Two methods can be used to produce expansion anchors. Method A costs $70,000 initially and will have a $18,000 salvage value after 3 years. The operating cost with this method will be $29,000 in year 1, increasing by $4000 each year. Method B will have a first cost of $130,000, an operating cost of $6000 in year 1, increasing by $6000 each year, and a $40,000 salvage value after its 3-year life. At an interest rate of 14% per year, what are the present worth of Method A and Method B? O a. PWA $-133,655 PWB $-129,647 O b. PWA $-133,655 PWB $-125,178 O C.PWA = $-150,260 PWB $-129,647 O d. PWA $-150,260 PWB = $-125,178

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Two methods can be used to produce expansion anchors. Method A costs $70,000 initially and will have a $18,000 salvage value after 3 years. The
operating cost with this method will be $29,000 in year 1, increasing by $4000 each year. Method B will have a first cost of $130,000, an operating
cost of $6000 in year 1, increasing by $6000 each year, and a $40,000 salvage value after its 3-year life. At an interest rate of 14% per year, what
are the present worth of Method A and Method B?
a. PWA = $-133,655
PWB = $-129,647
PWB = $-125,178
PWB = $-129,647
PWB = $-125,178
O b. PWA $-133,655
=
O C.PWA $-150,260
O d. PWA = $-150,260
=
Transcribed Image Text:Two methods can be used to produce expansion anchors. Method A costs $70,000 initially and will have a $18,000 salvage value after 3 years. The operating cost with this method will be $29,000 in year 1, increasing by $4000 each year. Method B will have a first cost of $130,000, an operating cost of $6000 in year 1, increasing by $6000 each year, and a $40,000 salvage value after its 3-year life. At an interest rate of 14% per year, what are the present worth of Method A and Method B? a. PWA = $-133,655 PWB = $-129,647 PWB = $-125,178 PWB = $-129,647 PWB = $-125,178 O b. PWA $-133,655 = O C.PWA $-150,260 O d. PWA = $-150,260 =
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