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:

you need to show the graph

step-by-step explanation:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!