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1) $1,800,000 2) $288,000 3) $450,000 4) $352,800 1) $1,800,000 2) $288,000 3) $

ID: 2843655 • Letter: 1

Question

1) $1,800,000 2) $288,000 3) $450,000 4) $352,800
1) $1,800,000 2) $288,000 3) $450,000 4) $352,800 Question 2
$50,000 that was invested in 1990 was worth $134,100 in 2000. What annual interest rate did the investment earn in that 10 year period? Assume continuous compounding. Question 22 options: 1) 118.06% 2) 10.91% 3) 108.20% 4) 9.87% Save $50,000 that was invested in 1990 was worth $134,100 in 2000. What annual interest rate did the investment earn in that 10 year period? Assume continuous compounding. Question 22 options: 1) 118.06% 2) 10.91% 3) 108.20% 4) 9.87% $50,000 that was invested in 1990 was worth $134,100 in 2000. What annual interest rate did the investment earn in that 10 year period? Assume continuous compounding. $50,000 that was invested in 1990 was worth $134,100 in 2000. What annual interest rate did the investment earn in that 10 year period? Assume continuous compounding. 1) 118.06% 2) 10.91% 3) 108.20% 4) 9.87% 1) 118.06% 2) 10.91% 3) 108.20% 4) 9.87% 1) $1,800,000 2) $288,000 3) $450,000 4) $352,800 Suppose that the total salary of all the employees at a factory in a certain city is $300,000. Of this salary, sixty% is spent in the city. Of the money spent in the city, sixty% is again spent in the city. If this continues indefinitely, how much total money is spent in the city, that is, how much of an effect does the original total salary have on the local economy? $50,000 that was invested in 1990 was worth $134,100 in 2000. What annual interest rate did the investment earn in that 10 year period? Assume continuous compounding.

Explanation / Answer

Question 2

A = Pe^rt

A = 134, 100

P = 50,000

r = ?

t = 10


Solve for r


134100 = 50000e^10r

134100/50000 = e^10r

ln(134100/50000) = lne^10r

10r = ln(134100/50000)

r = 1/10 * ln(134100/50000)


Can't find my calculator or I'd simplify it for you, but multiply your answer for r by 100% and that will be your interest.