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mel-nik [20]
2 years ago
8

Drivers education - Unit 3

Engineering
1 answer:
melamori03 [73]2 years ago
4 0

The following scenarios are pertinent to driving conditions that one may encounter. See the following rules of driving.

<h3>What do you do when the car is forced into the guardrail?</h3>

Best response:

  • I'll keep my hands on the wheel and slow down gradually.
  • The reason I keep my hands on the steering wheel is to avoid losing control.
  • This will allow me to slowly back away from the guard rail.
  • The next phase is to gradually return to the fast lane.
  • Slamming on the brakes at this moment would result in a collision with the car behind.

Scenario 2: When driving on a wet road and the car begins to slide

Best response:

  • It is not advised to accelerate.
  • Pumping the brakes is not recommended.
  • Even lightly depressing and holding down the brake pedal is not recommended.
  • The best thing to do is take one foot off the gas pedal.
  • There should be no severe twists at this time.

Scenario 3: When you are in slow traffic and you hear the siren of an ambulance behind

Best response:

  • The best thing to do at this moment is to go to the right side of the lane and come to a complete stop.
  • This helps to keep the patient in the ambulance alive.
  • It also provide a clear path for the ambulance.
  • Moving to the left is NOT recommended.
  • This will exacerbate the situation. If there is no place to park on the right shoulder of the road, it is preferable to stay in the lane.

Learn more about rules of driving. at;

brainly.com/question/8384066

#SPJ1

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chubhunter [2.5K]

Answer:

B probably

Explanation:

Because the prompt doesn't specify what sort of violation it could be anything maybe when they release the metals during the day and so on.

5 0
3 years ago
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
Define Ancestor, Descendant, Siblings, Height, Depth, Root and Leaf for the
mr_godi [17]

Answer:

Ancestor are our great grandparents

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3 years ago
What are the ropes of secretaries and treasures in a meeting​
Nastasia [14]

Answer:

To prepare and issue notices and agendas of all meetings in consultation with the chairman, and to ensure that any background papers are available well before the meeting. To attend and take the minutes of every committee meeting. To circulate minutes to all committee members, and to conduct the correspondence

Explanation:

I think you want to say roles.

4 0
3 years ago
What is the greatest speedup (in terms of throughput) possible with pipelining? You do not need to worry about the register timi
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Answer:

The greatest speedup possible with pipelining is 2.5 times

Explanation:

As we know

Span of stage = Maximum span of any span

As stage B has the maximum duration of time, So

Duration of time of each stage = Time duration of Stage B = 8 ns

Now calculate the total time without pipeline = Duration of Stage A + Duration of Stage B + Duration of Stage C + Duration of Stage D = 5 ns + 8 ns + 4 ns + 3 ns = 20 ns

So the greatest speedup can be calculated as follow

Greatest speedup = Total time without pipeline / Duration of time of each stage = 20 ns / 8 ns = 2.5 times

3 0
3 years ago
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