from 2007 to 2012 is only 5 years, so we can see this as a compound interest with a rate of 1.2% per annum for 5 years, so
![~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$113505\\ r=rate\to 1.2\%\to \frac{1.2}{100}\dotfill &0.012\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per year, thus once} \end{array}\dotfill &1\\ t=years\dotfill &5 \end{cases} \\\\\\ A=113505\left(1+\frac{0.012}{1}\right)^{1\cdot 5}\implies A=113505(1.012)^5\implies A\approx 120481](https://tex.z-dn.net/?f=~~~~~~%20%5Ctextit%7BCompound%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%5Cleft%281%2B%5Cfrac%7Br%7D%7Bn%7D%5Cright%29%5E%7Bnt%7D%20%5Cquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cdotfill%20%26%5C%24113505%5C%5C%20r%3Drate%5Cto%201.2%5C%25%5Cto%20%5Cfrac%7B1.2%7D%7B100%7D%5Cdotfill%20%260.012%5C%5C%20n%3D%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%20%5Ctextit%7Bper%20year%2C%20thus%20once%7D%20%5Cend%7Barray%7D%5Cdotfill%20%261%5C%5C%20t%3Dyears%5Cdotfill%20%265%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D113505%5Cleft%281%2B%5Cfrac%7B0.012%7D%7B1%7D%5Cright%29%5E%7B1%5Ccdot%205%7D%5Cimplies%20A%3D113505%281.012%29%5E5%5Cimplies%20A%5Capprox%20120481)
Answer:
about 1 inch
Step-by-step explanation:
Answer:
t greater than or less than -10
Step-by-step explanation:
Answer:
39 men, 13 women
Step-by-step explanation:
you're saying you just need the answer so i hope this works. good luck :)