【单选题】【消耗次数:1】
立体声双声道采样频率为44.1kHz,量化位数为8位,一分钟这样的音乐所需要的存储量可按_____公式计算。
44.1×1000×16×2×60/8字节
44.1×1000×16×2×60/16字节
44.1×1000×8×2×60/8字节
44.1×l000×16×2×60/16字节
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相关题目
【单选题】 立体声双声道采样频率为44.1kHz,量化位数为8位,一分钟这样的音乐所需要的存储量可按( )公式计算。
①  44.1×1000×16×2×60/8字节
②  44.1×1000×16×2×60/16字节
③  44.1×1000×8×2×60/8字节
④  44.1×l000×16×2×60/16字节
【判断题】 1MB等于1024字节。
①  正确
②  错误
【判断题】 Mifare1 S50卡片EEPROM存储器共1K字节,划分成16个扇区,每个扇区分成4个数据块,每个数据块16字节。
①  正确
②  错误
【单选题】 1分钟立体声、16位采样位数、22.05KHz采样频率声音的不压缩的数据量约为( )。
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【判断题】 字符串“string”占用7字节的存储空间。
①  正确
②  错误
【单选题】 两分钟双声道,16位采样位数,22.05kHz采样频率声音的不压缩数据量是_______
①  5.05MB
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【单选题】 一个汉字的内码长度为2字节,其每个字节的最高二进制位的值分别为________。
①  0,0
②  1,1
③  1,0
④  0,1
【判断题】 接收方的糊涂窗口综合征指的是接收方每腾出1字节空间就发送1字节的确认,使得网络效率很低
①  正确
②  错误
【单选题】 两分钟双声道、 16bit量化位数、 22.05kHz采样频率声音的不压缩数据量是
①  5.05MB
②  10.09MB
③  10.34MB
④  10.58MB
【单选题】 3分钟单声道、16位采样位数、22.05KHz采样频率声音的不压缩的数据量约为( )。
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④  15.57MB
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②  Hurrying to the classroom, she saw nobody there.
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①  until
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①  makes
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①  since
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