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MA_775_DIABLO [31]
2 years ago
13

A combination lock has a code consisting of 3 numbers, each of which can be 0 to 39, with numbers repeated. Jillian says that th

ere are only 120 possible codes. Is Jillian correct? If not, explain her error.
Mathematics
2 answers:
levacccp [35]2 years ago
6 0

Answer:

Jillian is not correct

Jillian has summed up the possible number of codes for each digit instead of multiplying it

Step-by-step explanation:

The lock has a code that consists of 3 number

It is given that the number in the lock code can be used between 0 to 39

Thus, out of total 40 set of numbers (i.e 0-39), the numbers can be repeated.

This means for all three code numbers the opportunity of choosing a number is same  i.e. between  0-39

Now, the first digit of the code can be any number between  0-39

Like wise the second and third  digit of the code can be any number between  0-39

Thus. the possible number of codes with repetition allowed are

40 * 40* 40\\64000

Hence, Jillian is not correct

Jillian has summed up the possible number of codes for each digit instead of multiplying it

dezoksy [38]2 years ago
4 0
39*3=117 not 120 so that is why she is wrong

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