Find the indicated area under the curve of the standard normal distribution; then covert it to a percentage and fill in the blan
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Answer:
Part 1; The volume of the box Thomas wants to make is 224 = 2·w² + 12·w
Part 2; The zeros for the equation of the function, are w = -14, or w = 8
Part 3
The width of the box is 8 inch
The length of the box, is 14 inches
The height of the box, is given as 2 inches
Part 4
Please find attached the graph of the function
Step-by-step explanation:
Part 1
The volume of the box Thomas wants to make, V = 224 in.³
The dimensions he cuts out from the length and width = 2 in² each
The length of the box = 6 inches + The width of the box
Let <em>l</em> represent the length of the box and let <em>w</em> represent the width of the box, we have;
l = 6 + w
The height of the box, h = The length of the cut out square = 2 inches
The volume of the box, V = Length, l × Width, w × Height, h
∴ V = l × w × h
l = 6 + w, h = 2
∴ V = (6 + w) × w × 2
V = 2·w² + 12·w,
The equation of the volume of the box, V = 2·w² + 12·w, where, V = 224
∴ 224 = 2·w² + 12·w
Part 2
The zeros of the equation for the volume of the box, V = 2·w² + 12·w, where, V = 224 are found as follows;
V = 224 = 2·w² + 12·w
∴ 2·w² + 12·w - 224 = 0
Dividing by 2 gives;
(2·w² + 12·w - 224)/2 = w² + 6·w - 112 = 0
∴ (w + 14) × (w - 8) = 0
The zeros for the equation of the function, are w = -14, or w = 8
Part 3
We reject the value, w = -14, therefore, the width of the box, w = 8 inch
The length of the box, l = 6 + w
∴ l = 6 + 8 = 14
The length of the box, l = 8 inches
The height of the box, <em>h</em>, is given as h = 2 inches
Part 4
The graph of the function created with MS Excel is attached
