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grin007 [14]
2 years ago
12

Una empresa industrial agrupó sus fábricas de acuerdo con el valor de la producción anual en millones de C$ de cada una; se obtu

vo la siguiente distribución...... Valor de producción: 41-45, 46-50, 51-55, 56-60, 61-65, 66-70.......Numeros de fábricas: 7, 10, 11, 9, 8, 7 ..... Completar distribución de frecuencias y determine e intérprete la producción anual promedio de las fábricas y la desviación estándar de las producciones.
Ayuda plis:(!
Engineering
2 answers:
bulgar [2K]2 years ago
5 0

Answer:

Explanation:

An industrial company grouped its factories according to the value of annual production in millions of C $ of each; the following distribution was obtained ...... Production value: 41-45, 46-50, 51-55, 56-60, 61-65, 66-70 ....... Number of factories: 7 , 10, 11, 9, 8, 7 ..... Complete the frequency distribution and determine and interpret the average annual production of the factories and the standard deviation of the productions.

vodka [1.7K]2 years ago
5 0
Daddy chilll no capppp
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3 years ago
1.0•10^-10 standard form
Drupady [299]

Answer:

1.0 * 10^{-10} = 0.0000000001

Explanation:

Given

1.0 * 10^{-10}

Required

Convert to standard form

1.0 * 10^{-10}

From laws of indices

a^{-x} = \frac{1}{a^x}

So, 1.0 * 10^{-10} is equivalent to

1.0 * 10^{-10} = 1.0 * \frac{1}{10^{10}}

1.0 * 10^{-10} = 1.0 * \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}

1.0 * 10^{-10} = 1.0 * \frac{1}{10000000000}

1.0 * 10^{-10} = 1.0 * 0.0000000001

1.0 * 10^{-10} = 0.0000000001

Hence, the standard form of 1.0 * 10^{-10} is 0.0000000001

3 0
3 years ago
If the slotted arm rotates counterclockwise with a constant angular velocity of thetadot = 2rad/s, determine the magnitudes of t
astraxan [27]

Answer:

Magnitude of velocity=10.67 m/s

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Explanation:

The solution of the problem is given in the attachments

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3 years ago
A polyethylene rod exactly 10 inches long with a cross-sectional area of 0.04 in2 is used to suspend a weight of 358 lbs-f (poun
Nadya [2.5K]

Answer:

Final length of the rod = 13.90 in

Explanation:

Cross Sectional Area of the polythene rod, A = 0.04 in²

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Viscosity, \eta = 1*10^{9} psi -sec

Weight = 358 lbs - f

time, t = 1 hr = 3600 sec

Stress is given by:

\sigma = \frac{Force}{Area} \\\sigma = \frac{358}{0.04} \\\sigma = 8950 psi

Based on Maxwell's equation, the strain is given by:

strain = \sigma ( \frac{1}{E} + \frac{t}{\eta} )\\Strain = 8950 ( \frac{1}{25000} + \frac{3600}{10^{9} } )\\Strain = 0.39022

Strain = Extension/(original Length)

0.39022 = Extension/10

Extension = 0.39022 * 10

Extension = 3.9022 in

Extension = Final length - Original length

3.9022 =  Final length - 10

Final length = 10 + 3.9022

Final length = 13.9022 in

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7 0
3 years ago
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