A local bakery opened with new equipment worth $80,000. Ten years later, the owners’ accountants told them that their equipment had depreciated at a rate of 6 percent per year. Which equation would you use to create a sequence modeling this situation?(1 point)

Question

Responses

an=80,000(0.94)n−1
a subscript n baseline equals 80,000 left parenthesis 0.94 right parenthesis superscript n minus 1 baseline

an=80,000(0.06)n
a subscript n baseline equals 80,000 left parenthesis 0.06 right parenthesis superscript n baseline

an=80,000(0.94)n
a subscript n baseline equals 80,000 left parenthesis 0.94 right parenthesis superscript n baseline

an=80,000(0.06)n−1
a subscript n baseline equals 80,000 left parenthesis 0.06 right parenthesis superscript n minus 1 baseline
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Answer 1

To model the depreciation of the bakery's equipment, we need to use the formula that accounts for a constant rate of depreciation. Since the equipment depreciates at a rate of 6 percent per year, the value of the equipment after \( n \) years can be expressed using the formula:

\[ a_n = a_0 (1 - r)^n \]

In this case:

• \( a_0 \) is the initial value of the equipment, which is $80,000.
• \( r \) is the depreciation rate (6 percent or 0.06).
• \( n \) is the number of years.

So the formula for the remaining value after \( n \) years becomes:

\[ a_n = 80,000(1 - 0.06)^n = 80,000(0.94)^n \]

Among the given responses, the correct equation to model this situation is:

\[ a_n = 80,000(0.94)^n \]

So the correct choice is:

an=80,000(0.94)n