L*w=6*20
6*15= 90m
20*15=300m
l*w= 90*300=27000m <-- area
The answer is the second option y= -2x + 11
Simple the answer is: n=14
The ancestors to the equation is c=20
Answer:
450,000 codes
Step-by-step explanation:
Number of digits in the access code = 6
Conditions:
- First digit cannot be 2
- Last digit must be odd
Since, there are 10 digits in total from 0 to 9 and first digit cannot be 2, there are 9 ways to fill the place of first digit.
Last digit must be odd. As there are 5 odd digits from 0 to 9, there are 5 ways to fill the last digit.
The central 4 digits can be filled by any of the 10 numbers. So, each of them can be filled in 10 ways.
According to the fundamental rule of counting, the total possible codes would be the product of all the possibilities of individual digits.
Therefore,
Number of possible codes = 9 x 10 x 10 x 10 x 10 x 5 = 450,000 codes
Hence, 450,000 different codes are possible for the lock box
