Volume = base area x height.
Required amount of water = volume of tank - amount of water currently in the tank. = (45 x 25) - (45 x 12) = 45(25 - 12) = 45 x 13 = 598 cm^3
Answer:386
Step-by-step explanation:
We have given
Smoothing parameter ![\left ( \alpha \right )=0.56](https://tex.z-dn.net/?f=%5Cleft%20%28%20%5Calpha%20%5Cright%20%29%3D0.56)
Forecasted demand![\left ( F_t\right )=385](https://tex.z-dn.net/?f=%5Cleft%20%28%20F_t%5Cright%20%29%3D385)
Actual demand![\left ( D_t\right )=386](https://tex.z-dn.net/?f=%5Cleft%20%28%20D_t%5Cright%20%29%3D386)
And Forecast is given by
![F_{t+1}=\alpha D_t+\left ( 1-\alpha \right )F_t](https://tex.z-dn.net/?f=F_%7Bt%2B1%7D%3D%5Calpha%20D_t%2B%5Cleft%20%28%201-%5Calpha%20%5Cright%20%29F_t)
![F_{t+1}=0.56\cdot 386+\left ( 1-0.56\right )385=385.56\approx 386](https://tex.z-dn.net/?f=F_%7Bt%2B1%7D%3D0.56%5Ccdot%20386%2B%5Cleft%20%28%201-0.56%5Cright%20%29385%3D385.56%5Capprox%20386)
![F_{t+2}=0.56\cdot 386+\left ( 1-0.56\right )385.56=385.806\approx 386](https://tex.z-dn.net/?f=F_%7Bt%2B2%7D%3D0.56%5Ccdot%20386%2B%5Cleft%20%28%201-0.56%5Cright%20%29385.56%3D385.806%5Capprox%20386)
![F_{t+3}=0.56\cdot 386+\left ( 1-0.56\right )385.806=385.914\approx 386](https://tex.z-dn.net/?f=F_%7Bt%2B3%7D%3D0.56%5Ccdot%20386%2B%5Cleft%20%28%201-0.56%5Cright%20%29385.806%3D385.914%5Capprox%20386)
![F_{t+4}=0.56\cdot 386+\left ( 1-0.56\right )385.914=385.962\approx 386](https://tex.z-dn.net/?f=F_%7Bt%2B4%7D%3D0.56%5Ccdot%20386%2B%5Cleft%20%28%201-0.56%5Cright%20%29385.914%3D385.962%5Capprox%20386)
You just need to solve for the slope of the line.
= 12/8
= 3/2
= 1.5
I believe the answer is 2 but I could be wrong sorry