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otez555 [7]
3 years ago
5

Alexandra invested $800 in an account paying an interest rate3.1% compounded continuously. Assuming no deposits or withdrawals a

re made, how much money , to the nearest hundred dollars, would be in the account after 20 years
Mathematics
1 answer:
s2008m [1.1K]3 years ago
8 0

\bf ~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$800\\ r=rate\to 3.1\%\to \frac{3.1}{100}\dotfill &0.031\\ t=years\dotfill &20 \end{cases} \\\\\\ A=800e^{0.031\cdot 20}\implies A=800e^{0.62} \\\\\\ A\approx 1487.14\implies A=\stackrel{rounded~up}{1500}

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Now that we have our slope, we can start filling out the slope-intercept form equation.

y=mx+b

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Since we don't know the y-intercept, we can use one of the given points and solve for b.

6=(-13/12)(-12)+b                              [multiply (-13/12) and -12]

6=13+b                                              [subtract both sides by 13]

b=-7

With the y-intercept, we can complete our equation.

y=-13/12x-7

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