course mth 163
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23:54:44 `q001. Note that this assignment has 12 questions If you are given $1000 and invest it at 10% annual interest, compounded annually, then how much money will you have after the first year, how much after the second, and how much after the third? Is the the change in the amount of money the same every year, does the change increase year by year, does the change decrease year by year or does it sometimes increase and sometimes decrease?
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RESPONSE --> r = .10 p(0) = 1000 n = 0 P (n+1) = 1 = r) P(n) P(1) = (1.10)(1000) P(1) = 1100 r = .10 P(1) = 1100 n = 1 P(2) = 1.10(1100) P(2) = 1210 r = .10 P(2) = 1210 n = 2 P(3) = 1.10(1210) p(3) = 1331 After the 1st year, you would have $1100. After the second year, you would have $1210. After the third year, you would have $1331. The change in the amount of money is NOT the same every year, but always increases.
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23:55:04 During the first year the interest will be 10% of $1000, or $100. This makes the total at the end of the first year $1100. During the second year the interest will be 10% of $1100, or $110. At the end of the second year the total will therefore be $1100 + $110 = $1210. During the third year the interest will be 10% of $1210, or $121. At the end of the sphere year the total will therefore be $1210 + $121 = $1331. The yearly changes are $100, $110, and $121. These changes increase year by year.
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RESPONSE --> OK
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23:58:51 `q002. In the preceding problem you obtained amounts $1100, $1210 and $1331. What number would you multiply by $1000 to get $1100? What number we do multiply by $1100 to get $1210? What number would we multiply by $1210 to get $1331? What is the significance of this number and how could we have found it from the original information that the amount increases by 10 percent each year?
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RESPONSE --> The number one would need to multiply in order to increase 100 to 1100: 1100 to 1210: and 1210 to 1331 is 1.1. All one would have to do is to multiply the principle amount by a growth of 10% or 1.1 to obtain the next year principle and then continue with the new principle multiplied by the growth rate to get the next years' principle...continue this as long as the growth rate stays constant.
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23:59:02 To get $1100 you have to multiply $1000 by 1100 / 1000 = 1.1. To get $1210 you have to multiply $1210 by 1210 / 1100 = 1.1. To get $1331 you have to multiply $1331 by 1331 / 1210 = 1.1. If the amount increases by 10 percent, then you end up with 110 percent of what you start with. 110% is the same as 1.1.
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RESPONSE --> OK
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00:07:09 `q003. Given the recurrence relation P(n) = 1.10 * P(n-1) with P(0) = 1000, substitute n = 1, 2, and 3 in turn to determine P(1), P(2) and P(3). How is this equation related to the situation of the preceding two problems?
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RESPONSE --> In this question, one would multiply 1.1 - 1) or .1 by the principle of 1000 for the 1st year principle which would be 100... p(1) = 100 p(2) = 10 p(3) = 1 This question works the same way mathemathically. In this case, you are multiplying by a number smaller than 1, therefore the principle decreases instead of increasing.
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00:10:34 Substituting 1 into P(n) = 1.10 * P(n-1) we obtain P(1) = 1.10 * P(1-1), or P(1) = 1.10 * P(0). Since P(0) = 1000 we get P(1) = 1.1 * 1000 = 1100. Substituting 2 into P(n) = 1.10 * P(n-1) we obtain P(2) = 1.10 * P(1). Since P(1) = 1100 we get P(1) = 1.1 * 1100 = 1210. Substituting 3 into P(n) = 1.10 * P(n-1) we obtain P(3) = 1.10 * P(2). Since P(2) = 1000 we get P(3) = 1.1 * 1210 = 1331.
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RESPONSE --> I did this one completely incorrect. This problem works the same as the two previous. I don't see that much difference in the way that the solution is found. The substitution of the 1, 2, and 3 are basically the same way that I found the solution in the first problem.
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00:12:52 `q004. If you are given $5000 and invest it at 8% annual interest, compounded annually, what number would you multiply by $5000 to get the amount at the end of the first year? Using the same multiplier, find the results that the end of the second and third years.
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RESPONSE --> You would multipy the $500 by 1.08 to get 5400 which would be the amount of money you would have at the end of the first year. 2nd year...5400 * 1.08 = $5832 3rd year...5832 * 1.08 = 6298.56
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00:13:02 If your money increases by 8% in a year, then it the end of the year you will have 108% as much as at the beginning. Since 108% is the same as 1.08, our yearly multiplier will be 1.08. If we multiply $5000 by 1.08, we obtain $5000 * 1.08 = $5400, which is the amount the end of the first your. At the end of the second year the amount will be $5400 * 1.08 = $5832. At the end of the third year the amount will be $5832 * 1.08 = $6298.56.
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RESPONSE --> OK
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00:14:53 `q005. How would you write the recurrence relation for a $5000 investment at 8 percent annual interest?
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. r = .08 P(0) = 5000 n = 0 P(n+1) = (1 + r) P(n)
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00:15:20 Just as the recurrence relation for 10 percent annual interest, as seen in the problem before the last, was P(n) = 1.10 * P(n-1), the recurrence relation for 8 percent annual interest is P(n) = 1.08 * P(n-1).
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RESPONSE --> OK
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ȶݕx|ič Student Name: assignment #015
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18:19:05 `q001. Note that this assignment has 12 questions If you are given $1000 and invest it at 10% annual interest, compounded annually, then how much money will you have after the first year, how much after the second, and how much after the third? Is the the change in the amount of money the same every year, does the change increase year by year, does the change decrease year by year or does it sometimes increase and sometimes decrease?
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RESPONSE -->
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18:19:08 During the first year the interest will be 10% of $1000, or $100. This makes the total at the end of the first year $1100. During the second year the interest will be 10% of $1100, or $110. At the end of the second year the total will therefore be $1100 + $110 = $1210. During the third year the interest will be 10% of $1210, or $121. At the end of the sphere year the total will therefore be $1210 + $121 = $1331. The yearly changes are $100, $110, and $121. These changes increase year by year.
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RESPONSE -->
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18:19:10 `q002. In the preceding problem you obtained amounts $1100, $1210 and $1331. What number would you multiply by $1000 to get $1100? What number we do multiply by $1100 to get $1210? What number would we multiply by $1210 to get $1331? What is the significance of this number and how could we have found it from the original information that the amount increases by 10 percent each year?
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RESPONSE -->
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18:19:12 To get $1100 you have to multiply $1000 by 1100 / 1000 = 1.1. To get $1210 you have to multiply $1210 by 1210 / 1100 = 1.1. To get $1331 you have to multiply $1331 by 1331 / 1210 = 1.1. If the amount increases by 10 percent, then you end up with 110 percent of what you start with. 110% is the same as 1.1.
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RESPONSE -->
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18:19:14 `q003. Given the recurrence relation P(n) = 1.10 * P(n-1) with P(0) = 1000, substitute n = 1, 2, and 3 in turn to determine P(1), P(2) and P(3). How is this equation related to the situation of the preceding two problems?
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RESPONSE -->
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18:19:18 Substituting 1 into P(n) = 1.10 * P(n-1) we obtain P(1) = 1.10 * P(1-1), or P(1) = 1.10 * P(0). Since P(0) = 1000 we get P(1) = 1.1 * 1000 = 1100. Substituting 2 into P(n) = 1.10 * P(n-1) we obtain P(2) = 1.10 * P(1). Since P(1) = 1100 we get P(1) = 1.1 * 1100 = 1210. Substituting 3 into P(n) = 1.10 * P(n-1) we obtain P(3) = 1.10 * P(2). Since P(2) = 1000 we get P(3) = 1.1 * 1210 = 1331.
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RESPONSE -->
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18:19:21 `q004. If you are given $5000 and invest it at 8% annual interest, compounded annually, what number would you multiply by $5000 to get the amount at the end of the first year? Using the same multiplier, find the results that the end of the second and third years.
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RESPONSE -->
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18:19:22 If your money increases by 8% in a year, then it the end of the year you will have 108% as much as at the beginning. Since 108% is the same as 1.08, our yearly multiplier will be 1.08. If we multiply $5000 by 1.08, we obtain $5000 * 1.08 = $5400, which is the amount the end of the first your. At the end of the second year the amount will be $5400 * 1.08 = $5832. At the end of the third year the amount will be $5832 * 1.08 = $6298.56.
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RESPONSE -->
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18:19:25 `q005. How would you write the recurrence relation for a $5000 investment at 8 percent annual interest?
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RESPONSE -->
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18:19:29 Just as the recurrence relation for 10 percent annual interest, as seen in the problem before the last, was P(n) = 1.10 * P(n-1), the recurrence relation for 8 percent annual interest is P(n) = 1.08 * P(n-1).
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RESPONSE -->
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18:53:20 `q006. If you are given amount $5000 and invest it at annual rate 8% or .08, then after n years how much money do you have? What does a graph of amount of money vs. number of years look like?
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RESPONSE --> I believe the question written out in an exponential function would be P(n) = $5000(0.08)^n. The graph would be an asymptote with a positive curve increasing with more of a slope as it reaches ""n"".
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18:55:08 After 1 year the amount it $5000 * 1.08. Multiplying this by 1.08 we obtain for the amount at the end of the second year ($5000 * 1.08) * 1.08 = $5000 * 1.08^2. Multiplying this by 1.08 we obtain for the amount at the end of the third year ($5000 * 1.08^2) * 1.08 = $5000 * 1.08^3. Continuing to multiply by 1.08 we obtain $5000 * 1.08^3 at the end of year 3, $5000 * 1.08^4 at the end of year 4, etc.. It should be clear that we can express the amount at the end of the nth year as $5000 * 1.08^n. If we evaluate $5000 * 1.08^n for n = 0, 1, 2, ..., 10 we get $5000, $5400, 5832, 6298.56, 6802.45, 7346.64, 7934.37, 8569.12, 9254.65, 9995.02, 10,794.62. It is clear that the amount increases by more and more with every successive year. This result in a graph which passes through the vertical axis at (0, 5000) and increases at an increasing rate.
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RESPONSE --> I did the math for ten years, but after looking at the question again, it ask for ""n"" years, therefore my response, but if the question was mistakenly written and you wanted ten years, I have the $ after ten years as 11658.19.
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18:56:18 `q007. With a $5000 investment at 8 percent annual interest, how many years will it take to double the investment?
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RESPONSE --> It would take 9 years to double your investment. At the end of 9 years, one would have 10794.62 which is a little more than doubling your investment.
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18:57:39 Multiplying $5000 successively by 1.08 we obtain amounts $5400, 5832, 6298.56, 6802.45, 7346.64, 7934.37, 8569.12, 9254.65, 9995.02, 10,794.62 at the end of years 1 thru 10. We see that the doubling to $10,000 occurs very shortly after the end of the ninth year. We can make a closer estimate. If we calculate $5000 * 1.08^x for x = 9 and x = 9.1 we get about $10,072. So at x = 9 and at x = 9.1 the amounts are $9995 and $10072. The first $5 of the $77 increase will occur at about 5/77 of the .1 year time interval. Since 5/77 * .1 = .0065, a good estimate would be that the doubling time is 9.0065 years. If we evaluate $5000 * 1.08^9.0065 we get $10,000.02.
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RESPONSE --> This is correct, although I didn't break it down this closely, I do understand the decimal that you would have to use to get it closer.
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19:06:02 `q008. If you are given amount P0 and invest it at annual rate r (e.g., for the preceding example r would be 8%, which in numerical form is .08), then after n years how much money do you have?
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RESPONSE --> I write this expression as: P(n) = (0.08)^n * p(0)
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19:07:23 If the annual interest rate is .08 then each year we would multiply the amount by 1.08, the amount after n years would be P0 * 1.08^n. If the rate is represented by r then each year then each year we multiply by 1 + r, and after n years we have P0 * (1 + r)^n.
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RESPONSE --> yes, after year after year we would have to multiply by (1 + r)^n to obtain the amount.
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19:25:39 `q009. If after an injection of 800 mg an antibiotic your body removes 10% every hour, then how much antibiotic remains after each of the first 3 hours? How long does it take your body to remove half of the antibiotic?
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RESPONSE --> After the first hour there would be 720 mg of the antibiotic left in your system. After the second hour there would be 648 mg of the antibiotic left in your system. After the third hour there would be 583.2 mg of the antibiotic left in your system. It would take between 6 hours and 7 hours to remove half of the antibiotic.
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19:27:02 If 10 percent of the antibiotic is removed each hour, then at the end of the hour the amount left will be 90 percent of what was present at the beginning of the hour. Thus after 1 hour we have .90 * 800 mg, after a second hour we have .90 of this, or .90^2 * 800 mg, and after a third hour we have .90 of this, or .90^3 * 800 mg. The numbers are 800 mg * .90 = 720 mg, then .90 of this or 648 mg, then .90 of this or 583.2 mg.
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RESPONSE --> I actually went thru another method to come up with the answers that were listed. Your method is quicker and will give you the answer with less steps. I have changed my method of this type of problem.
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19:33:00 `q010. In the preceding problem, what function Q(t) represents the amount of antibiotic present after t hours? What does a graph of Q(t) vs. t look like?
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RESPONSE --> Q(t) = (800) * (.9)^t is the equation that I think you would use. The graph would be an asymptote that would have less slope as it reached the x axis.
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19:33:20 After t hours we will have 800 * .9^t mg left. So Q(t) = 800 * .9^t. The amounts for the first several years are 800, 720, 648, 583.2, etc.. These amounts decrease by less and less each time. As a result the graph, which passes through the vertical axes at (0,800), decreases at a decreasing rate. We note that no matter how many times we multiply by .9 our result will always be greater than 0, so the graph will keep decreasing at a decreasing rate, approaching the horizontal axis but never touching it.
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RESPONSE --> OK
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21:22:57 `q011. Suppose that we know that the population of fish in a pond, during the year after the pond is stocked, should be an exponential function of the form P = P0 * b^t, where t stands for the number of months after stocking. If we know that the population is 300 at the end of 2 months and 500 at the end of six months, then what system of 2 simultaneous linear equations do we get by substituting this information into the form of the function?
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RESPONSE --> I guess writing something for the answer is better than giving no solution at all, but in this case, I think that I will continue and see what the correct answer is before I comment. I do have an answer, but it doesn't seem correct for the simple fact that if I was given these linear equation, I don't know that I could answer them. My equations are in the quadratic form.
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21:25:22 We substitute the populations 300 then 500 for P and substitute 2 months and 6 months for t to obtain the equations 300 = P0 * b^2 and 500 = P0 * b^6.
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RESPONSE --> This is the equations that I had, but since they were in quadratic form and the question ask for linear, I felt that I was incorrect...My answer was as follows: P = P(0) * b^t 300 = P(0) * b^2 500 = P(0) =*b^6 I thought there would have to be some reduction to get rid of the powers in order for the answer to be correct.
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22:16:27 `q012. We obtain the system 300 = P0 * b^2 500 = P0 * b^6 in the situation of the preceding problem. If we divide the second equation by the first, what equation do we obtain? What do we get when we solve this equation for b? If we substitute this value of b into the first equation, what equation do we get? If we solve this equation for P0 what do we get? What therefore is our specific P = P0 * b^t function for this problem?
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RESPONSE --> dividing the second equation by the first, I get: .6 = b^4 Solving for b, I get .8801 Substituting this value of b into the first eqution, I get: 300 = p(0) *.8801^2 300 = P(0) * .7746 P(0) = 387 I will have to look at your answer to self critique myself on the last part of this question.
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22:19:22 Dividing the second equation by the first the left-hand side will be left-hand side: 500/300, which reduces to 5/3, and the right-hand side will be right-hand side: (P0 * b^6) / (P0 * b^2), which we rearrange to get (P0 / P0) * (b^6 / b^2) = 1 * b^(6-2) = b^4. Our equation is therefore b^4 = 5/3. To solve this equation for b we take the 1/4 power of both sides to obtain (b^4)^(1/4) = (5/3)^(1/4), or b = 1.136, to four significant figures. Substituting this value back into the first equation we obtain 300 = P0 * 1.136^2. Solving this equation for P0 we divide both sides by 1.136^2 to obtain P0 = 300 / (1.136^2) = 232.4, again accurate to 4 significant figures. Substituting our values of P0 and b into the original form P = P0 * b^t we obtain our function P = 232.4 * 1.136^t.
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RESPONSE --> I was completely wrong, just by making a stupid mistake. I didn't divide by the reciprocle of the 4th power, I took the 4th root of 5/3 or .6. This is a critical error that cause all following answers to be incorrect. I see that you have to divide by the reciprocal of the 4th power in order to get this answer. The rest falls in place.
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