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Sagot :
To find the ending balance of an investment compounded continuously, we use the formula for continuous compounding:
[tex]\[ A = P \cdot e^{(rt)} \][/tex]
where:
- [tex]\( A \)[/tex] is the final amount,
- [tex]\( P \)[/tex] is the principal amount (initial investment),
- [tex]\( e \)[/tex] is the base of the natural logarithm (approximately equal to 2.71828),
- [tex]\( r \)[/tex] is the annual interest rate (expressed as a decimal),
- [tex]\( t \)[/tex] is the time in years.
In this problem, we are given:
- [tex]\( P = 1000 \)[/tex] (the initial investment amount),
- [tex]\( r = 0.16 \)[/tex] (the annual interest rate as a decimal),
- [tex]\( t = 5 \)[/tex] years.
Plugging these values into the formula, we have:
[tex]\[ A = 1000 \cdot e^{(0.16 \cdot 5)} \][/tex]
Calculating the exponent, we get:
[tex]\[ 0.16 \cdot 5 = 0.8 \][/tex]
So the formula becomes:
[tex]\[ A = 1000 \cdot e^{0.8} \][/tex]
Using the value of [tex]\( e^{0.8} \)[/tex], we find:
[tex]\[ A = 1000 \cdot 2.225540928492468 \][/tex]
Multiplying these numbers together, we get:
[tex]\[ A = 2225.540928492468 \][/tex]
Thus, the ending balance after 5 years is approximately [tex]$2,225.54 when rounded to two decimal places and formatted in the $[/tex]#,###.## format.
So, the final answer is:
[tex]\[ \boxed{\$2,225.54} \][/tex]
[tex]\[ A = P \cdot e^{(rt)} \][/tex]
where:
- [tex]\( A \)[/tex] is the final amount,
- [tex]\( P \)[/tex] is the principal amount (initial investment),
- [tex]\( e \)[/tex] is the base of the natural logarithm (approximately equal to 2.71828),
- [tex]\( r \)[/tex] is the annual interest rate (expressed as a decimal),
- [tex]\( t \)[/tex] is the time in years.
In this problem, we are given:
- [tex]\( P = 1000 \)[/tex] (the initial investment amount),
- [tex]\( r = 0.16 \)[/tex] (the annual interest rate as a decimal),
- [tex]\( t = 5 \)[/tex] years.
Plugging these values into the formula, we have:
[tex]\[ A = 1000 \cdot e^{(0.16 \cdot 5)} \][/tex]
Calculating the exponent, we get:
[tex]\[ 0.16 \cdot 5 = 0.8 \][/tex]
So the formula becomes:
[tex]\[ A = 1000 \cdot e^{0.8} \][/tex]
Using the value of [tex]\( e^{0.8} \)[/tex], we find:
[tex]\[ A = 1000 \cdot 2.225540928492468 \][/tex]
Multiplying these numbers together, we get:
[tex]\[ A = 2225.540928492468 \][/tex]
Thus, the ending balance after 5 years is approximately [tex]$2,225.54 when rounded to two decimal places and formatted in the $[/tex]#,###.## format.
So, the final answer is:
[tex]\[ \boxed{\$2,225.54} \][/tex]
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