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Eduardwww [97]
3 years ago
12

Lambda Computer Products competed for and won a contract to produce two prototype units of a new type of computer that is based

on laser optics rather than on electronic binary bits. The first unit produced by Lambda took 5,000 hours to produce and required $250,000 worth of material, equipment usage, and supplies. The second unit took 3,250 hours and used $187,500 worth of materials, equipment usage, and supplies. Labor is $20 per hour. Use Exhibit 6.5. a. Lambda was asked to present a bid for 10 additional units as soon as the second unit was completed. Production would start immediately. What would this bid be
Business
1 answer:
deff fn [24]3 years ago
5 0

Answer:

$1,330,102.50

Explanation:

first unit produced by lambda took 5,000 hours to produce and required $250,000 worth of material, equipment usage, and supplies

the second unit took 3,250 hours and used $187,500 worth of materials, equipment usage, and supplies

learning rate = time needed to produce second unit / time needed to produce first unit = 3,250 hours / 5,000 hours = 65%

materials and equipment usage rate = $187,500 / $250,000 = 75%

using the attached table of cumulative values, we can determine the cumulative improvement factors needed to solve this question:

Lambda's accumulated cost for producing 10 more computers

  • work hours = 3,250 x 4.341 (65% and 10 units) x $20 per hour = $282,165
  • materials and equipment = $187,500 x 5.589 (75% and 10 units) = $1,047,937.50
  • total = $282,165 + $1,047,937.50 = $1,330,102.50

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3 years ago
Consider the following marginal cost function. a. Find the additional cost incurred in dollars when production is increased from
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Explanation:

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\int^{150}_{100}\ C'(x)\ dx\\\\=\int^{150}_{100} (4000-0.4x)\ dx\\\\=[4000x-\dfrac{0.4x^2}{2}]^{150}_{100}\\\\=[4000(150)-\dfrac{0.4(150)^2}{2}-4000(100)+\dfrac{0.4(100)^2}{2}]\\\\=[600000-4500-400000+2000]\\\\=197500

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\int^{550}_{500}\ C'(x)\ dx\\\\=\int^{550}_{500} (4000-0.4x)\ dx\\\\=[4000x-\dfrac{0.4x^2}{2}]^{550}_{500}\\\\=[4000(550)-\dfrac{0.4(550)^2}{2}-4000(500)+\dfrac{0.4(500)^2}{2}]\\\\=[2200000-60500-2000000+50000]\\\\=189,500

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3 years ago
How to write a letter to your brother in abroad telling him your plans after senior high school<br>​
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Answer:

easy

Explanation:

1. address

2. greeting

3. reason for this letter

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5. conclusion

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