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hjlf
3 years ago
10

​A manager wants to know if the mean productivity of two workers is the same. For a random selection of 30 hours in the past​ mo

nth, he compares the number of items produced by each worker in that hour.A. The data should be treated as paired samples. Each pair consists of an hour in which the productivity of the two workers is compared.B. The data should be treated as paired samples. Each pair consists of two hours in which the number of items produced is compared.C. The data should be treated as independent samples because the samples are random.
Mathematics
1 answer:
vekshin13 years ago
7 0

Answer:

A.The data should be treated as paired samples. Each pair consists of an hour in which the productivity of the two workers is compared.

Explanation:

If the mean productivity of two workers is the same.

For a random selection of 30 hours in the past month, the manager compares the number of items produced by each worker in that hour.

There are two samples and the productivity of the two men is paired for each hour.

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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

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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}

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