X=number of the chairs y=the total cost of chairs
150/50=3$ y=3x
Solve both equations for y:

Set the expressions equal to each other:

So, for every nonzero number a the system of equations has
one solution.
Answer: The average daily inventory is 200 cases.
Step-by-step explanation:
Since we have given that
N(t)=600-20√30t
We need to find the average daily inventory.
![\dfrac{1}{b-a}\int\limits^a_b {600-20\sqrt{30t}} \, dt\\\\=\dfrac{1}{30}\int\limits^{30}_0 {600-20\sqrt{30t}} \, dt \\\\=\dfrac{1}{30}[600t-\dfrac{20(30t)^\frac{3}{2}}{45}|_0^{30}\\\\=\dfrac{1}{30}[18000-\dfrac{20\times 30^3}{45}]\\\\=\dfrac{1}{30}[18000-12000]\\\\=200](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bb-a%7D%5Cint%5Climits%5Ea_b%20%7B600-20%5Csqrt%7B30t%7D%7D%20%5C%2C%20dt%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B30%7D%5Cint%5Climits%5E%7B30%7D_0%20%7B600-20%5Csqrt%7B30t%7D%7D%20%5C%2C%20dt%20%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B30%7D%5B600t-%5Cdfrac%7B20%2830t%29%5E%5Cfrac%7B3%7D%7B2%7D%7D%7B45%7D%7C_0%5E%7B30%7D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B30%7D%5B18000-%5Cdfrac%7B20%5Ctimes%2030%5E3%7D%7B45%7D%5D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B30%7D%5B18000-12000%5D%5C%5C%5C%5C%3D200)
Hence, the average daily inventory is 200 cases.