keeping in mind that anything raised at the 0 power, is 1, with the sole exception of 0 itself.
![\bf ~~~~~~~~~~~~\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} \qquad \qquad \cfrac{1}{a^n}\implies a^{-n} \qquad \qquad a^n\implies \cfrac{1}{a^{-n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{(r^{-7}b^{-8})^0}{t^{-4}w}\implies \cfrac{1}{t^{-4}w}\implies \cfrac{1}{t^{-4}}\cdot \cfrac{1}{w}\implies t^4\cdot \cfrac{1}{w}\implies \cfrac{t^4}{w}](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bnegative%20exponents%7D%0A%5C%5C%5C%5C%0Aa%5E%7B-n%7D%20%5Cimplies%20%5Ccfrac%7B1%7D%7Ba%5En%7D%0A%5Cqquad%20%5Cqquad%0A%5Ccfrac%7B1%7D%7Ba%5En%7D%5Cimplies%20a%5E%7B-n%7D%0A%5Cqquad%20%5Cqquad%20a%5En%5Cimplies%20%5Ccfrac%7B1%7D%7Ba%5E%7B-n%7D%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Ccfrac%7B%28r%5E%7B-7%7Db%5E%7B-8%7D%29%5E0%7D%7Bt%5E%7B-4%7Dw%7D%5Cimplies%20%5Ccfrac%7B1%7D%7Bt%5E%7B-4%7Dw%7D%5Cimplies%20%5Ccfrac%7B1%7D%7Bt%5E%7B-4%7D%7D%5Ccdot%20%5Ccfrac%7B1%7D%7Bw%7D%5Cimplies%20t%5E4%5Ccdot%20%5Ccfrac%7B1%7D%7Bw%7D%5Cimplies%20%5Ccfrac%7Bt%5E4%7D%7Bw%7D%20)
List out the multiples of 8
8, 16, 24, 32, 40, 48, 56, 64, 72, 80
I'm going to highlight in bold the values between 20 and 50
8, 16, 24, 32, 40, 48, 56, 64, 72, 80
So the favorable outcomes, aka the outcomes we want, are: {24, 32, 40, 48}
These four values are all divisible by 8. Also, these values are between 20 and 50.
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.
Cant really see it the picture
