Question

TASK TIME A. Greet/register the patient 3 minutes/patient B. Optometrist conducts eye exam 20 minutes/patient C. Frame/lenses selection 24 minutes/patient D. Glasses made (process can run 6 pairs of glasses at the same time) 66 minutes/patient E. Final fitting 5 minutes/patient

Answer 1

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Solution - Answer (a): The current maximum output would depend on slowest operations and that is B. Optometrist conducting eye exam (25min per patient) Hence total time available for task(B)= 10 hours * 60 min - (2 minute for first patient to greet + frame selection 20 min + glasses made 60 min + final fitting 5 min) = 600 -87 = 513 minutes. Hence, maximum capacity = 513/25 = 20.52 ~ 20 patients Answer (b): Since, the slowest process is B. Optometrist conducting eye test, if a person were to be added it must be there. The added capacity at no other place would increase total capacity of the patients processed in a day. Answer (c): In the mentioned model, the process of D and E, totaling to 65 minutes would be eliminated. And hence, in day...

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