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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 sarah
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
Ms. Sue
We have several great math tutors here. I am not one of them since I know little math beyond the elementary level.
Answered by
Reiny
notice for Option 1: 1100 1210 1331
1210 / 1100 = 1331 / 1210 = 1.1
so we have a geometric progression, making the
function exponentinal
term(n) = 1100 (1.1)^(n-1) , where n is the number of years
in Option 2: 1100 1200 1300
we have 1200 - 1100 = 1300 - 1200 = 100
we have a common difference which gives us an arithmetic progression or a linear function
term(n) = 1100 + 100(n-1)
= 1000 + 100n , where n is the number of years
so you want sum(20) for each case
for option 1:
sum(20) = 1100( 1.1^20 - 1)/ .1 = 63002.50
for option 2:
sum(20) = (20/2)(2(1100) + 19(100)) = 41000.00
1210 / 1100 = 1331 / 1210 = 1.1
so we have a geometric progression, making the
function exponentinal
term(n) = 1100 (1.1)^(n-1) , where n is the number of years
in Option 2: 1100 1200 1300
we have 1200 - 1100 = 1300 - 1200 = 100
we have a common difference which gives us an arithmetic progression or a linear function
term(n) = 1100 + 100(n-1)
= 1000 + 100n , where n is the number of years
so you want sum(20) for each case
for option 1:
sum(20) = 1100( 1.1^20 - 1)/ .1 = 63002.50
for option 2:
sum(20) = (20/2)(2(1100) + 19(100)) = 41000.00
Answered by
Anonymous
Whoa
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