​GB/T1800.2-2020将线性尺寸的标准公差分为 等级。 A: 12 B: 20 C: 30 D: 28

下列属于国家标准的是： A: GB1353—2018玉米 B: SC/T 3110—1019 冻虾仁 C: QB/T2019—2020 低钠盐 D: T/SDAS 113—2020山东小磨香油

2020春夏四大T台色彩数据中，( )占据总体最大份额。

‍GB/T1800.2-2020将线性尺寸的标准公差分为 等级。‌ A: 12 B: 18 C: 20 D: 28 E: 30

下面代码的执行结果是： ls = [“2020”, “2020”, “Python”] ls.append(2020) ls.append([2020, “2020”]) print(ls): [‘2020’, ‘2020’, ‘Python’, 2020, 2020, ‘2020’]|[‘2020’, ‘2020’, ‘Python’, 2020, [2020, ‘2020’]]|[‘2020’, ‘2020’, ‘Python’, 2020, [‘2020’]]|[‘2020’, ‘2020’, ‘Python’, 2020]

下面代码的执行结果是‪‬‪‬‪‬‪‬‪‬‮‬‫‬‭‬‪‬‪‬‪‬‪‬‪‬‮‬‭‬‪‬‪‬‪‬‪‬‪‬‪‬‮‬‭‬‫‬‪‬‪‬‪‬‪‬‪‬‮‬‭‬‫‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‪‬( )‪‬‪‬‪‬‪‬‪‬‮‬‫‬‭‬‪‬‪‬‪‬‪‬‪‬‮‬‭‬‪‬‪‬‪‬‪‬‪‬‪‬‮‬‭‬‫‬‪‬‪‬‪‬‪‬‪‬‮‬‭‬‫‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‪‬ ‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬‬。ls = ["2020", "20.20", "Python"] ls.append(2020)ls.append([2020, "2020"])print(ls) A: ['2020', '20.20', 'Python', 2020, 2020, '2020'] B: ['2020', '20.20', 'Python', 2020, [2020, '2020']] C: ['2020', '20.20', 'Python', 2020] D: ['2020', '20.20', 'Python', 2020, ['2020']]

下面代码的执行结果是：‪‪‪‪‪‪‫‪‪‪‪‪‪‫‪‪‪‪‪‫‪‪‪‪‪‫‪‪‪‪‪‪‫‪‪‪‪‪‪‫ ls = ["2020", "20.20", "Python"] ls.append(2020) ls.append([2020, "2020"]) print(ls) A: ['2020', '20.20', 'Python', 2020, ['2020']] B: ['2020', '20.20', 'Python', 2020, 2020, '2020'] C: ['2020', '20.20', 'Python', 2020] D: ['2020', '20.20', 'Python', 2020, [2020, '2020']]

以下代码的输出结果是______。ls = ["2020", "20.20", "Python"]ls.append(2020)ls.append([2020, "2020"])print(ls) A: ['2020', '20.20', 'Python', 2020] B: ['2020', '20.20', 'Python', 2020, [2020, '2020']] C: ['2020', '20.20', 'Python', 2020, 2020, '2020'] D: ['2020', '20.20', 'Python', 2020, ['2020']]

下面代码的执行结果是：（ ）ls= [“2020”, “20.20”, “Python”]ls.append(2020)ls.append([2020, “2020”])print(ls) A: [‘2020’, ‘20.20’, ‘Python’, 2020] B: [‘2020’, ‘20.20’, ‘Python’, 2020, [2020, ‘2020’]] C: [‘2020’, ‘20.20’, ‘Python’, 2020, [‘2020’]] D: [‘2020’, ‘20.20’, ‘Python’, 2020, 2020, ‘2020’]r

以下代码的输出结果是( )。 ls = ["2020", "20.20", "Python"] ls.append([2020, "2020"]) [br][/br]print(ls) A: ['2020', '20.20', 'Python', [2020, '2020']] B: ['2020', '20.20', 'Python', 2020, ['2020']] C: ['2020', '20.20', 'Python', 2020,] D: ['2020', '20.20', 'Python', 2020, '2020']