Determine the possible number of 4-digit id numbers made up of the digits 0 to 9 with leading zero's disallowed. [note: some examples of leading zero's are 0431, 0001 and .]

There are: 9*10*10*10 = 9000 possible number of 4-digit ID numbers made up of the digits 0 to 9 with leading zero's disallowed

Step-by-step explanation:

The ID numbers in this problem are made of 4-digits.

So:

D1 - D2 - D3 - D4

Leading zero's is disallowed, so D1 cannot be zero. It means that there are only 9 possible values for D1. For D2, D3 and D4, there are no restrictions. So, each one of these can have 10 digits.

In all, our possibilies in each position are this

9 - 10 - 10 - 10

It means that for each one of the 9 possibilies in D1, there are 10 possibilities in D2. For each of the 10 in D2, there are 10 in D3. And for each of the 10 in D3, there are 10 in D4.

So there are: 9*10*10*10 = 9000 possible number of 4-digit ID numbers made up of the digits 0 to 9 with leading zero's disallowed

8 2/3 is the answer

step-by-step explanation: just use a calculator

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %