Assignment 15

course Mth 163

July 13 11:30 am*********************************************

Question: `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 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:

First year:

$1000*10%= $100 interest or $1100 total

Second Year:

$1100*10%= $110 interest or $1210 total

Third year:

$1210*10%= $121 interest or $1331 total

Interest increases every year because the total sum of money each year increases.

Confidence rating: 3

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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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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:

You multiply them all by 1.1. The 1 is for the original amount (in other words, 100% of the original amount) and then .1 is for 10% extra of the original amount.

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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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:

P(n) = 1.10 * P(n-1)

N=1

P(1)= 1.10*P(1-1)

P= 1.10*1000 = 1100

N=2

P(2)= 1.10*p(2-1)

P= 1.10*1100= 1210

N=3

P(3)= 1.1*p(3-1)

P= 1.10*1210= 1331

Confidence rating: 3

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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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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:

For the total amount, you would multiply each year by 1.08.

First year:

$5000*0.08= $400 interest or $5400 total

Second year:

$5400*0.08= $432 interest or $5832 total

Third year:

$5832*0.08= $466.56 interest or $6298.56 total

Confidence rating: 3

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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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Question: `q005. How would you write the recurrence relation for a $5000 investment at 8 percent annual interest?

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Your solution:

P(n)= 1.08*p(n-1)

It is just like the equation before, but 1.08 for 8% instead of 1.10 for 10%.

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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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:

I had to look at the solution on this one because I wasn’t quite sure what the question was asking. I believe I can put it into my own words, though. For every year, you can take the original equation of 5000 multiplied by 1.08, and for each consecutive year, do the equation to the power of whatever number year it is. For example, for the 6th year, you would do 5000*1.08^6. The graph continuously climbs in an upward way because the interest and total amount continues to grow as the years go by.

Confidence rating: 3

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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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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:

5000*1.08= 5400

5000*1.08^2= 5832

5000*1.08^3= 6298.56

5000*1.08^4= 6802.45

5000*1.08^5= 7346.64

5000*1.08^6= 7934.37

5000*1.08^7= 8569.12

5000*1.08^8= 9254.65

5000*1.08^9= 9995.02

It is almost doubled by the end of year 9, but actually would take a little longer than just year 9.

Confidence rating: 3

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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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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:

P(0)*1.08^nth year

Confidence rating: 2

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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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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:

800*.9(for the amount remaining)= 720 mg

720*.9= 648

648*.9= 583.2  would be left after the first 3 hours

583.2*.9= 524.88

524.88*.9= 472.392

472.392*.9= 425.1528

425.1528*.9= 382.63752

583.2 would be left after the first 3 hours.

It would take between 6 and 7 hours for half of the antibiotics to be gone.

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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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:

Q(t)= 800*.9^nth

As the hours increase, the amount decreases.

Confidence rating: 2

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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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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:

P(0)*b^2= 300

P(0)*b^6= 500

Confidence rating: 3

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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.

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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:

P(0)*b^2= 300

Divided by

P(0)*b^6= 500

B^4= 500/300

B^4/(1/4)= 1.6667^(1/4)

Confidence rating: 1

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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.

&#This looks good. Let me know if you have any questions. &#