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S=500(1.02)^(t)
The given equation models the amount of money in dollars, 
S, in a savings account 
t years after the initial deposit was made. If the interest rate remains constant, which of the following equations models the amount of money in the savings account 
d decades after the initial deposit was made?
Choose 1 answer:
(A) 
S=500(1.02)^((d)/( 10))
(B) 
S=500(1.02)^(10 d)
(C) 
S=500(1.22)^(10 d)
(D) 
S=500(1.002)^(d)

S=500(1.02)t S=500(1.02)^{t} \newlineThe given equation models the amount of money in dollars, S S , in a savings account t t years after the initial deposit was made. If the interest rate remains constant, which of the following equations models the amount of money in the savings account d d decades after the initial deposit was made?\newlineChoose 11 answer:\newline(A) S=500(1.02)d10 \boldsymbol{S}=500(1.02)^{\frac{d}{10}} \newline(B) S=500(1.02)10d S=500(1.02)^{10 d} \newline(C) S=500(1.22)10d S=500(1.22)^{10 d} \newline(D) S=500(1.002)d S=500(1.002)^{d}

Full solution

Q. S=500(1.02)t S=500(1.02)^{t} \newlineThe given equation models the amount of money in dollars, S S , in a savings account t t years after the initial deposit was made. If the interest rate remains constant, which of the following equations models the amount of money in the savings account d d decades after the initial deposit was made?\newlineChoose 11 answer:\newline(A) S=500(1.02)d10 \boldsymbol{S}=500(1.02)^{\frac{d}{10}} \newline(B) S=500(1.02)10d S=500(1.02)^{10 d} \newline(C) S=500(1.22)10d S=500(1.22)^{10 d} \newline(D) S=500(1.002)d S=500(1.002)^{d}
  1. Understand Equation and Time Conversion: Understand the original equation and the time conversion from years to decades.\newlineThe original equation is S=500(1.02)tS = 500(1.02)^t, where tt is the time in years. Since 11 decade is equal to 1010 years, we need to find the equivalent expression where tt is replaced by the number of decades, dd, multiplied by 1010.
  2. Convert Time to Decades: Convert the time variable from years to decades.\newlineTo model the amount of money in the savings account after dd decades, we need to replace tt with 10d10d in the original equation because each decade consists of 1010 years.\newlineThe new equation becomes S=500(1.02)10dS = 500(1.02)^{10d}.
  3. Check Answer Choices: Check the answer choices to see which one matches the equation derived in Step 22.\newlineThe correct equation that models the amount of money in the savings account after dd decades is S=500(1.02)10dS = 500(1.02)^{10d}, which matches answer choice (B).

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