#$&*
course Mth163
July 1,2013 7:41pm
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
015.
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Question: `q001. Note that this assignment has 14 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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Your solution:
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After the first year you would have $1000+$100 =($1000*.1)/1 so you would have $1100 after year one
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You would have $1100 after one year, but
$1000+$100 =($1000*.1)/1
is a false statement.
($1000*.1)/1 = 100, not $1000+$100 .
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After year two years 1100*.1=110/2=$1,155
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1100*.1=110/2=$1,155
is a completely incorrect statement, though your intent is clear.
The statement is incorrect becase equality is transitive. If
1100*.1=110/2=$1,155
then the following must also be true:
1100*.1=$1,155 (but this is clearly false)
1100*.1=110/2 (this also is clearly false)
110/2=$1,155 (and this is clearly false as well).
None of these statements are true, and the falsehood of any one of these statements makes the original statement 1100*.1=110/2=$1,155 false.
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Suppose that at the beginning of a certain year you have $1800, and the amount increases by 10%.
How much do you therefore have at the end of that year?
Does the answer depend on just which year this happens to be?
Now suppose that at the beginning of a certain year you have $1100, and the amount increases by 10%.
How much do you therefore have at the end of that year?
Does the answer depend on just which year this happens to be?
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Year three you would have (1155*.1)/3=38.5 1155+38.5=$1193
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This statement is also mathematically incorrect.
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If you started with $1155, your interest would be 10% of $1155. You don't divide your interest by 3 just because this happens to be the third year.
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As the years progressed the amount of money made was still over all increasing but the rate at which it did so was decreasing
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Your results aren't correct, but if they were then you would be right about the amounts increasing at a decreasing rate.
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confidence rating #$&*: 2
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Given Solution:
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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Self-critique (if necessary):
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I thought the annual compound rate was 1 so after two years wouldn’t that be 2??? Or after every year it just stays one???
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@&
Every year you add 10% to what you had at the beginning of the year.
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Self-critique Rating:2
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Question: `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 $1331?
What is the significance of this number and how could we have found it from the original information that the amount increases by 10 percent each year?
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Your solution:
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1100/100=1.1
1210/1100=1.1
1331/1210=1.1
The significance of this number is that is constant
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confidence rating #$&*: 3
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Given Solution:
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 $1331 by 1331 / 1210 = 1.1.
If the amount increases by 10 percent, then you end up with 110 percent of what you start with. 110% is the same as 1.1.
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Self-critique (if necessary):
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If the amount increases by 10% then you end up with 110 percent of what you started with and 110*.1=11. This does not make sense to why you have 1.1 from 10% and 110
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Self-critique Rating:1
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Nothing in the above indicates that 110 should be multiplied by .1.
The preceding statement simply says that 110% is the same as 1.1, which is true, just as each of the following is true:
2% = .02
12% = .12
30% = .30
70% = .70
140% = 1.40
370% = 3.70.
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Question: `q003. Given the recurrence relation P(n) = 1.10 * P(n-1) with P(0) = 1000, substitute n = 1, 2, and 3 in turn to determine P(1), P(2) and P(3). How is this equation related to the situation of the preceding two problems?
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Your solution:
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P(0) = P(1)= 1.10 * P(1000-1)=1098.9
P(1) = P(2)= 1.10 * P(1098.9-1)=1207.69
P(2) = P(3)= 1.10 * P(1207.69-1)=1327.36
The answers are very close to the same of the pervious
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confidence rating #$&*: 1
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Given Solution:
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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Self-critique (if necessary):
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You times your preset answer by the pervious one
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Question: `q004. If you are given $5000 and invest it at 8% annual interest, compounded annually, what number would you multiply by $5000 to get the amount at the end of the first year?
Using the same multiplier, find the results that the end of the second and third years.
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Your solution:
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5000*.8=4000, so 5000/4000=1.25
1.25*5000=end of first year =6200
6200*1.25=7750
7750*1.25=9687
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confidence rating #$&*: 2
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Given Solution:
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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Self-critique (if necessary):
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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. Why would you have 108% as much???? I over all in one is 8% and then you say in that same year from being to end its really 108%
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You still have all the money you started with, which is 100% of your starting amount.
You add 8% of that amount, which gives you 108% of your starting amount.
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Question: `q005. How would you write the recurrence relation for a $5000 investment at 8 percent annual interest?
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Your solution:
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P(n) = 1.10 * P(n-1)
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confidence rating #$&*: 3
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Given Solution:
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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Self-critique (if necessary):
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I still do not understand where you got 108%
because when i divide 5000/4000
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No calculation related to $5000 at 8% results in an amount of $4000.
I beleive this goes back to your having used .8 for 8%.
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@&
In any case you would not divide 8% of 5000 by 5000 to get the current result. This division would give you .08, or 8%; i.e., it would take you right back to where you started.
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Self-critique Rating:1
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Question: `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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Your solution:
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$5000 * 1.08 = $5400
$5400 * 1.08 = $5832.
$5832 * 1.08 = $6298.56.
My Graph is linear
Increasing at an increasing rate
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confidence rating #$&*: 2
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Given Solution:
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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Self-critique (if necessary):
Self-critique Rating:OK
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Question: `q007. With a $5000 investment at 8 percent annual interest, how many years will it take to double the investment?
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Your solution:
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confidence rating #$&*::
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Given Solution:
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 $10,000.02.
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Self-critique (if necessary):
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Okay since it has been 1 year you would add that to .08 therefore making 1.08???
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@&
1.08 is the factor by which you multiply the amount at the beginning of each year to get the amount at the end of that year.
If you do this for n years, you multiply by 1.08 a total of n times, so you can get the result after n year by multiplying the original amount by 1.08^n.
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Question: `q008. If you are given amount P0 and invest it at annual rate r (e.g., for the preceding example r would be 8%, which in numerical form is .08), then after n years how much money do you have?
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Your solution:
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(PO)r
PO(1.08^n)
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confidence rating #$&*: 3
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Given Solution:
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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Self-critique (if necessary):
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Self-critique Rating:OK
Question: `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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Your solution:
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800-240=560mg (800*.1)
5 Hours (800*.5)
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confidence rating #$&*: 3
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Given Solution:
If 10 percent of the antibiotic is removed each hour, then at the end of the hour the amount left will be 90 percent of what was present at the beginning of the hour. Thus after 1 hour we have .90 * 800 mg, after a second hour we have .90 of this, or .90^2 * 800 mg, and after a third hour we have .90 of this, or .90^3 * 800 mg.
The numbers are 800 mg * .90 = 720 mg, then .90 of this or 648 mg, then .90 of this or 583.2 mg.
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Self-critique (if necessary):
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I did not raise my hours by powers
Any amount of time can be raised to a power
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It isn't the number of hours that it being raised to a power.
It is the percent remaining after a year that is raised to a power.
The power is the number of years.
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Self-critique Rating:3
Question: `q010. In the preceding problem, what function Q(t) represents the amount of antibiotic present after t hours? What does a graph of Q(t) vs. t look like?
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Your solution:
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Q(t)=d(decimal)^t*mg
. t(amount of time passed)
Our graph is decreasing at a decreasing rate
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confidence rating #$&*: 3
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Given Solution:
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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Self-critique (if necessary):
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Self-critique Rating:OK
Question: `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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Your solution:
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[300=PO(b^2]-[500=PO(b^6)]
200=PO(b^-4)
@&
If you subtract b^6 from b^2 you don't get b^-4.
b^6 =- b^2 = b^2 ( b^4 - 1),
but
b^6 - b^2 is not b^-4.
b^2 / b^6 = b^-4.
But you can't subtract part of the expression (500 from 300) while you divide the other part (b^2 / b^6).
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200=PO(1/b*4)
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b^4 is not b * 4, or 4 b.
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(4b)200=PO(1/b*4)(4b)
200(4b)=PO
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confidence rating #$&*: 2
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Given Solution:
We substitute the populations 300 then 500 for P and substitute 2 months and 6 months for t to obtain the equations
300 = P0 * b^2 and
500 = P0 * b^6.
Self-critique :OK
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Question: `q012. We obtain the system
300 = P0 * b^2
500 = P0 * b^6
in the situation of the preceding problem.
If we divide the second equation by the first, what equation do we obtain?
What do we get when we solve this equation for b?
If we substitute this value of b into the first equation, what equation do we get?
If we solve this equation for P0 what do we get?
What therefore is our specific P = P0 * b^t function for this problem?
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Your solution:
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(500 = P0 * b^6)/( 300 = P0 * b^2)
5/3(PO)(b2/b1)^4
@&
Your first line did not include anything about b2 or b1. It did include b^6 and b^2.
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confidence rating #$&*:
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Given Solution:
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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Self-critique (if necessary):
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What does the PO stand for???
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P0 stands for the initial priniciple. This notation is consistent with that used in the worksheets and elsewhere in these document.
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Alright this makes sense
5/3=(PO)(b2/b1)^4 If I had put the equal sign to begin with
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P0 will not appear in this expression. It appeared in the numerator and denominator of the expression it came from and it therefore divides out.
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300 = P0 * 1.136^2.
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The 1.136 seems to have just appeared here, without having been connected to any of the preceding.
Are you sure you know how that number arose out of this analysis?
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divide both sides by 1.136^2
P0 = 300 / (1.136^2) = 232.4
P = P0 * b^t
P = 232.4 * 1.136^t. to check
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Self-critique Rating:3
If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
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Question: `q013. If you invest $4000 at 5% interest, compounded annually:
• How much money do you have after 8 years?
• How much money do you have after t years?
• At the end of which year does the amount of money more than double the original amount?
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Your solution:
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$4000*1.05^8=5909.82
$4000*1.05^t=amount earned after t years
No
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confidence rating #$&*:2
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Question: `q014. If you invest a certain unknown amount of money at an unknown interest rate, compounded annually, and if after 5 years you have $1100 and after 10 years the amount is $1300, then:
• What function of the form P = P_0 * b^t models the amount as a function of time?
• What was the original amount?
• What was the interest rate?
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Your solution:
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$1100 = P0 * b^5
$1300 = P0 * b^10
Divide the second by the first
13/11=PO(b2/b1)^2
13/11=PO(b)^2
˝^(13/11)=PO((b)^2)^1/2
.b=1.09 (interest rate)
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If you divide
$1300 = P0 * b^10
by
$1100 = P0 * b^5
P0 divides out and you get
13/11 = b^5.
This will not give you b = 1.09.
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Plug back in to find PO
$1100 = P0 (1.09^5)
Divide each side by 1.54 which is (1.09^5)
714.29=PO (original amount )
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If that 1.09 had been right, this would also be right.
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confidence rating #$&*: 3
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Self-critique rating:N/A
@&
By the end of the exercise you were close to putting everything together. So basically you're not doing badly with the precalculus, but other issues that need to be resolved are making your efforts less effective than they should be.
You made numerous errors related to lower-level mathematics such as percents and basic algebra.
The entire SOL-driven elementary and secondary mathematics curriculum is to blame for this, but that doesn't mean you don't need get these things straight.
Please submit a revision or question form in response to my notes, so we can begin to resolve some of these issues.
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