am still working on the geometric sequencing. I find that if I do come across a real world problem, I can use deductive and / or inductive reasoning and see if I can use either one of these methods to arrive at a logical answer. The method for problem number 35 will be easy to use if it contains an arithmetic sequence. If I want to figure out how much money I will accrue if I put money into a savings account, I will be able to apply these methods to my questions. These methods will help me arrive at a logical answer. Reference Bluman, A. G. (2011). Mathematics in our world (1st ed. Ashford University Custom). United States: McGraw-Hill.