Answer:
x = 2000 cameras
Step-by-step explanation:
C(x) Total cost in producing x units
C- = C(x) /x Average cost of producing x units x > 0
Cannon Precision Instrument
C (x) Total monthly cost for producing x units of M1 cameras
is C(x) = 0.0025x² + 80x + 10000
Then average cost of producing x cameras M1 is
C-(x) = ( 0.0025x² + 80x + 10000) /x
C-(x) = 0.0025x + 80 + 10000/x
Taking derivatives on both sides of the equation
C-´(x) = 0.0025 - 10000/x²
Then
C-´(x) = 0
( 0.0025x² - 10000 ) / x² = 0
0.0025x² - 10000 = 0
x² = 10000 /0.0025 x² = 4000000
x = 2000 cameras
Answer B! Hope this helps
The answer is 1 because 8.2/8688 = <span>0.00094383057</span>
It would be any number times 0 so yeah I hope this helps
