Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Initially you have two working phones: an older “2+1” smartphone and a newer “me

ID: 1132824 • Letter: I

Question

Initially you have two working phones: an older “2+1” smartphone and a newer “mexus 6”. Unfortunately, during a peaceful dinner with your family, the beautiful “mexus 6” hits the floor, so its screen is completely shuttered, and the phone is unusable. For each phone you have several options in mind. For “2+1”, you can either do nothing (keep it) or sell it for $100. Since your “mexus 6” was purchased within last 30 days, you are still eligible for purchasing an insurance that covers drops and leakages, with premium $120 for a year plus $40 additional cost for each accident (insurance deductible is $40 and loss is not completely covered by the insurance payment). Overall, there are several options with “mexus 6”: do nothing (throw it away); purchase the insurance and use it to repair the phone; repair without insurance, which costs $150; sell a repaired phone for $250 (when no insurance was purchased); sell a repaired phone with insurance for $300. Also, you may either buy or not a new different phone for $400.

1. Suppose that you are willing to end up with exactly one working phone. How many action scenarios can lead you to that outcome? (A scenario is a combination of decisions regarding three phones: "2+1", "mexus 6", and the new one.)

2.Given that you end up with exactly two working phones (one major, one backup), what is the smallest total cost of the scenario achieving that (i.e., what is the smallest amount of money that you need to spend if you want to have two phones)?

Explanation / Answer

In this whole scenario, we have three phones i.e. "2+1", "Mexus 6" & "New one".

1. Lets discuss all the possible actions to end up with exactly one working phone with respect to each phone in following points:

2.To end up with two working phones by spending smallest total cost, the best possible scenario can be that you keep "2+1" and repair "Mexus 6" for $150. So in this case, the cost to have two working phone would be $150 which is smallest.