Answer:
The minimum number of bays to achieve an anticipated production of 100 cars per hour day is 3.
Explanation:
The numbers seem to be doubled.
We have the following informations:
The effective capacity of a Fast Lube service bay is 6.0.
The efficience is 0.85.
We have that
Expected output = effective capacity * efficience
Expected output = 6*0.85
Expected output = 5.1
The expected output is 5.1 cars per hour.
In an 8 hour day, that is: 8*5.1 = 40.8 cars.
This means that 1 bay can do 40.8 cars.
Now, we solve a rule of three to find how many bays we need for 100 cars.
1 bay - 40.8 cars
x bays - 100 cars
[tex]40.8x = 100[/tex]
[tex]x = \frac{100}{40.8}[/tex]
[tex]x = 2.45[/tex]
The number of bays is a discrete number, this means that the minimum number of bays to achieve an anticipated production of 100 cars per hour day is 3.
The minimum number of bays is 3 to achieve an anticipated production of 100 cars per hour a day.
Output is the amount of a thing that is produced with the help of a given level of inputs, the resulting result is called the output.
Computation of a number of days:
Given that,
The effective capacity of Lube bay = 6.06.0 cars/- Hour.
Efficiency = 0.850.85.
The formula for Expected Output(EO) :
[tex]\text{EO} = \text{Effective Capacity}\times \rm{Efficiency}\\\\[/tex]
Applying the values in the formula:
[tex]\text{EO} = \text{Effective Capacity}\times \rm{Efficiency}\\\\\text{EO} = 6\times0.85\\\\\text{EO} = 5.1 Per Hour.[/tex]
Then, 8 hours a day :
[tex]8\times5.1 = 40.8\text {cars}[/tex]
Now, it is given that 1 bay = 40.8 cars,
let, x bay= 100 cars.
Then,
[tex]40.8\times x=100\\\\x=2.45.[/tex]
Hence, The minimum number of days is 3 to achieve the minimum target of 100.
Learn more about output, refer:
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