NO LINKS!! An investment of P dollars increased to A dollars in t years. If interest was compounded continuously, find the interest rate. (Round your answer to the nearest whole number.)

A= 1822,. P= 1000,. t= 15

NO LINKS An investment of P dollars increased to A dollars in t years If interest was compounded continuously find the interest rate Round your answer to the ne class=

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Answer:

  • 4%

Step-by-step explanation:

Use continuous compound equation:

  • [tex]A = P*e^{rt}[/tex], where A- future amount, P- invested amount, t- time, r- rate

Given

  • A = $1822
  • P = $1000,
  • t = 15 years.

Plug in the values and solve for r

  • [tex]1822 = 1000*e^{r*15}[/tex]
  • [tex]e^{15r}=1822/1000[/tex]
  • [tex]e^{15r}=1.822[/tex]
  • [tex]ln\ e^{15r}=ln\ 1.822[/tex]
  • [tex]15r=0.6[/tex]
  • [tex]r=0.6/15[/tex]
  • [tex]r=0.04 \ rounded[/tex]

Answer:

4% (nearest whole number)

Step-by-step explanation:

Continuous Compounding Formula

[tex]\large \text{$ \sf A=Pe^{rt} $}[/tex]

where:

  • A = Final amount.
  • P = Principal amount.
  • e = Euler's number (constant).
  • r = Annual interest rate (in decimal form).
  • t = Time (in years).

Given values:

  • A = $1,822
  • P = $1,000
  • t = 15 years

Substitute the given values into the formula and solve for r:

[tex]\implies \sf 1822=1000 \cdot e^{15r}[/tex]

[tex]\implies \sf \dfrac{1822}{1000}=e^{15r}[/tex]

[tex]\implies \sf 1.822=e^{15r}[/tex]

[tex]\implies \sf \ln 1.822=\ln e^{15r}[/tex]

[tex]\implies \sf \ln 1.822=15r \ln e[/tex]

[tex]\implies \sf \ln 1.822=15r[/tex]

[tex]\implies \sf r=\dfrac{\ln 1.822}{15}[/tex]

[tex]\implies \sf r=0.039995653...[/tex]

To convert into a percentage, multiply by 100:

[tex]\implies \sf r=0.039995653... \times 100[/tex]

[tex]\implies \sf r=3.99995653...\%[/tex]

[tex]\implies \sf r=4\%\; (nearest\;whole\;number)[/tex]

Therefore, the interest rate is 4% (nearest whole number).

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