course Mth 163 tRStudent Name:
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15:07:21 `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 --> A=p(1+r/n)^nt a=total amount p= money invested t=time in years n=total number of periods r=rate of interest A=1000(1+10/1)^1*1 A=1000(11)^1 A=11,000 You would have 11,000 after one year according to this formula I found for compound interest. I would have just multiplied $1000 by 10% and that gives you 100, so you would have $1,100 after one year. I think that the number would rise by this amount every year. So the second year, you would have $110 added to the origina $1000 = $1110, the third year would be more, etc. increasing every year.
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15:08:38 11-05-2006 15:08:38 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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NOTES -------> That's what I thought. I wonder why the formula gave me the wrong answer?
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15:27:05 `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 --> You would have to multiply $1000 by 1.1 to get $1100. We multiply $1100 by 1.1 to get $1210. We multiply that by the same (1.1) to get $1331. The significance is that it is equal to ten percent. But I'm not sure how we could have found it by knowing that the amount increased by 10 % each year.
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15:27:57 11-05-2006 15:27:57 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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NOTES -------> Oh, I see.
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15:34:47 `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 --> P(1)=1.1*P(1-1) P(1)=1.1*P(0) P(1)=1.1*1000 P(1)=1100 P(2)=1210 P(3)=1331 This equation allows you to find the amount of money earned at ten percent interest per year, like in the last two problems.
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15:35:05 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 --> ok
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15:43:29 `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 multiply $5000 by 1.08 to get the amount at the end of the first year, which is $5400. At the end of the second year you would have $5832. At the end of the third, $6298.56
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15:43:55 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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15:44:56 `q005. How would you write the recurrence relation for a $5000 investment at 8 percent annual interest?
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RESPONSE --> P(n)=1.08*P(n-1)
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15:45:04 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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15:48:36 `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 --> You have A=5000(1.08)(years?) I do know that the graph would be increasing at a increasing rate.
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15:50:46 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 --> O.k. I understand. I had part of the equation right. I see how by raising the multiplied amount to the power of the number of years we would get the correct answer. That makes sense.
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15:54:39 `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 --> Well, in nine years you would have almost double $9995.02 (approx). In ten years you would have $10,794.62 (approx.) So in 9.1 years you would have $10,072.24 (approx.)
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15:55:54 11-05-2006 15:55:54 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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NOTES -------> O.K. I understand. Thanks
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15:57:13 `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 --> P0*.08^n
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16:00:43 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 --> I am a little confused. I see how I made a mistake by putting .08 instead of 1.08 and I know why, but the rate is 1.08, right, so doesn't that make us putting that into the equation twice with (1+r)? I know you would have to add each year into the equation, that is why you put (1+r).
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16:19:14 `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 you would have 720, the second hour 648 and after the third 583.2. 4 hrs. = 524.88 5 hrs. = 472.392 6 hrs. = 425.153 7 hrs. = 382.638 So half of the anitbiotic would be removed after 6+ hours. Between six and seven hours 42.515 is removed.
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16:19:58 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 --> O.K.
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16:22:02 `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 --> The graph would be decreasing at a decreasing rate.
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16:23:47 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 --> Oops, I forgot to put my equation. I had the same thing on my paper 800*.9^t.
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16:29:42 `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 --> 300=P0*b^2 500=P0*b^6
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16:29:49 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 --> ok
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18:05:56 `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 --> 1.67 = P0*3b is what you get when you divide the second equation by the first. 1.67=P0*3b divide each side by 3. .5567=P0b Multiply each side by P0. P0.5567=b If we substitue b into the first equation we get 300=2PO*.5567^2 300=2P0*.3099 300=.6198P0 484.027=P0 So the function is 1.67=484.027*3(P0.5567) I think.
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18:10:28 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 --> Boy, I was way off. I divided the 6 and 2 (wrote them too big, they didn't look like exponents). That was my first mistake and the rest followed. I understand everything except the last part. We don't substitue our values of b and PO into the equation we got when we divided to get the one equation?
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