An access code consists of of a letter followed by four digits. any letter can be used, the first digit cannot be 0, and the last digit must be even. (a) find the number of possible access codes. (b) what is the probability of randomly electing the correct access code on the first try? (c) what is the probability of not selecting the right access code on the first try?

(a)117000  (b)1/117000 (c)116999/117000

Step-by-step explanation:

(a)There are 26 letters, hence 26 options for the first character

There are 9 options for the second character, because, you can place 1,2,3,4,5,6,7,8,or 9.

For the third character you can have 10 options, since, you can 1 through 9 and 0.

The 4th one has 10 options too since the same terminology is used for the 3rd digit as the 4th one.

For the last digit, You have 5 options,0,  2, 4, 6, and 8.

Hence, the answer is 26*9*10*10*5 = 117000

(b)There is only 1 correct code, so the chance of getting it right is 1/117000.

(c)There are 117000-1 ways oh getting it wrong, so 116999/117000.

step-by-step explanation:

here's what happens:

1) the graph of f   = x^2 is shifted 2 units to the right.

2) the resulting graph is reflected in the x-axis, so that the parabola now opens down.

3) this final result is transformed up by 5 units.

answer: no, but if it does it's a obvious pattern.

step-by-step explanation: