Respuesta :
Answer:
overall performance 0.000297
Explanation:
cost of inputs:
labor 10,500 hours x 25 each = 262,500
suspension and engine 500 kits x 1,000 each = 500,000
energy 100,000 kilowatt-hous x 5 each = 500,000
total input resource cost 1,262,500
performance:
375 / 1,262,500= 0.00029703
rounding to 6 decimal places: 0.000297
The overall performance at Munson Performance Auto, Inc. will be 0.000297 auto per dollar of input if it modifies around 375 autos every year.
How to calculate overall performance?
For calculation of the overall performance, all the input costs need to be calculated by using the information given above as,
[tex]\rm Labor\ Cost = Cost\ per\ Hour\ x\ Total\ Hours\\\\\rm Labor\ Cost = 10500\ x\ 25\\\\\rm Labor\ Cost = \$262500[/tex]
Calculating further,
[tex]\rm Kits\ Cost= 500\ x\ 1000\\\\\rm Kits\ Cost= \$500000[/tex]
And,
[tex]\rm Energy\ Costs = 100000\ x\ 5\\\\\rm Energy\ Costs = \$500000[/tex]
So, the total costs of the Munson Performance Inc. will be $1,262,500. Now calculating the overall performance costs,
[tex]\rm Overall\ Performance = \dfrac{No.\ of\ Autos}{Overall\ Costs} \\\\\rm Overall\ Performance = \dfrac{375}{1262500}\\\\\rm Overall\ Performance = 0.000297[/tex]
Hence, the overall performance of the Munson Performance Auto, Inc. will be 0.00297 autos per every dollar spent if it modifies 375 autos every year.
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