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Mumz [18]
3 years ago
9

You want to determine whether the race of the defendant has an impact on jury verdicts. You assign participants to watch a trial

with either a Hispanic or Caucasian defendant, and you measure on a 1 (not at all) to 7 (very) scale how confident they feel the defendant is guilty. This is a confidence scale represents a _________ scale of measurement
Engineering
1 answer:
Andru [333]3 years ago
6 0

Answer:

The confidence scale represents an ordinal scale of measurement

Explanation:

An ordinal scale or level of measurement is used to measure attributes that can be ranked or ordered, but the interval between the attributes do not have quantitative significance. In this case, the measurement was done on a scale of 1 - 7, with a "1" being; not all that race of defendant has an impact on jury verdicts and a "7" being "very" meaning that race indeed has impact on jury verdicts. Another example can be a survey carried out on the level of customer satisfaction on a particular product, with "1" most dissatisfied and "10 " representing most satisfied. In the first example, it is wrong to say that the difference between 1 being "not at all" and maybe 3 is the same as the difference between 5 and 7 which have different connotations, because the numbers are merely for tagging and not to quantify.

Other levels of measurement include:

1. Nominal: this is the simplest level of measurement and it is simply used to categorize the attributes. Example is taking a survey on gender in the categories of male, female and transgender.

2. Interval: the interval scale is used when the distance between two attributes have meanings but there is no true zero value associated with the scale.

3. Ratio: this combines all the other three levels of measurement and is used to categorize, used to show ranking, has meaningful distances between the attributes and the scale has a true zero point. Example is the measurement of temperature using the celcius scale thermometer, where there is a true zero point at 0°C and the distance between 5°C and 10°C is the same as the distance between 10°C and 15°C.

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A hot air balloon is used as an air-vehicle to carry passengers. It is assumed that this balloon is sealed and has a spherical s
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Answer:

a. \dfrac{D_{1}}{ D_{2}}  =  \left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )^{-3\times n} which is constant therefore, n = constant

b. The temperature at the end of the process is 109.6°C

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The work done on the surrounding atmospheric air = 10.6 MJ

Explanation:

p₁ = 100 kPa

T₁ = 27°C

D₁ = 10 m

v₂ = 1.2 × v₁

p ∝ α·D

α = Constant

v_1 = \dfrac{4}{3} \times  \pi \times r^3

\therefore v_1 = \dfrac{4}{3} \times  \pi \times  \left (\dfrac{10}{2}  \right )^3 = 523.6 \ m^3

v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³

Therefore, D₂ = 10.63 m

We check the following relation for a polytropic process;

\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}}   \right )^{n} = \left (\dfrac{T_{1}}{T_{2}}   \right )^{\dfrac{n}{n-1}}

We have;

\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times  \pi \times  \left (\dfrac{D_2}{2}  \right )^3}{\dfrac{4}{3} \times  \pi \times  \left (\dfrac{D_1}{2}  \right )^3}   \right )^{n} = \left (\dfrac{   \left{D_2}  ^3}{ {D_1}^3}   \right )^{n}

\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{   \left{D_2}  }{ {D_1}}   \right )^{3\times n} =  \left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )^{-3\times n}

\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2  \right )^{n} = \left (\dfrac{   \left{D_2}  ^3}{ {D_1}^3}   \right )^{n}

log  \left (\dfrac{D_{1}}{ D_{2}}\right )  =  -3\times n \times log\left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )

n = -1/3

Therefore, the relation, pVⁿ = Constant

b. The temperature T₂ is found as follows;

\left (\dfrac{628.32 }{523.6}   \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}}   \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}}   \right )^{\dfrac{1}{4}}

T₂ = 300.15/0.784 = 382.75 K = 109.6°C

c. W_{pdv} = \dfrac{p_1 \times v_1 -p_2 \times v_2 }{n-1}

p_2 = \dfrac{p_{1}}{ \left (\dfrac{V_{2}}{V_{1}}   \right )^{n} } =  \dfrac{100\times 10^3}{ \left (1.2) \right  ^{-\dfrac{1}{3} } }

p₂ =  100000/0.941 = 106.265 kPa

W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3  \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J

The work done by the balloon boundaries = 10.81 MJ

Work done against atmospheric pressure, Pₐ, is given by the relation;

Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J

The work done on the surrounding atmospheric air = 10.6 MJ

4 0
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An artist is creating a gigantic mobile using old cars. On one end of the mobile, a 2100 lb Plymoth is suspended 30 feet from th
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Answer:

  c  45 feet from the fulcrum

Explanation:

The moment at the fulcrum must be the same for each car. If the distance is d, then the artist must have ...

  (2100 lb)(30 ft) = (1400 lb)(d)

  2100·30/1400 ft = d = 45 ft

The Volkswagen should be 45 ft from the fulcrum.

7 0
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