Consider an expensive part with a relability of 97.4%. If the part fals, it will cost the firm $8,000. a. What is the expected failure cost per part? The probability of failure is K (round your rospanse to two decimal ploces). The expected failure cost per part is $ (round your respanso to the noarest perny). b. On each part, a rather unfeliable backup can be installed that has a reliablity of just 40.00 What is the maximum amount that the firm should be willing to pay per part to install the backup? The new probabifty of talure is Phound your response to two decimal places). The new expected cost of failure is ? (round your response to the nearest penny). The firm should pay as much as $ for the backup (round your response to the nearost pany). c. Suppose that a second 40.00 teliable backup part could be installed, so that if both the original and the first backup part fal, then the second backup part wat be used. If that second backup part costs $120.00, should it be instaled? Support your answer. The new probability of falure is plound your response to Mwo decimal places.). The new expected cost of failure is $ (round your response to the nearest penny). The firm should pay as much as? for the backup (round your response to the nearest penny). Should the firm instat the second backup? Yes No