The sampling method used to examine a population when the auditor wants to estimate a continuous amount (or value) of the popula
1 answer:
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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
(Image Attached)
Answer:
the y coordinate is 5
Step-by-step explanation:
In the figure attached, the directed line segment is shown.
J is located at (-3, 1) and K is located at (-8, 11)
run: x2 -x1 = -8 - (-3) = -5
rise: y2 - y1 = 11 - 1 = 10
Taking J as reference, the coordinate of the point that divides the directed line segment from J to k into a ratio of 2:3 is:
c = 2/(2+3) = 0.4
(x1 + c*run, y1 + c*rise)
(-3 + 0.4*-5, 1+0.4*10)
(-5, 5)
