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Yes, the computation is correct: the passwords will have an entropy of $log_2(26^2\cdot10)\simeq {12.7}$$\log_2(26^2\cdot10)\approx {12.7}$, which means that they are weaker than a randomly chosen secret key with 13 bits.
Yes, the computation is correct: the passwords will have an entropy of $log_2(26^2\cdot10)\simeq {12.7}$, which means that they are weaker than a randomly chosen secret key with 13 bits.
Yes, the computation is correct: the passwords will have an entropy of $\log_2(26^2\cdot10)\approx {12.7}$, which means that they are weaker than a randomly chosen secret key with 13 bits.
Yes, the computation is correct: the passwords will have an entropy of $log_2(26^2\cdot10)\simeq2^{12.7}$$log_2(26^2\cdot10)\simeq {12.7}$, which means that they are weaker than a randomly chosen secret key with 13 bits.
Yes, the computation is correct: the passwords will have an entropy of $log_2(26^2\cdot10)\simeq2^{12.7}$, which means that they are weaker than a randomly chosen secret key with 13 bits.
Yes, the computation is correct: the passwords will have an entropy of $log_2(26^2\cdot10)\simeq {12.7}$, which means that they are weaker than a randomly chosen secret key with 13 bits.
Yes, the computation is correct: the passwords will have an entropy of $log_2(26^2\cdot10)\simeq2^{12.7}$, which means that they are weaker than a randomly chosen secret key with 13 bits.
