A savings account with a current balance of $5,000 earns 2.75% annual interest,

compounded continuously. To the nearest dollar, what will be the account balance in 10
years?

To find the account balance after 10 years, we can use the formula for continuous compound interest:

A = P * e^(rt)

Where:
A = final account balance
P = initial balance (principal)
r = annual interest rate (as a decimal)
t = time in years
e = Euler's number (approximately 2.71828)

Let's plug in the values for this scenario:

P = $5,000
r = 2.75% = 0.0275 (as a decimal)
t = 10 years

Now we can calculate the final account balance:

A = $5,000 * e^(0.0275 * 10)

To the nearest dollar, let's calculate the value of e^(0.0275 * 10):

e^(0.0275 * 10) ≈ 1.312

Now we can calculate the final account balance:

A ≈ $5,000 * 1.312

A ≈ $6,560 (rounded to the nearest dollar)

Therefore, the account balance will be approximately $6,560 after 10 years.

a=?,P=5,00,r=2.75,t=10

A=P(1+rt)
A=5,000(1+2.75*10)
A=5,000(1+27.5)
A=5,000(28.5)
A=142,500

5000(1+0.0275*10)

=6375