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Sveta_85 [38]
3 years ago
15

An engineer measures a sample of 1200 shafts out of a certain shipment. He finds the shafts have an average diameter of 2.45 inc

h and a standard deviation of 0.07 inch. Assume that the shaft diameter follows a Gaussian distribution. What percentage of the diameter of the total shipment of shafts will fall between 2.39inch and 2.60 inch?
Engineering
1 answer:
Vadim26 [7]3 years ago
7 0

Answer: 78.89%

Explanation:

Given : Sample size : n=  1200

Sample mean : \overline{x}=2.45

Standard deviation : \sigma=0.07

We assume that it follows Gaussian distribution (Normal distribution).

Let x be a random variable that represents the shaft diameter.

Using formula, z=\dfrac{x-\mu}{\sigma}, the z-value corresponds to 2.39 will be :-

z=\dfrac{2.39-2.45}{0.07}\approx-0.86

z-value corresponds to 2.60 will be :-

z=\dfrac{2.60-2.45}{0.07}\approx2.14

Using the standard normal table for z, we have

P-value = P(-0.86

=P(z

Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%

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telo118 [61]

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

-0.40= 1.005(ln T_2 - 5.68697)- 0.5968

Solving for T2 we have:

T_2 = 5.8828

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

w_in = c_p(T_2 - T_1)+q_out

w_in = 1.005(358.8 - 295)+120

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = \frac{q_out}{T_1}

\frac{120kJ/kg.k}{295K}

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

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3 years ago
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Answer:

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Explanation:

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Imagine you compare the effectiveness of four different types of stimulant to keep you awake while revising statistics using a o
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Answer:

The correct answer is At least two of the stimulants will have different effects on the mean time spent awake.

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The null hypothesis exposed leads to consider an alternative hypothesis that differs from the behavior of the 4 stimulants. In this case it is related to the effects on patients, since in practice at least one will present differences compared to the others, influencing the sleep time of consumers.

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You are evaluating the lifetime of a turbine blade. The blade is 4 cm long and there is a gap of 0.16 cm between the tip of the
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Answer:

Explanation:

Given conditions

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2)The yield strength of the blade is 175 MPa

3)The Young’s modulus for the blade is 50 GPa

4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.

5)The temperature of the blade is 800°C.

6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)

where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K

Young Modulus, E = Stress, \sigma /Strain, ∈

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\epsilon_i = \frac{100\times 10^{6} Pa}{50\times 10^{9} Pa}

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\frac{\delta \epsilon}{\delta t} = (1 \times {10}^{-5})\sigma^4 exp^(\frac{-2eV}{kT} )

\frac{\epsilon_{initial} - \epsilon _{primary}}{t_{initial}-t_{final}} = 1 \times 10^{-5}(100)^{4}exp(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )(800+273)K} )

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\epsilon_{initial} - \epsilon _{primary}} = 0.002 - 0.003 = -0.001

\frac{-0.001}{-t_{final}} = 1 \times 10^{-5}(100)^{4}\times 10^{(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )1073K} )}

solving the above equation,

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gregori [183]

Answer:

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