a realtor uses a lock box to store the keys to a house that is for sale. the access code for the lock consist of five digits. the first digit cannot be 1 and the last digit must be even. how many different codes are avaible

45,000 codes

Step-by-step explanation:

Given:

Code of 5 digits

Condition

Required

Calculate the number of codes available

Digits = {0,1,29}

n(Digits) = 10

Let the format of the code be represented as follows;

ABCDE

From the conditions given

A can't be 1;

This means that A can be any of 0,2,3,49

This implies that A can be any of the above 9 digits

n(A) = 9

There's no condition attached to BCD;

This means that B can be any of 10 digits

This means that C can be any of 10 digits

This means that D can be any of 10 digits

n(B) = n(C) = n(D) = 10

Lastly, E must be an even number;

This means that E can be any of 0,2,4,6,8

This implies that E can be any of the above 5 digits

n(E) = 5

So,

Number of available codes = n(A) * n(B) * n(C) * n(D) * n(E)

Number of available codes = 9 * 10 * 10 * 10 *5

Number of available codes = 45,000

Hence, there are 45,000 available codes

step-by-step explanation:   divide the numerator by the denominator to get the decimal

it would still be 193

if divide by 100 then 1.93

i am not sure