Answer:
896
Step-by-step explanation:
Let's talk first about how many 3 digit numbers there are. The first 3 digit number is 100 and the last is 999. So there are 999-100+1 numbers that are 3 digits long. That simplifies to 900.
Now let's find how many of those have a sum for the digits being 1, then 2 ? Then take that sum away from the 900 to see how many 3 digit numbers have the sum of their digits being more than 2.
3 digit numbers with sum of 1:
The first and only number is 100 since 1+0+0=1.
We can't include 010 or 001 because these aren't really three digits long.
3 digit numbers with sum of 2:
The first number is 101 since 1+0+1=2.
The second number is 110 since 1+1+0=2.
The third number is 200 since 2+0+0=2.
That's the last of those. We could only use 0,1, and 2 here.... Anything with a 3 in it would give us something larger than or equal to 3.
So there are 900-1-3 numbers who are 3 digits long and whose sum of digits is greater than 2.
This answer simplifies to 896.
To figure out how much she spent all we have to do is add up $0.85, $4.50 and $1.50
$0.85 + $4.50 + $1.50 = $6.85
Therefore she spent $6.85
Hope this helps
The answer is
x > 16/7
The > has a line under<span />
-6 is the answer since it’s closer to zero
Answer:
a) 2048
b)164
c)54
d) 11
Step-by-step explanation:
a) A coin has two faces.
Outcome possible =2^11= 2048
b) 8 heads possible outcome = 11!/8!3! = 3971688/241/920= 164 ways
c)2 heads outcome= 11!/2!8!= 39716800/725760=54
d) 7 heads= 11!/7!6! =39716800/3628800 = 11
