Investment Problems An investment of P dollars is deposited in a savings account that is compounded monthly with an annual interest rate of r , where r is expressed as a decimal. The amount of money A in the account after t years is given by A = P ( 1 + r / 12 ) 12 t . Use this equation to complete the following exercises. 63. Determine the time it takes an investment of $1000 to increase to $1100 dollars if it is placed in an account that is compounded monthly with an annual interest rate of 1%( r = 0.01).
Investment Problems An investment of P dollars is deposited in a savings account that is compounded monthly with an annual interest rate of r , where r is expressed as a decimal. The amount of money A in the account after t years is given by A = P ( 1 + r / 12 ) 12 t . Use this equation to complete the following exercises. 63. Determine the time it takes an investment of $1000 to increase to $1100 dollars if it is placed in an account that is compounded monthly with an annual interest rate of 1%( r = 0.01).
Solution Summary: The author explains how the time taken to increase the investment of 1000 to 1100 is 9.53 years.
Investment Problems An investment of P dollars is deposited in a savings account that is compounded monthly with an annual interest rate of r, where r is expressed as a decimal. The amount of money A in the account after t years is given by
A
=
P
(
1
+
r
/
12
)
12
t
. Use this equation to complete the following exercises.
63. Determine the time it takes an investment of $1000 to increase to $1100 dollars if it is placed in an account that is compounded monthly with an annual interest rate of 1%(r = 0.01).
The amount of money in an account is given by A = P(1+r)*, where
Pis the principal invested, r is the interest rate (as a decimal), and t is
the time of the investment.
Find the amount at the end of (t=) 2 years if (P=) $600 is invested at 5%
(r=0.05). Use appropriate units and round to the nearest cent.
hp
fg
f10
f11
f12
AA
8
60
Chapter 1 Solutions
Single Variable Calculus Format: Unbound (saleable)
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