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Ronch [10]
3 years ago
15

The dot plot shows the number of cupcakes bought by each person who came to a bake sale.

Engineering
1 answer:
Maslowich3 years ago
6 0

The true statement about the dot plot is 1 has 4 and 0 dots.

Explanation:

  • after creating a Bar chart 4 is the right answer.
  • The highest among st all the other plots is 1 but 4 shows 3.
  • Taking an average from all the data points 3 comes to the right answer.
  • Median the central Mid-point is 3.
  • Mode also comes to 3.
  • It is skewed on the right due to the 1st one.
  • Skewed shows the data point either increasing or in decreasing.
  • There a to Bi- histograms which has two ups and downs.
  • The true statement has to be as Mean=Median=Mode is 3.

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Which option identifies the free resource Judi can use in the following scenario?
lakkis [162]

Answer:

Computer programming for three dimensional designs

Explanation:

  • Doll is not a 2D creation . it's a 3D creation
  • So on creating the design on 3D scale it's more effective to determine what can be added more.
4 0
2 years ago
A rigid tank contains an ideal gas at 40°C that is being stirred by a paddle wheel. The paddle wheel does 240 kJ of work on the
Sergeu [11.5K]

To solve the problem it is necessary to consider the concepts and formulas related to the change of ideal gas entropy.

By definition the entropy change would be defined as

\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{P_2}{P_1})

Using the Boyle equation we have

\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{v_1T_2}{v_2T_1})

Where,

C_p = Specific heat at constant pressure

T_1= Initial temperature of gas

T_2= Final temprature of gas

R = Universal gas constant

v_1= Initial specific Volume of gas

v_2= Final specific volume of gas

According to the statement, it is an isothermal process and the tank is therefore rigid

T_1 = T_2, v_2=v_1

The equation would turn out as

\Delta S = C_p ln1-ln1

<em>Therefore the entropy change of the ideal gas is 0</em>

Into the surroundings we have that

\Delta S = \frac{Q}{T}

Where,

Q = Heat Exchange

T = Temperature in the surrounding

Replacing with our values we have that

\Delta S = \frac{230kJ}{(30+273)K}

\Delta S = 0.76 kJ/K

<em>Therefore the increase of entropy into the surroundings is 0.76kJ/K</em>

8 0
3 years ago
La iluminación de la superficie de un patio amplio es 1600 lx cuando el ángulo de elevación del sol 53°. Calcular la iluminación
gregori [183]

Answer:

 I = 1205.69 Lx

Explanation:

The irradiation or intensity of the solar radiation on the earth is maximum for the vertical fire, with a value I₀

          I = I₀ sin θ

in this case with the initial data we can calculate the initial irradiance

         I₀ = \frac{I}{sin  \ \theta }

         I₀ = 1600 /sin 53

         I₀ = 2003.42 lx

for when the angle is θ = 37º

         I = 2003.42 sin 37

         I = 1205.69 Lx

6 0
3 years ago
on the same scale for stress, the tensile true stress-true strain curve is higher than the engineeringstress-engineering strain
Bess [88]

Answer:

The condition does not hold for a compression test

Explanation:

For a compression test the engineering stress - strain curve is higher than the actual stress-strain curve and this is because the force needed in compression is higher than the force needed during Tension.  The higher the force in compression leads to increase in the area therefore for the same scale of stress the there is more stress on the Engineering curve making it higher than the actual curve.

<em>Hence the condition of : on the same scale for stress, the tensile true stress-true strain curve is higher than the engineering stress-engineering strain curve.</em><em> </em>does not hold for compression test

5 0
2 years ago
What is the measurement below?
Bess [88]

Explanation:

इसिसिसिसैस्स्स्स्स्स्स्स्स्स्सूस्सोस्स्स्स्स्स

8 0
3 years ago
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