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.
Answer:
$54.20
Step-by-step explanation:
minute=m
total monthly cost = p
39.99 + .49m = p
39.99 + .49(29) =p
39.99 + 14.21 =p
54.2 =p
The sum of a geometric sequence is:
s(n)=a(1-r^n)/(1-r) in our case:
s(n)=3(1-4^n)/(1-4)
s(n)=-(1-4^n)
s(n)=(4^n)-1 and s=1023
(4^n)-1=1023
4^n=1024
n ln4=ln1024
n=(ln1024)/(ln4)
n=5
Answer:
C: 3.0
Step-by-step explanation:
You can write a ratio as a fraction (12/4). Write it as a fraction then just do the math.
12/4 = 3
