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3 years ago

11

In a cumulative relative frequency distribution, the last class will have a cumulative relative frequency equal to a.one. b.zero

. c.the total number of elements in the data set. d.None of these alternatives is correct.

1 answer:

kirza4 [7]3 years ago

7 0

Answer: equal to one

Explanation:

Because is the sum of the relative frequencies.

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10. Convert the following:<br> a. 37.4 mL into ML<br> b. 689 km/hr into m/s<br> c. 34.5 m² into mm²

Snezhnost [94]

A. When we convert 37.4 mL to ML, the result obtained is 3.74×10¯⁸ ML

B. When we convert 689 km/hr to m/s, the result obtained is 191.39 m/s

C. When we convert 34.5 m² to mm², the result obtained is 3.45×10⁷ mm²

<h3>A. How to convert millimeters (mL) to megaliter (ML)</h3>

Volume (mL) = 37.4 mL
Volume (ML) =?

1 mL = 1×10¯⁹ ML

Therefore,

37.4 mL = 37.4 × 1×10¯⁹

37.4 mL = 3.74×10¯⁸ ML

Thus, 37.4 mLis equivalent to 3.74×10¯⁸ ML

<h3>B. How to convert 689 km/hr to m/s</h3>

Conversion scale

3.6 Km/hr = 1 m/s

Therefore,

689 km/hr = 689 / 3.6

689 km/hr = 191.39 m/s

Thus, 689 km/hr is equivalent to 191.39 m/s

<h3>C. How to convert 34.5 m² to mm²</h3>

Conversion scale

1 m² = 1×10⁶ mm²

Therefore,

34.5 m² = 34.5 × 1×10⁶

34.5 m² = 3.45×10⁷ mm²

Thus, 34.5 m² is equivalent to 3.45×10⁷ mm²

Learn more about conversion:

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Scissors are considered a compound machine because

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They make work easier

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Liquid droplets that fall from the sky are?<br> A) rain<br> B)snow<br> C) sleet <br> D) hail

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Read 2 more answers

At a certain location, wind is blowing steadily at 9 m/s. Determine the mechanical energy of air per unit mass and the power gen

Misha Larkins [42]

Answer:

The specific mechanical energy of the air in the specific location is 40.5 J/kg.
The power generation potential of the wind turbine at such place is of 2290 kW
The actual electric power generation is 687 kW

Explanation:

The mechanical energy of the air per unit mass is the specific kinetic energy of the air that is calculated using: $\frac{1}{2} V^2$ where V is the velocity of the air.
The specific kinetic energy would be: $\frac{1}{2}(9\frac{m}{s})^2=40.5\frac{m^2}{s^2}=40.5\frac{m^2 }{s^2}\frac{kg}{kg}=40.5\frac{N*m }{kg}=40.5\frac{J}{kg}$ .
The power generation of the wind turbine would be obtained from the product of the mechanical energy of the air times the mass flow that moves the turbine.
To calculate mass flow it is required first to calculate the volumetric flow. To calculate the volumetric flow the next expression would be: $\frac{V\pi D_{blade}^2}{4} =\frac{9\frac{m}{s}\pi(80m)^2}{4} =45238.9\frac{m^3}{s}$
Then the mass flow is obtain from the volumetric flow times the density of the air: $m_{flow}=1.25\frac{kg}{m^3}45238.9\frac{m^3}{s}=56548.7\frac{kg}{s}$
Then, the Power generation potential is: $40.5\frac{J}{kg} 56548.7\frac{kg}{s} =2290221W=2290.2kW$
The actual electric power generation is calculated using the definition of efficiency: $\eta=\frac{E_P}{E_I}}$ , where η is the efficiency, $E_P$ is the energy actually produced and, $E_I$ is the energy input. Then solving for the energy produced: $E_P=\eta*E_I=0.30*2290kW=687kW$

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