Answer: Level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.
Step-by-step explanation:
Since we have given that
Height = 9 inches
Diameter = 6 inches
Radius = 3 inches
So, 
Volume of cone is given by

By differentiating with respect to time t, we get that

Now, the liquid drips out the bottom of the filter at the constant rate of 4 cubic inches per second, ie 
and h = 2 inches deep.

Hence, level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.
I believe the answer to this is:
Mean: <u>8.7</u> minutes
Median: <u>9</u> minutes
Hope this helps! :D
The maximum volume of the box is 40√(10/27) cu in.
Here we see that volume is to be maximized
The surface area of the box is 40 sq in
Since the top lid is open, the surface area will be
lb + 2lh + 2bh = 40
Now, the length is equal to the breadth.
Let them be x in
Hence,
x² + 2xh + 2xh = 40
or, 4xh = 40 - x²
or, h = 10/x - x/4
Let f(x) = volume of the box
= lbh
Hence,
f(x) = x²(10/x - x/4)
= 10x - x³/4
differentiating with respect to x and equating it to 0 gives us
f'(x) = 10 - 3x²/4 = 0
or, 3x²/4 = 10
or, x² = 40/3
Hence x will be equal to 2√(10/3)
Now to check whether this value of x will give us the max volume, we will find
f"(2√(10/3))
f"(x) = -3x/2
hence,
f"(2√(10/3)) = -3√(10/3)
Since the above value is negative, volume is maximum for x = 2√(10/3)
Hence volume
= 10 X 2√(10/3) - [2√(10/3)]³/4
= 2√(10/3) [10 - 10/3]
= 2√(10/3) X 20/3
= 40√(10/27) cu in
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1/5 is equal to 2/10 so the one with 3/10 has more