Respuesta :

The correct answer would be B.

The lower e is used for continuous compounding and it is raised by the interest rate times the amount of time

Formula  that describes the value of an initial investment of [tex]\$300[/tex] growing at an interest rate of [tex]6\%[/tex] compounded continuously is equals to [tex]A(t) = 300e^{.06t}[/tex].

What is compounded continuously?

" Compounded continuously is defined as the interest calculation and reinvestment of the amount over infinite period."

Formula used

[tex]A(t) = P e^{rt}[/tex]

[tex]A(t) =[/tex] Final amount

[tex]P =[/tex] Principal amount

[tex]t =[/tex]  time period interest is applied

[tex]r=[/tex] rate of interest

According to the question,

Given,

Principal amount [tex]= \$300[/tex]

Rate of interest [tex]= 6\%[/tex]

As per the given condition interest compounded continuously,

Substitute the value in the formula of interest compounded continuously  we get,

[tex]A(t) = 300 e^{\frac{6}{100} t}\\\\\implies A(t) = 300 e^{.06 t}[/tex]

Hence, Option (B) is the correct answer.

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