The Investment Problem. Suppose that Rod invests $1,000 at 6% compounded daily and Sheila invests $1,000 at 7% (per year) simple interest. In how many years will Rod’s investment be worth more than Sheila’s investment? Complete the following table to answer your question (the amounts for the first 2 years are given; you may not have to do all 10 years):

Question

Year 6% compounded daily 7% simple
1 $1,061.83 $1,070.00
2 $1,127.49 $1,140.00
3
4
5
6
7
8
9
10

Answer 1

You didn't give the table. You can't copy and paste; you must type it in.

Answer 2

Here is the table:
Year 6% Compounded daily 7% simple
1 $1,061.83 $1,070.00
2 $1,127.49 $1,140.00
3
4
5
6
7
8
9
10

Answer 3

For 6% compounded daily, do this.
[(0.06/365)+ 1]^365 = 1.06183 and that x 1000 = $1061.83 for year 1.
For year 2, just take 1061.83*1.06183 = $1127.49.
For year 3, it will be 1127.49 x 1.06183 = ?? etc.

For 7% simple,
For year 1 it is $1,000 x 0.07 = 70.00 and 1,000 + 70.00 = $1070.00
For year 2 it is 1070.00 x 0.07 = 74.90 and 1,070.00 + 70.90 = 1,140.90. etc.