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Nataliya [291]
2 years ago
11

Answer the statistical measures and create a box and whiskers plot for the following set of data. 4,4,5,5, 6, 6, 8, 10, 11, 11,

12, 14 Min: Q1: Med: Q3: Max: Create the box plot by dragging the lines: 2 3 4 b 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20​

Mathematics
1 answer:
vredina [299]2 years ago
6 0

Answer:

Step-by-step explanation:

First you take your smallest number, which in this case is 4 - then you find your maximum number which is 14. Then we can list these numbers on a piece of paper and cross a number out from each side until you reach the middle number which would be 8. Then you go from the min to the middle number which would be 5. Then you repeat between the max number and the middle number which is 11. So your minumum would be 4, max would be 14, Q1 would be 5 Median would be 8 and Q3 would be 11. Let me know if I got it right! I haven't done it for a while, but I am pretty sure this is correct..

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5 0
3 years ago
A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constru
Viktor [21]

Answer:

1) S = 2\cdot w\cdot l - 8\cdot x^{2}, 2) The domain of S is 0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}. The range of S is 0 \leq S \leq 2\cdot w \cdot l, 3) S = 176\,in^{2}, 4) x \approx 4.528\,in, 5) S = 164.830\,in^{2}

Step-by-step explanation:

1) The function of the box is:

S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)

S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w

S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x

S = 2\cdot w\cdot l - 8\cdot x^{2}

2) The maximum cutout is:

2\cdot w \cdot l - 8\cdot x^{2} = 0

w\cdot l - 4\cdot x^{2} = 0

4\cdot x^{2} = w\cdot l

x = \frac{\sqrt{w\cdot l}}{2}

The domain of S is 0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}. The range of S is 0 \leq S \leq 2\cdot w \cdot l

3) The surface area when a 1'' x 1'' square is cut out is:

S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}

S = 176\,in^{2}

4) The size is found by solving the following second-order polynomial:

20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}

20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}

8\cdot x^{2} - 164\,in^{2} = 0

x \approx 4.528\,in

5) The equation of the box volume is:

V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x

V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x

V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}

V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^{2} + 4\cdot x^{3}

V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}

The first derivative of the function is:

V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}

The critical points are determined by equalizing the derivative to zero:

12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0

x_{1} \approx 4.952\,in

x_{2}\approx 1.548\,in

The second derivative is found afterwards:

V'' = 24\cdot x - 78\,in

After evaluating each critical point, it follows that x_{1} is an absolute minimum and x_{2} is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is:

x \approx 1.548\,in

The surface area of the box is:

S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1.548\,in)^{2}

S = 164.830\,in^{2}

4 0
2 years ago
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