The bookkeeper recorded a deposit of \( \$ 100 \) as \( \$ 10 \). On the bank reconciliation, this will be a(n): A. addition of \( \$ 90 \) to the balance per bank. B. addition of \( \$ 90 \) to the balance per books. C. subtraction of \( \$ 90 \) to the balance per books. D. subtraction of \( \$ 90 \) from the balance per bank.

\( \$ 5000 \) are invested in a bank account at an interest rate of 5 percent per year. Find the amount in the bank after 13 years if interest is compounded annually. Find the amount in the bank after 13 years if interest is compounded quarterly. Find the amount in the bank after 13 years if interest is compounded monthly. Finally, find the amount in the bank after 13 years if interest is compounded continuously.

If you want to earn \( 4 \% \) annual simple interest on an investment, how much should you pay for a note that will be worth \( \$ 18,500 \) in 10 months? (Round your answer to two decimal places.) \( \$ \) Read It

We deposit \( \$ 17000 \) into an account earning \( 4 \% \) interest compounded semiannually. How many years will it take for the account to grow to \( \$ 42500 \) ? Round to 2 decimal places. \( \square \) years

A car was valued at \( \$ 31,000 \) in the year 1991 . The value depreciated to \( \$ 13,000 \) by the year 2007 . A) What was the annual rate of change between 1991 and 2007? \( r=\square \) Round the rate of decrease to 4 decimal places. B) What is the correct answer to part A written in percentage form? \( r=\square \). C) Assume that the car value continues to drop by the same percentage. What will the value be in the year 2010 ? value \( =\$ \square \) Round to the nearest 50 dollars.

\( \$ 5000 \) are invested in a bank account at an interest rate of 5 percent per year. Find the amount in the bank after 13 years if interest is compounded annually. 9428 Find the amount in the bank after 13 years if interest is compounded quarterly. 9539 Find the amount in the bank after 13 years if interest is compounded monthly. Finally, find the amount in the bank after 13 years if interest is compounded continuously.