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Create a dictionary ‘ODD’ of odd numbers between 1 and 10, where the key is the decimal number and the value is the corresponding number in words. Perform the following operations on the dictionary:i) display the keysii) display the valuesiii) check if 7 is present or not​

Answer» 0:'One',3.0:'Three',5.0:'Five',7.0:'Seven',9.0:'Nine'}>>> ODD.keys()dict_keys([1.0, 3.0, 5.0, 7.0, 9.0])>>> ODD.values()dict_values(['One', 'Three', 'Five', 'Seven', 'Nine'])7.0 in ODDTrue


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