(126.72)8=(？)2(126.72)16=(？)2
举一反三
- (3B6D.14)()16()=(?)()2()(1100100101.0101)()2()=(?)()8()(239.375)()10()=(?)()2()(365.5)()8()=(?)()10()(115.4375)()10()=(?)()2()=(?)()8()=(?)()16()(2A.3C)()16()=(?)()10()=(?)()2()(126.75)()8()=(?)()16()=(?)()2()(1001101.10101)()2()=(?)()16()=(?)()8
- 【填空题】完成下列数制转换: ( 25 ) 8 = () 2 ( CF ) 16 = () 2 ( 1101011 ) 2 = () 8 (1 101011011 ) 2 = () 16 ( 856 ) 10 = () 2 = () 8 = () 16
- (1)()(()101101())()2()=()(()____()____())()10()=()(()____()____())()8()=()(()____()____())()16()(2)()(()11110010())()2()=()(()____()____())()10()=()(()____()____())()8()=()(()____()____())()16()(3)()(()10100.1011())()2()=()(()_______())()10()=()(()___()___())()8()=()(()____()____())()16()(4)()(25.75)()10()=()(()____()____())()16()(5)()(110.125)()10()=()(()____()____())()16
- 采样频率为 44.1KHZ,16 位量化的双声道立体声,其 1 分钟的数据量约为( )。 A: 44.1×16÷8×2×60bit B: 44100×16÷8×2×60Byte C: 44100×16÷8×2×60bit D: 44.1×16÷8×2×60Byte
- 完成下列数制转换： (1)(10010111)2=(97)16=(151)10 (2)(0.01011111)2=(0.5F)16=(0.37109375)10 (3)（101110.001）2=(46.375)10 (4)(254.76)10=(376.604)8 (5)(4DE.C8)16=(1246.78125)10 (6)(23F.45)16=(001000111111.01000101)2 (7)（47）10＝（）2＝（）8421BCD＝（）16 (8)（1A）16＝（）2＝（）10＝（）8