Answer:
no
because it doesn't work with the first equation
You could roll a 6-sided die 3 times to simulate this. This is a good simulation because it has one side for each possible tool, and each roll is independent, just as selecting the tool with replacement would be.
Answer:
9 and 4/5
Step-by-step explanation:
Distributive property, commutative property, and associative property (I believe. I tried my best but I’m not completely sure that I’m correct. Good luck!)
now, if we take 2000 to be the 100%, what is 2200? well, 2200 is just 100% + 10%, namely 110%, and if we change that percent format to a decimal, we simply divide it by 100, thus 
.
so, 1.1 is the decimal number we multiply a term to get the next term, namely 1.1 is the common ratio.
![\bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\a_1=2000\\r=1.1\\n=4\end{cases}\\\\\\S_4=2000\left[ \cfrac{1-(1.1)^4}{1-1.1} \right]\implies S_4=2000\left(\cfrac{-0.4641}{-0.1} \right)\\\\\\S_4=2000(4.641)\implies S_4=9282](https://tex.z-dn.net/?f=%20%5Cbf%20%5Cqquad%20%5Cqquad%20%5Ctextit%7Bsum%20of%20a%20finite%20geometric%20sequence%7D%5C%5C%5C%5CS_n%3D%5Csum%5Climits_%7Bi%3D1%7D%5E%7Bn%7D%5C%20a_1%5Ccdot%20r%5E%7Bi-1%7D%5Cimplies%20S_n%3Da_1%5Cleft%28%20%5Ccfrac%7B1-r%5En%7D%7B1-r%7D%20%5Cright%29%5Cquad%20%5Cbegin%7Bcases%7Dn%3Dn%5E%7Bth%7D%5C%20term%5C%5Ca_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5Cr%3D%5Ctextit%7Bcommon%20ratio%7D%5C%5C----------%5C%5Ca_1%3D2000%5C%5Cr%3D1.1%5C%5Cn%3D4%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5CS_4%3D2000%5Cleft%5B%20%5Ccfrac%7B1-%281.1%29%5E4%7D%7B1-1.1%7D%20%5Cright%5D%5Cimplies%20S_4%3D2000%5Cleft%28%5Ccfrac%7B-0.4641%7D%7B-0.1%7D%20%20%5Cright%29%5C%5C%5C%5C%5C%5CS_4%3D2000%284.641%29%5Cimplies%20S_4%3D9282%20)
