Q:

If you can show your work so I can learn that would be great (all three questions go together). If you have any questions please ask1. You and your best friend want to take a vacation to Peru. You have done some research and discovered that it will cost $2400 for the plane tickets, all-inclusive hotel and resort, and souvenirs. You have already saved $1750. If you invest this money in a savings account with a 2.3% interest rate compounded annually, how long will it take to earn enough money to go on the trip? Use the compound interest formula A = P (1 + i)n, where A is the accumulated amount, P is the principal, i is the interest rate per year, and n is the number of years. Round your final answer to the nearest tenth2. You are planning to go on this trip in 2 years. How much money will you need to invest at a 2.3% interest rate compounded annually in order to have $2400 in 2 years? Use the compound interest formula A = P (1 + i)n. (Round final answer to the nearest cent, but otherwise don’t round any intermediate values)3. Now say you only have $1600 to invest and the highest interest rate you can find is 3.55% compounded annually. If you decide to wait 7 years to go on the trip, how much money will you have to spend on the trip? Use the compound interest formula A = P (1 + i)n. (Round final answer to the nearest cent, but otherwise don’t round any intermediate values)

Accepted Solution

A:
1) 13.9 years
2) $2293.30
3) $2042.54

Explanation:
1) Using the equation
[tex]A=p(1+i)^n[/tex], plugging in our values, we have:

[tex]2400=1750(1+0.023)^n \\ \\2400=1750(1.023)^n \\ \\\frac{2400}{1750}=1.023^n[/tex]

Using logarithms to solve this, we have:
[tex]\log_{1.023}(\frac{2400}{1750})=n \\ \\13.9=n[/tex]

2)  This time, we substitute different values into our equation:
[tex]A=p(1+i)^n \\ \\2400=p(1+0.023)^2 \\ \\2400=p(1.023)^2 \\ \\\frac{2400}{1.023^2}=p \\ \\2293.30 = p[/tex]

3)  This time we change all of the values in our equation:
[tex]A=p(1+i)^n \\ \\A=1600(1+0.035)^7 \\ \\A=1600(1.035)^7 \\ \\A=2042.54[/tex]