a safe has a 4-digit lock code that does not include zero as a digit and no digit is repeated. What is the probability that the lock code consists of all even didgits?

Step-by-step explanation:

9 all possible digits (1, 2, 3, 3, 4, 5, 6, 7, 8, 9)

4 digits and no digits repeated so

9•8•7•6      - number of all possible codes

4 even digits (2, 4, 6, 8)

4 digits and no digits repeated so

4•3•2•1       - number of all codes wich consists of all even digits

What is the probability that the lock code consists of all even didgit:

third option: 6x+8

step-by-step explanation:

we can see that the given question is a sum of series and the values of n vary from 0 to 3

the first term will be:

when n=0, [(0)*x+2] => (0+2) => 2

when n=1, [(1)*x+2] => (x+2) => x+2

when n=2, [(2)*x+2] => 2x+2

when n=4, [(3)*x+2] => 3x+2

so adding up all the terms will give us:

=2+x+2+2x+2+3x+2

=x+2x+3x+2+2+2+2

=6x+8

so,

third option is the correct answer ..

the answer is d. (1.2, -4.7) on edgn

hope this !