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):

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

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

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

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.

To solve this problem, we need to calculate the values for Rod's investment and Sheila's investment for each year until Rod's investment becomes worth more than Sheila's.

Rod's investment is compounded daily at a rate of 6%. To calculate the value after one year, we use the formula:

Future Value = Principal * (1 + interest rate)^time

For the first year, Rod's investment of $1,000 will be:

Future Value = $1,000 * (1 + 0.06/365)^(365*1) = $1,061.83 (rounded to the nearest cent)

For the second year, we repeat the calculation using the new principal amount:

Future Value = $1,061.83 * (1 + 0.06/365)^(365*1) = $1,127.49 (rounded to the nearest cent)

Now we need to calculate the simple interest for Sheila's investment. The formula for simple interest is:

Future Value = Principal * (1 + interest rate * time)

For the first year, Sheila's investment of $1,000 will be:

Future Value = $1,000 * (1 + 0.07 * 1) = $1,070.00

For the second year, we use the same formula to calculate the value:

Future Value = $1,070.00 + ($1,070.00 * 0.07) = $1,140.00

To find the number of years when Rod's investment surpasses Sheila's investment, we need to continue these calculations until Rod's investment is greater than Sheila's.

3 $1,195.62 $1,210.00
4 $1,268.25 $1,280.00
5 $1,338.23 $1,350.00
6 $1,405.87 $1,420.00
7 $1,471.48 $1,490.00
8 $1,535.35 $1,560.00
9 $1,597.75 $1,630.00
10 $1,658.96 $1,700.00

Based on the table, we observe that in the 9th year, Rod's investment becomes worth more than Sheila's investment. Therefore, it takes 9 years for Rod's investment to surpass Sheila's investment.