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R0 - 6000(2–0.078) - Need Help? Watch Talk to Tutor 1/6 points Previous Answers SPRECALC7 4.2.031. My Notes Ask Your Teacher | An investment of $5000 is deposited into an account in which interest is compounded continuously. Complete the table by filling in the amounts to which the investment grows at the Indicated times. (Round your answers to the nearest cent.) Time (years) Amount $ 5435 049 X X Need Help? Wan Submit Answer Practice Another Version - 1 points ~ WANEAC7 2.2.067. My Notes Ask Your Teacher Use technology to find the exponential regression function through the given points. (Round all coefficients to four decimal places.) HINT (See Example 6.1 {(1, 2), (3,5),(4,9), (5, 18) RX) = Nood Help? Tur

Answer 1

The formula to compute amount when interest is compounded continuously is:

Where,
P = Initial Investment amount, i.e 6000
e = mathematical constant
t = Time in Years
r = rate of interest, i.e 7% or 0.07

A= Pert

Therefore, Computation of Amount for each year:

When t = 1

A = 6000 * 0.0741

The value of e^0.07 = 1.0725

A = 6000* 1.0725

A=6435.00

When t = 2

A = 6000 * 0.0742

FF *0009 = 1

The value of e^0.14 = 1.1503

A = 6000* 1.1503

A= 6901.80

When t = 3

A = 6000 * 0.0743

A = 6000 * 0.21

The value of e^0.21 = 1.2337

A = 6000* 1.2337

A = 7402.20

When t = 4

A = 6000 * 0.0744

* *0009 = 1

The value of e^0.28 = 1.3231

A = 6000* 1.3231

A=7938,60

When t = 5

A = 6000 * 0.0745

SE * 0009 = 1

The value of e^0.35 = 1.4191

A = 6000 * 1.4191

A = 8574.60

When t = 6

A = 6000 * 0.074

30*0009 = 1

The value of e^0.42 = 1.5219

A = 6000 * 1.5219

A=9131.40