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kirill [66]
3 years ago
14

Macrohard have conducted a multiple linear regression analysis to predict the loading time (y) in milliseconds (thousandths of s

econds) for Macrohard Workstation files based on the size of the file (x1) in kilobytes and the speed of the processor used to view the file (x2) in megahertz. The analysis was based on a random sample of 400 Macrohard Workstation users. The file sizes in the sample ranged from 110 to 5,000 kilobytes and the speed of the processors in the sample ranged from 500 megahertz to 4,000 megahertz. The multiple linear regression equation corresponding to Macrohard's analysis is:y^i = 246.10 + 0.36x1i - 0.043x2iAccording to Macrohard's multiple regression equation, it is most reasonable to conclude that:a. holding x1 constant, a one megahertz increase in x2 will result in an increase of 0.36 in the predicted value of yb. a 5,600 kilobyte file is predicted to take 2,004.1 milliseconds to load on a 6,000 megahertz processorc. a 75 kilobyte file is predicted to take 257.2 milliseconds to load on a 370 megahertz processord. holding x1 constant, a one megahertz increase in x2 will result in a decrease of 0.043 in the predicted value of y
Engineering
1 answer:
Contact [7]3 years ago
7 0

Answer:

Hook's law holds good up to. A elastic limit. B. plastic limit. C.yield point. D.Breaking point

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A book of the Law was found in the Temple, which was being repaired, during the___
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2

Explanation:

5 0
3 years ago
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In the flash distillation of salt water, the salt is totally nonvolatile (this is the equilibrium statement). Show a McCabe-Thie
Gala2k [10]

Answer:

attached below

Explanation:

4 0
3 years ago
An uninsulated, thin-walled pipe of 100-mm diameter is used to transport water to equipment that operates outdoors and uses the
Viefleur [7K]

Answer:

4.6 mm

Explanation:

Given data includes:

thin-walled pipe diameter = 100-mm =0.1 m

Temperature of pipe T_p = -15° C = (-15 +273)K =258 K

Temperature of water T_w = 3° C = (3 + 273)K = 276 K

Temperature of ice T_i = 0° C = (0 +273)K =273 K

Thermal conductivity (k) from the ice table = 1.94 W/m.K  ;  R = 0.05

convection coefficient Lh_l =2000 W/m².K

The energy balance can be expressed as:

q_{conduction} =q_{convention}

where;

q_{conduction} = \frac{2\pi LK(T_i-T_p)}{In(R/r)}       -------------   equation (1)

q_{convention} = \pi DLh_l(T_w-T_i)  ------------ equation(2)

Equating both equation (1) and (2); we have;

\frac{2\pi LK(T_i-T_p)}{In(R/r)} = \pi DLh_l(T_w-T_i)

Replacing the given data; we have:

\frac{2\pi (1)(1.94)(273-258)}{In(0.05/r)} = \pi (0.1)*2000(276-273)

\frac{182.84}{In(\frac{0.05}{r}) } = 1884.96

In(\frac{0.05}{r})*1884.96 = 182.84

In(\frac{0.05}{r}) = \frac{182.84}{1884.96}

In(\frac{0.05}{r}) =0.0970

\frac {0.05}{r} =e^{0.0970}

\frac {0.05}{r} =1.102

r=\frac{0.05}{1.102}

r = 0.0454

The thickness (t) of the ice layer can now be calculated as:

t = (R - r)

t = (0.05 - 0.0454)

t = 0.0046 m

t = 4.6 mm

6 0
3 years ago
What is referred to as "Pyroelectric" materials?
aliina [53]

Answer and Explanation:

Pyroelectric material

Pyroelectric materials have special property of generating potential difference (although it is very less ) when these material are treated with heat or when celled down.

The potential difference generated is for very less time

The generation of potential difference is due to change in position of atoms after heating or cooling.

5 0
3 years ago
g A part made from annealed AISI 1018 steel undergoes a 20 percent cold-work operation. (a) Obtain the yield strength and ultima
nikdorinn [45]

Answer:

A) - Yield strength before operation = 32 kpsi

- Ultimate Strength before operation = 49.5 kpsi

- Yield strength after operation = 61.854 kpsi

- Ultimate Strength after operation = 61.875 kpsi

- Percentage increase of yield strength = 93.29%

- Percentage increase of ultimate strength = 25%

B) ratio before operation = 1.55

Ratio after operation = 1

Explanation:

From online values of the properties of this material, we have;

Yield strength; S_y = 32 kpsi

Ultimate Strength; S_u = 49.5 kpsi

Modulus; m = 0.25

Percentage of cold work; W_c = 0.2

S_o = 90 kpsi

A) Let's calculate the strain(ε) from the formula;

A_o/A = 1/(1 - W_c)

A_o/A = 1/(1 - 0.2)

A_o/A = 1.25

Thus, strain is;

ε = In(A_o/A)

ε = In(1.25)

ε = 0.2231

Yield strength after the cold work operation is;

S'_y = S_o(ε)^(m)

Plugging in the relevant values;

S'_y = 90(0.2231)^(0.25)

S'_y = 61.854 kpsi

Percentage increase of yield strength = S'_y/(S'_y - S_u) × 100% = (61.854 - 32)/32 × 100% = 93.29%

Ultimate strength after the cold work operation is;

S'_u = S_u/(1 - W_c)

S'_u = 49.5/(1 - 0.2)

S'u = 61.875 kpsi

Percentage increase of ultimate strength = S'_u/(S'_u - S_u) × 100% = (61.875 - 49.5)/49.5 × 100% = 25%

B) Ratio of ultimate strength and yield strength before cold work operations is;

S_u/S_y = 49.5/32

S_u/S_y = 1.547

Ratio of ultimate strength and yield strength after cold work operations is;

S'_u/S'_y = 61.875/61.854 = 1

The ratio after the operation is less than before the operation, thus the ductility reduced.

6 0
3 years ago
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