$~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$5150\\ P=\textit{original amount deposited}\dotfill & \$4500\\ r=rate\to 7.5\%\to \frac{7.5}{100}\dotfill &0.075\\ t=years \end{cases}$

$5150=4500e^{0.075\cdot t} \implies \cfrac{5150}{4500}=e^{0.075t}\implies \cfrac{103}{90}=e^{0.075t} \\\\\\ \log_e\left( \cfrac{103}{90} \right)=\log_e(e^{0.075t})\implies \log_e\left( \cfrac{103}{90} \right)=0.075t \\\\\\ \ln\left( \cfrac{103}{90} \right)=0.075t\implies \cfrac{\ln\left( \frac{103}{90} \right)}{0.075}=t\implies\stackrel{\textit{about 1 year and 291 days}}{ 1.8\approx t}$

4 0

1 year ago

Jaime wrote 4.4_0.33=1.1 is his answer reasonable?why're why not?

shusha [124]

No you have to make sure to line up the decimals when subtracting them. Do not just line of the numbers, this wills get you the wrong answers. Hope this helps. If you need anything else just call for a helping hand.

7 0

3 years ago

Liam puts $2000 in the bank with a 3% annual interest rate compounded annually. if Liam does not touch his money how much money

inna [77]

D. He will have $2,121.80.

After one year at 3% interst, $2000 becomes $2000 x 1.03 = $2,060.
During the second year, that $2060 x 1.03 becomes $2121.80

8 0

3 years ago

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