now, there are 12 months in a year, so 18 months is really 18/12 of a year, thus
![~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$4000\\ P=\textit{original amount deposited}\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ t=years\to \frac{18}{12}\dotfill &\frac{3}{2} \end{cases} \\\\\\ 4000=P[1+(0.05)(\frac{3}{2})]\implies 4000=P(1.075) \\\\\\ \cfrac{4000}{1.075}=P\implies 3720.93\approx P](https://tex.z-dn.net/?f=~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%281%2Brt%29%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5Cdotfill%20%26%20%5C%244000%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5C%5C%20r%3Drate%5Cto%205%5C%25%5Cto%20%5Cfrac%7B5%7D%7B100%7D%5Cdotfill%20%260.05%5C%5C%20t%3Dyears%5Cto%20%5Cfrac%7B18%7D%7B12%7D%5Cdotfill%20%26%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%204000%3DP%5B1%2B%280.05%29%28%5Cfrac%7B3%7D%7B2%7D%29%5D%5Cimplies%204000%3DP%281.075%29%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B4000%7D%7B1.075%7D%3DP%5Cimplies%203720.93%5Capprox%20P)
Alright, let's do this! If $25 is one fifth of how much the game system cost then, multiply 25 times 5. And you should get your answer.
I hope this helps.
Answer:
7.6
Step-by-step explanation:
m<DBA is 68 degrees
This is easy trust me everything will be fine first drag it to the bottom right the same way you do it on the Hub and then go 4 units below and you’re all set.
I'm not sure what exactly you did wrong, but I agree with you that the sample size is too small, so the correct answer will probably be the fourth options. Hope that this gives you some confidence, and 'm sorry not to be able to help you any further...
