An efficiency study of the morning shift at a certain factory indicates that an average worker who arrives on the job at 7:00 A.M. will have assembled ‘y’ number of Television sets ‘x’ hour later. y=f(x)= -x^3+5x^2+ 12x How many sets will such a worker have assembled by 04:00 PM? How many sets will such a worker assemble between 11:00 AM and 12:00 noon?
Answers
50 seats ok got it bye bye
Given : An efficiency study of the morning shift at a certain factory indicates that an average worker who arrives on the job at 7:00 A.M. will have assembled ‘y’ number of Television sets ‘x’ hour later.
y=f(x)= -x³+5x²+ 12x
To find : How many sets will such a worker have assembled by 04:00 PM? How many sets will such a worker assemble between 11:00 AM and 12:00 noon?
Solution:
y=f(x)= -x³+5x²+ 12x
x = number of hours worked from 7:00AM
y = number of Televisions assembled
7:00AM to 04:00 PM = 9 hrs
=> x = 9
=> y = f(9) = -9³+5(9)²+ 12(9)
=> y = f(9) = -729 + 425+ 108
=> y = f(9) = -196
number of Television sets can not be -ve hence looks mistake in data
assemble between 11:00 AM ( 4 hrs from 7:00 AM )and 12:00 noon ( 5 hrs from 7:00 AM )
= f(5) - f(4)
= -5³+5(5)²+ 12(5) - ( -4³+5(4)²+ 12(4))
= -125 + 125 + 60 - ( -64 + 80 + 48)
= 60 - 64
= - 4
looks mistake in equation
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