Complete the table to determine the balance A for $12,000 invested at rate r for t years, compounded continuously. (Round your answers to the nearest cent.)

r = 1.5%

find
T1-1.2
T10-2.9
T20-8.4
T30-24.3
T40-70.4
T50-203.9

To determine the balance A for $12,000 invested at a rate of 1.5% for t years, compounded continuously, we can use the formula:

A = P * e^(rt)

Where:
A = Balance after t years
P = Principal amount (initial investment)
e = Euler's number (approximately 2.71828)
r = Interest rate (in decimal form)
t = Time period (in years)

Let's calculate the balances for the given time periods:

T1:
A = 12000 * e^(0.015 * 1)
A ≈ 12000 * e^(0.015)
A ≈ 12000 * 1.01511305483
A ≈ 12181.36

T10:
A = 12000 * e^(0.015 * 10)
A ≈ 12000 * e^(0.15)
A ≈ 12000 * 1.16183424228
A ≈ 13942.01

T20:
A = 12000 * e^(0.015 * 20)
A ≈ 12000 * e^(0.3)
A ≈ 12000 * 1.34985880758
A ≈ 16198.31

T30:
A = 12000 * e^(0.015 * 30)
A ≈ 12000 * e^(0.45)
A ≈ 12000 * 1.6497212707
A ≈ 19796.65

T40:
A = 12000 * e^(0.015 * 40)
A ≈ 12000 * e^(0.6)
A ≈ 12000 * 1.82211880039
A ≈ 21865.43

T50:
A = 12000 * e^(0.015 * 50)
A ≈ 12000 * e^(0.75)
A ≈ 12000 * 2.00855369232
A ≈ 24102.65

So, the complete table is as follows:

T1: 12181.36
T10: 13942.01
T20: 16198.31
T30: 19796.65
T40: 21865.43
T50: 24102.65

To find the balance A for $12,000 invested at a continuously compounded interest rate of 1.5% for t years, we use the formula:

A = P * e^(rt)

where:
- A is the final balance
- P is the principal amount (initial investment)
- e is the mathematical constant approximately equal to 2.71828
- r is the interest rate (expressed as a decimal)
- t is the number of years

To complete the table and find the balances for each given time period, we can plug in the values of r, P, and t into the formula and calculate A.

For T1 (1 year), r = 1.5%, t = 1 year:
A = $12,000 * e^((0.015)(1))
A ≈ $12,000 * e^(0.015)
A ≈ $12,000 * 1.01515
A ≈ $12,181.80
Therefore, T1-1.2 ≈ $12,181.80 (rounded to the nearest cent).

For T10 (10 years), r = 1.5%, t = 10 years:
A = $12,000 * e^((0.015)(10))
A ≈ $12,000 * e^(0.15)
A ≈ $12,000 * 1.16183
A ≈ $13,941.96
Therefore, T10-2.9 ≈ $13,941.96 (rounded to the nearest cent).

For T20 (20 years), r = 1.5%, t = 20 years:
A = $12,000 * e^((0.015)(20))
A ≈ $12,000 * e^(0.3)
A ≈ $12,000 * 1.34986
A ≈ $16,198.32
Therefore, T20-8.4 ≈ $16,198.32 (rounded to the nearest cent).

For T30 (30 years), r = 1.5%, t = 30 years:
A = $12,000 * e^((0.015)(30))
A ≈ $12,000 * e^(0.45)
A ≈ $12,000 * 1.57353
A ≈ $18,882.36
Therefore, T30-24.3 ≈ $18,882.36 (rounded to the nearest cent).

For T40 (40 years), r = 1.5%, t = 40 years:
A = $12,000 * e^((0.015)(40))
A ≈ $12,000 * e^(0.6)
A ≈ $12,000 * 1.82212
A ≈ $21,865.44
Therefore, T40-70.4 ≈ $21,865.44 (rounded to the nearest cent).

For T50 (50 years), r = 1.5%, t = 50 years:
A = $12,000 * e^((0.015)(50))
A ≈ $12,000 * e^(0.75)
A ≈ $12,000 * 2.11700
A ≈ $25,404.00
Therefore, T50-203.9 ≈ $25,404.00 (rounded to the nearest cent).

Please note that all values are rounded to the nearest cent.