Fraction: 16/75
Decimal: 0.21
How many 6-digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, if repetitions of digits are allowed?
sveta [45]
There are 6 digits. Each digit can take ten different numbers except for the first digit since it cannot be zero.
So:
9 x 10 x 10 x 10 x 10 x 10
900000 numbers.
Another way of thinking about this is to just count up to 999,999. Obviously there are 999,999 different numbers here. But since our number has to have 6 digits in them, we have to delete 99,999 numbers. Thus there are 900,000 different numbers.
Answer:
No you can not
Step-by-step explanation:
You cant have a probability greater than 1 because then the probability would be false
Minus ka from both sides
u=ba-ka
undistribute a
u=(a)(b-k)
diivde both sides by (b-k)
Answer:
$48
Step-by-step explanation:
1. 192/12 = 16
2. 16 * 3 = 48
3. Your answer is 48
