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.
Step-by-step explanation:
The ratio's can be expressed as fractions and you may compare those.
In order to compare fractions you need to bring them into equal denominator.
8 and 9 have LCM of 8*9 = 72 since they have no common factors.
<u>The equivalent fractions with common denominator are:</u>
- 5/8 = 9*5/(9*8) = 45/72
- 7/9 = 8*7/(8*9) = 56/72
<u>Now we can compare the fractions:</u>
- 45/72 < 56/72, since 45 < 56
It would be 104? I don’t understand the question but I guessing you’re adding
Juan rode the slowest. It took him 5 minutes to go 2 blocks and another 3 minutes to go 2 more blocks. It was taking him longer minutes to go 2 blocks at a time. Took Juan 17 minutes to do 8 blocks and Rita did 8 blocks in 8 minutes she was faster. Hope this helps
