To determine the simple interest rate required for [tex]$15,000 to double in 8 years, we can use the simple interest formula:
\[ A = P(1 + rt) \]
where:
- \( A \) is the final amount
- \( P \) is the principal amount
- \( r \) is the interest rate
- \( t \) is the time in years
Given:
- The principal amount \( P = \$[/tex]15,000 \)
- The final amount [tex]\( A = 2 \times 15000 = \$30,000 \)[/tex]
- The time [tex]\( t = 8 \)[/tex] years
Rearranging the formula to solve for [tex]\( r \)[/tex]:
[tex]\[ A = P(1 + rt) \][/tex]
[tex]\[ 30000 = 15000(1 + r \times 8) \][/tex]
[tex]\[ 30000 = 15000 + 120000r \][/tex]
[tex]\[ 15000 = 120000r \][/tex]
[tex]\[ r = \frac{15000}{120000} \][/tex]
[tex]\[ r = 0.125 \][/tex]
To convert this to a percentage, we multiply by 100:
[tex]\[ r = 0.125 \times 100 = 12.5\% \][/tex]
Therefore, the simple interest rate needed for $15,000 to double in 8 years is 12.5%.
For the second part of your question, we need to arrange these fractions of a year in ascending order: 190 days, 5 months, 160 days, 6 months, 200 days.
First, we convert all the given time units to a common unit, fractions of a year. Assuming a year has 365 days:
- 190 days: [tex]\( \frac{190}{365} \approx 0.5205 \)[/tex] of a year
- 5 months: [tex]\( 5 \times \frac{30}{365} = \frac{150}{365} \approx 0.410 \)[/tex]
- 160 days: [tex]\( \frac{160}{365} \approx 0.4384 \)[/tex] of a year
- 6 months: [tex]\( 6 \times \frac{30}{365} = \frac{180}{365} \approx 0.4932 \)[/tex]
- 200 days: [tex]\( \frac{200}{365} \approx 0.5479 \)[/tex] of a year
Now, we arrange these fractions in ascending order:
- 5 months: [tex]\( 0.410 \)[/tex] of a year
- 160 days: [tex]\( 0.4384 \)[/tex] of a year
- 6 months: [tex]\( 0.4932 \)[/tex] of a year
- 190 days: [tex]\( 0.5205 \)[/tex] of a year
- 200 days: [tex]\( 0.5479 \)[/tex] of a year
In ascending order, these fractions of a year are:
5 months, 160 days, 6 months, 190 days, 200 days.
