1. Implement the function dict_intersect, which takes two dictionaries as parameters d1 and d2, and returns a new dictionary which contains only those keys which appear in both d1 and d2, whose values are a tuple of the corresponding values from d1 and d2. E. g., dict_intersect({'a': 'apple', 'b': 'banana'}, {'b': 'bee', 'c': 'cat'}) should return {'b': ('banana', 'bee')}

2. Implement the function consolidate, which accepts zero or more sequences in the star parameter seqs, and returns a dictionary whose keys consist of values found in those sequences, which in turn map to numbers indicating how many times each value appears across all the sequences.

E. g., consolidate([1,2,3], [1,1,1], [2,4], [1]) should return the dictionary {1: 5, 2: 2, 3: 1, 4: 1}.

1  

def dict_intersect(d1,d2): #create dictionary

  d3={} #dictionaries

  for key1,value1 in d1.items():       #iterate through the loop  

      if key1 in d2:   #checking condition

          d3[key1]=(d1[key1],d2[key1])   #add the items into the dictionary  

  return d 3

print(dict_intersect({'a': 'apple', 'b': 'banana'}, {'b': 'bee', 'c': 'cat'})) #display

2

def consolidate(*l1):  #create consolidate

  d3={} # create dictionary

  for k in l1:       #iterate through the loop

      for number in k:   #iterate through  the loop                               d3[number]=d3.get(number,0)+1   #increment the value

             return d 3 #return

print(consolidate([1,2,3], [1,1,1], [2,4], [1])) #display

Explanation:

1

Following are  the description of program

2

Following are  the description of program

answer; i believe the correct answer is (psychological pricing);