course MTH 163
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22:19:48 `questionNumber 150001 `q001. Note that this assignment has 12 questions If you are given $1000 and invest it at 10% annual interest, compounded annually, then how much money will you have after the first year, how much after the second, and how much after the third? Is the the change in the amount of money the same every year, does the change increase year by year, does the change decrease year by year or does it sometimes increase and sometimes decrease?
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RESPONSE --> t (p) i new p($) year $ $ pt.10p 0 1,000 100 1100 1 1100 110 1210 2 1210 121 1331 3 1331 133 1464 the changes increase yearly
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22:20:02 `questionNumber 150001 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 --> i love these kind of problems
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22:32:17 `questionNumber 150002 `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 --> we used 1.1 to multiply the interst rates we get the same values that are in the table in the preceding problem this number saves alot of time for the equations
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22:32:19 `questionNumber 150002 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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22:33:51 `questionNumber 150003 `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)=1100 P(2)=1210 p(3)=1331 THE RECURRENCE FORMULA GIVES US THE SAME INFORMATION, JUST MAKES IT EASIER TO FIND A HIGHER YEAR
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22:33:57 `questionNumber 150003 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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22:37:41 `questionNumber 150004 `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 --> I WOULD FIND THE FIRST SOULUTION TRADITIONALLY THEN I WOULD DIVIDE THE START NUMBER OF 5000 BY THE SOLUTION OF 5400 WHICH LEAVES ME WITH THE MULTIPLIER OF 1.08%. SO THE SECOND YEAR 5400*1.08=5832 THIRD YEAR 5832*1.08=6298 FOURTH 6298*1.08=6801
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22:38:10 `questionNumber 150004 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 --> YES, IF IFORGET TO PUT THE CHANGE AT THE END WILL I MISS THE PROBLEM ON MY TEST OR QUIZ?
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22:38:37 `questionNumber 150005 `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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22:38:39 `questionNumber 150005 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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22:39:12 `questionNumber 150006 `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 --> 5000*1.08^X
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22:39:35 `questionNumber 150006 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 --> WHOOPS I FORGOT TO ENTER IN THE REST OF THE PROBLEM
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22:41:38 `questionNumber 150007 `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 --> WEJ WOULD USE THE EQUATION5000*1.08^X WE WOULD THEN FIGURE OUT COORDIANTES AND FIND THE SLOPE BETWEEN THE COORDINATES. SO I GOT THAT THE YEAR THE INVESTMENT WOULD DOUBLE ITSELF WOULD BE THE 10TH YEAR AT THE BEGINNING ROUNDING MY NUMBERS
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22:42:01 `questionNumber 150007 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 --> OK SO WE DON'T ROUND
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22:44:13 `questionNumber 150008 `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=1.08^N=AMOUNT AFTER THE SAID AMOUNT OF YEARS P0=1+R WOULD TELL US HOW TO FIND THE RATE PO*(1+R)^N= WOULD TELL US THE AMOUNT WITH THE RATE AFTER X AMOUNT OF YEARS.
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22:44:17 `questionNumber 150008 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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22:51:46 `questionNumber 150009 `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 --> 800 *.10 80=720 720 *72 =648
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22:52:46 `questionNumber 150009 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 --> HOW DID YOU GET THE .90 I CAME UP WITH ALMOST THE SAME ANSWERS JUST IN A DIFFERENT WAY. WOULD THIS STILL COUNT AS CORRECT?
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22:53:31 `questionNumber 150010 `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 *.90^X THE GRAPH WOULD PASS THROUGH THE AXIS AT (0,800) IT WOULD DECEND
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22:54:02 `questionNumber 150010 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 --> I HAD THAT IT WOULD PASS THROUGH THE AXIS. I MUST HAVE HAD AN ERROR IN MY GRAPH.
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22:54:25 `questionNumber 150011 `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=POB^2 500=P0 B^6
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22:56:34 `questionNumber 150011 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 --> IF WE DIVIDE THE 2ND EQUATION BY THE FIRST WE WOULD GET B^4=5/3 SOVING FOR BE THE FINAL ANSWER WOULD BE 233 IF WE SOLVE FOR PO THE FORM WOULD BE P=P0*B^X OUR SPECIFIC FUNCTION WOULD BE P=232.4*1.136^X
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22:56:58 `questionNumber 150012 `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 --> AWESOME, I UNDERSTAND THIS PART TOTALLY THE NOTES WERE ALOT EASIER TO UNDERSTAND.
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22:57:07 `questionNumber 150012 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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