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A local bakery opened with new equipment worth $80,000. Ten years later, the owners’ accountants told them that their equipment...Question
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)
Responses
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
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
a subscript n baseline equals 80,000 left parenthesis 0.06 right parenthesis superscript n baseline
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
Responses
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
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
a subscript n baseline equals 80,000 left parenthesis 0.06 right parenthesis superscript n baseline
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
Answers
Answered by
GPT-4o mini
To model the depreciation of the bakery's equipment, you want to use an equation that reflects the value of the equipment decreasing by 6% per year. This means that after each year, 94% (100% - 6%) of the equipment's value remains.
The formula for the value after n years can be expressed as:
\[
a_n = a_0 (1 - r)^n
\]
In this scenario:
- \( a_0 \) is the initial value ($80,000)
- \( r \) is the rate of depreciation (0.06)
- \( n \) is the number of years
This can be calculated as:
\[
a_n = 80,000 (0.94)^n
\]
Where \( 0.94 \) represents the remaining value after 6% depreciation.
So, the correct equation to use is:
\[
\boxed{a_n = 80,000(0.94)^n}
\]
The formula for the value after n years can be expressed as:
\[
a_n = a_0 (1 - r)^n
\]
In this scenario:
- \( a_0 \) is the initial value ($80,000)
- \( r \) is the rate of depreciation (0.06)
- \( n \) is the number of years
This can be calculated as:
\[
a_n = 80,000 (0.94)^n
\]
Where \( 0.94 \) represents the remaining value after 6% depreciation.
So, the correct equation to use is:
\[
\boxed{a_n = 80,000(0.94)^n}
\]
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