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Belinda wants to invest $1000. The table below shows the value of her investment under two different options for two different...Asked by Anon
Belinda wants to invest $1000. The table below shows the value of her investment under two different options for three different years:
Number of years 1 2 3
Option 1 (amount in dollars) 1100 1210 1331
Option 2 (amount in dollars) 1100 1200 1300
Part A: What type of function, linear or exponential, can be used to describe the value of the investment after a fixed number of years using option 1 and option 2? Explain your answer. (2 points)
Part B: Write one function for each option to describe the value of the investment f(n), in dollars, after n years. (4 points)
Part C: Belinda wants to invest in an option that would help to increase her investment value by the greatest amount in 20 years. Will there be any significant difference in the value of Belinda's investment after 20 years if she uses option 2 over option 1? Explain your answer, and show the investment value after 20 years for each option. (4 points)
Number of years 1 2 3
Option 1 (amount in dollars) 1100 1210 1331
Option 2 (amount in dollars) 1100 1200 1300
Part A: What type of function, linear or exponential, can be used to describe the value of the investment after a fixed number of years using option 1 and option 2? Explain your answer. (2 points)
Part B: Write one function for each option to describe the value of the investment f(n), in dollars, after n years. (4 points)
Part C: Belinda wants to invest in an option that would help to increase her investment value by the greatest amount in 20 years. Will there be any significant difference in the value of Belinda's investment after 20 years if she uses option 2 over option 1? Explain your answer, and show the investment value after 20 years for each option. (4 points)
Answers
Answered by
Damon
option 1 is not linear because it does not change by the same amount every year. It does not have a constant slope.
option 2 increases by 100 per year, constant slope, straight line, linear
-----------------------
a = A y^n
1100 =A y
1210 = Ay^2
1331 = A y^3
---------------
1210/1100 = y = 1.1
check 1331/1210 = y = 1.1 sure enough
1100 = A (1.1)
A = 1000
so
a = 1000 * 1.1^n
and the linear option 2
a = 1000+100n
You can put the numbers in for part C
10% compounded is better than 10% simple :)
option 2 increases by 100 per year, constant slope, straight line, linear
-----------------------
a = A y^n
1100 =A y
1210 = Ay^2
1331 = A y^3
---------------
1210/1100 = y = 1.1
check 1331/1210 = y = 1.1 sure enough
1100 = A (1.1)
A = 1000
so
a = 1000 * 1.1^n
and the linear option 2
a = 1000+100n
You can put the numbers in for part C
10% compounded is better than 10% simple :)
Answered by
Baka
Nice cheating Anon, I guess I was right about you not being any different from the person from last time. :>
Answered by
no
no
Answered by
wimpy kid
was this right?
Answered by
Terral Stewart
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