The amount of a chemical solution is measured to be 2 liters. What is the percent error of the measurement?
2 answers:
The complete question in the attached figure
we know that
2 liters may be 1.5 to 1.9 rounded up to 2
or 2.1 to 2.4 rounded down to 2
the biggest percent error would be between 2 and 1.5, which is a volume error of 0.5 liters
2 - 1.5 = 0.5 liters
percent error = (absolute error / quantity) * 100
percent error = [0.5/2] * 100% = 0.25 * 100% = 25%
the answer is the option C) 25%
we know that
2 liters may be 1.5 to 1.9 rounded up to 2
or 2.1 to 2.4 rounded down to 2
the biggest percent error would be between 2 and 1.5, which is a volume error of 0.5 liters
2 - 1.5 = 0.5 liters
percent error = (absolute error / quantity) * 100
percent error = [0.5/2] * 100% = 0.25 * 100% = 25%
the answer is the option C) 25%
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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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Complete Question
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