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Shkiper50 [21]
3 years ago
8

In the dataset "RestRating", the 30 randomly selected patrons give the overall dining experience ratings (Outstanding, Very good

, Good, Average, or Poor) at a restaurant on a Saturday evening

Engineering
2 answers:
balu736 [363]3 years ago
8 0

Answer:

The dataset image is attached.

The question has three parts:

a) Find the frequency distribution and relative frequency distribution for these data.

b) Construct a percentage bar chart for these data

c) Construct a percentage pie chart for these data

Explanation:

a)

Frequency distribution tell us how often something occurs.

Frequency distribution

Experience Rating Frequency

OutStanding 14

Very Good 10

Good 5

Average 1

Poor 0

This relative frequency distribution table shows how something are distributed.

Relative Frequency Distribution

Experience Rating Frequency Relative Frequency

OutStanding 14 14/30 = 7/15 = 0.467

Very Good 10 10/30 = 1/3 = 0.333

Good 5 5/30 = 1/6 = 0.167

Average 1 1/30 = 0.033

Poor 0 0

Total 30 1.000

b) Chart image is attached

c) Chart image is attached

rusak2 [61]3 years ago
6 0

Answer: incomplete question

Explanation:

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(d) Suppose two students are memorizing a list according to the same model dL dt = 0.5(1 − L) where L represents the fraction of
Neporo4naja [7]

Answer:

Rate of learning =0

Explanation:

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8 0
3 years ago
A turbine operates at steady state, and experiences a heat loss. 1.1 kg/s of water flows through the system. The inlet is mainta
strojnjashka [21]

Answer:

\dot W_{out} = 399.47\,kW

Explanation:

The turbine is modelled after the First Law of Thermodynamics:

-\dot Q_{out} -\dot W_{out} + \dot m\cdot (h_{in}-h_{out}) = 0

The work done by the turbine is:

\dot W_{out} = \dot m \cdot (h_{in}-h_{out})-\dot Q_{out}

The properties of the water are obtained from property tables:

Inlet (Superheated Steam)

P = 10\,MPa

T = 520\,^{\textdegree}C

h = 3425.9\,\frac{kJ}{kg}

Outlet (Superheated Steam)

P = 1\,MPa

T = 280\,^{\textdegree}C

h = 3008.2\,\frac{kJ}{kg}

The work output is:

\dot W_{out} = \left(1.1\,\frac{kg}{s}\right)\cdot \left(3425.9\,\frac{kJ}{kg} -3008.2\,\frac{kJ}{kg}\right) - 60\,kW

\dot W_{out} = 399.47\,kW

5 0
3 years ago
Mnsdcbjksdhkjhvdskjbvfdfkjbcv hjb dfkjbkjfvvfebjkhbvefgjdf
Julli [10]

Answer:

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8 0
3 years ago
Read 2 more answers
What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a
Studentka2010 [4]

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

Explanation:

Given the data from the question

first we find the rpm for the spindle of the drilling tool, using the equation

Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

so we substitute

Ns = 12 × 120 / π2

Ns = 1440 / 6.2831

Ns = 229.18 rmp

Now we find the metal removal rate using the equation

MRR = (πD²/4) Fr × Ns

Fr is the feed rate( 0.02 ipr ),

so we substitute

MRR = ((π × 2²)/4) × 0.02 × 229.18

MRR = 14.3998 ≈ 14.4 in³/min

Therefore the metal removal rate is 14.4 in³/min

Next we find the allowance for approach of the tip of the drill

A = D/2

A = 2/2

= 1 in

now find the time required to drill the hole

Tm = (L + A) / (Fr × Ns)

Lis the the depth of the hole( 3.5 in)

so we substitute our values

Tm = (3.5 + 1) / (0.02 × 229.18  )

Tm = 4.5 / 4.5836

Tm = 0.98 min

Therefore the cutting time is 0.98 min

8 0
2 years ago
In order to impress your neighbors and improve your vision in traffic jams, you decide to mount a cylindrical periscope 2.0 m hi
kondaur [170]
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4 years ago
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