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Studentka2010 [4]
3 years ago
8

For installations where the nonlinear load is huge, most consulting engineers will specify ____-rated transformers.

Engineering
1 answer:
Scorpion4ik [409]3 years ago
8 0

Answer:

K

Explanation:

For installations where the nonlinear load is huge, most consulting engineers will specify K-rated transformers.

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Finally you will implement the full Pegasos algorithm. You will be given the same feature matrix and labels array as you were gi
Diano4ka-milaya [45]

Answer:

In[7] def pegasos(feature_matrix, labels, T, L):

   """

   .

   let learning rate = 1/sqrt(t),

   where t is a counter for the number of updates performed so far       (between 1   and nT inclusive).

Args:

       feature_matrix - A numpy matrix describing the given data. Each row

           represents a single data point.

       labels - A numpy array where the kth element of the array is the

           correct classification of the kth row of the feature matrix.

       T -  the maximum number of times that you should iterate through the feature matrix before terminating the algorithm.

       L - The lamba valueto update the pegasos

   Returns: Is defined as a  tuple in which the first element is the final value of θ and the second element is the value of θ0

   """

   (nsamples, nfeatures) = feature_matrix.shape

   theta = np.zeros(nfeatures)

   theta_0 = 0

   count = 0

   for t in range(T):

       for i in get_order(nsamples):

           count += 1

           eta = 1.0 / np.sqrt(count)

           (theta, theta_0) = pegasos_single_step_update(

               feature_matrix[i], labels[i], L, eta, theta, theta_0)

   return (theta, theta_0)

In[7] (np.array([1-1/np.sqrt(2), 1-1/np.sqrt(2)]), 1)

Out[7] (array([0.29289322, 0.29289322]), 1)

In[8] feature_matrix = np.array([[1, 1], [1, 1]])

   labels = np.array([1, 1])

   T = 1

   L = 1

   exp_res = (np.array([1-1/np.sqrt(2), 1-1/np.sqrt(2)]), 1)

   

   pegasos(feature_matrix, labels, T, L)

Out[8] (array([0.29289322, 0.29289322]), 1.0)

Explanation:

In[7] def pegasos(feature_matrix, labels, T, L):

   """

   .

   let learning rate = 1/sqrt(t),

   where t is a counter for the number of updates performed so far       (between 1   and nT inclusive).

Args:

       feature_matrix - A numpy matrix describing the given data. Each row

           represents a single data point.

       labels - A numpy array where the kth element of the array is the

           correct classification of the kth row of the feature matrix.

       T -  the maximum number of times that you should iterate through the feature matrix before terminating the algorithm.

       L - The lamba valueto update the pegasos

   Returns: Is defined as a  tuple in which the first element is the final value of θ and the second element is the value of θ0

   """

   (nsamples, nfeatures) = feature_matrix.shape

   theta = np.zeros(nfeatures)

   theta_0 = 0

   count = 0

   for t in range(T):

       for i in get_order(nsamples):

           count += 1

           eta = 1.0 / np.sqrt(count)

           (theta, theta_0) = pegasos_single_step_update(

               feature_matrix[i], labels[i], L, eta, theta, theta_0)

   return (theta, theta_0)

In[7] (np.array([1-1/np.sqrt(2), 1-1/np.sqrt(2)]), 1)

Out[7] (array([0.29289322, 0.29289322]), 1)

In[8] feature_matrix = np.array([[1, 1], [1, 1]])

   labels = np.array([1, 1])

   T = 1

   L = 1

   exp_res = (np.array([1-1/np.sqrt(2), 1-1/np.sqrt(2)]), 1)

   

   pegasos(feature_matrix, labels, T, L)

Out[8] (array([0.29289322, 0.29289322]), 1.0)

6 0
3 years ago
Consider three charged sheets, A, B, and C. The sheets are parallel with A above B above C. On each sheet there is surface charg
babunello [35]

Answer:

0.

Explanation:

To find the electrical force per unit area on each sheet we start defining our variables,

\sigma_A = -4.10^{-5}C/m^2

\sigma_B= -7.10^{-5}C/m^2

\sigma_C = -3.1^{-5}C/m^2

We find the electric field for each one, this formula is given by,

E= \frac{\sigma_i}{2\epsilon_0}

Substituting each value from the three charged sheets, we have

E_A= \frac{-4.10^{-5}C/m^2}{2\epsilon_0}(\hat{j})

E_B= \frac{-7.10^{-5}C/m^2}{2\epsilon_0}(-\hat{j})

E_C= \frac{-3.1^{-5}C/m^2}{2\epsilon_0}(\hat{j})

The electric field is

E_{NET}= E_A+E_B+E_C

E_{NET}=\frac{-4.10^{-5}C/m^2}{2\epsilon_0}(\hat{j})+\frac{-7.10^{-5}C/m^2}{2\epsilon_0}(-\hat{j})+ \frac{-3.1^{-5}C/m^2}{2\epsilon_0}(\hat{j})

E_{NET} = 0

Force on each sheet is,

F=E_{NET}\sigma ds

F=0

The total force is 0

5 0
3 years ago
The primary difference between LEED Certification for Buildings and LEED Professional Certifications is that professionals can c
TEA [102]

Answer:false

Explanation:

8 0
3 years ago
How is a Doctor Who is a generalist different a specialist?
bekas [8.4K]

Answer:

specialist focus in one very specific thing, generalist focus and many things, they are like a jack of all trades

7 0
3 years ago
A sewer pipe 8 inches in diameter can carry 0.662 ft3/s when flowing at a depth of 4 inches.
Rom4ik [11]

Answer:

a) the flow under full capacity is  q₂= 1.334 ft³/s

b) the velocity would be v= 3.793 ft/s

Explanation:

a) Since the pipe has 8 inches in diameter but 4 are covered with water flow ( half of a circle in area=A₁) , q₁=0.662 ft³/s then

q₁=A₁*v

then for the same velocity v but area A₂=2*A₁

flow under full capacity= q₂ = A₂*v= 2*A₁*v= 2*q₁=2*0.662 ft³/s= 1.334 ft³/s

b) when flowing at a depth of 4 inches

A₁= (1/2)*(π*D²/4) = π* (1/8)*(8 in)² = 8π in² * (1  ft²/ 144  in²) = π/18 ft² = 0.1745 ft²

then

v=q₁/A₁ = 0.662 ft³/s/0.1745 ft²= 3.793 ft/s

v= 3.793 ft/s

6 0
4 years ago
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