Mean is the average of all the numbers added together, then divided by the total amount of numbers. So 11+9+2+9+6+17 = 54 and there are 6 numbers so you divide 54 by 6 = 9
Answer:
B -2x - 12y
Step-by-step explanation:
-8x + 6x = -2x
12y - 24y = -12y
-2x - 12y
Let's see, if the first one has a Principal of $50, when it doubles the accumulated amount will then be $100,
recall your logarithm rules for an exponential,
![\bf \textit{Logarithm of exponentials}\\\\ log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)\\\\ -------------------------------\\\\ \qquad \textit{Compound Interest Earned Amount} \\\\ ](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7BLogarithm%20of%20exponentials%7D%5C%5C%5C%5C%0Alog_%7B%7B%20%20a%7D%7D%5Cleft%28%20x%5E%7B%7B%20%20b%7D%7D%20%5Cright%29%5Cimplies%20%7B%7B%20%20b%7D%7D%5Ccdot%20%20log_%7B%7B%20%20a%7D%7D%28x%29%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Cqquad%20%5Ctextit%7BCompound%20Interest%20Earned%20Amount%7D%0A%5C%5C%5C%5C%0A)
![\bf A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\to &\$100\\ P=\textit{original amount deposited}\to &\$50\\ r=rate\to 8\%\to \frac{8}{100}\to &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annnually, thus once} \end{array}\to &1\\ t=years \end{cases} \\\\\\ 100=50\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 100=50(1.08)^t \\\\\\ \cfrac{100}{50}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t) \\\\\\ ](https://tex.z-dn.net/?f=%5Cbf%20A%3DP%5Cleft%281%2B%5Cfrac%7Br%7D%7Bn%7D%5Cright%29%5E%7Bnt%7D%0A%5Cquad%20%0A%5Cbegin%7Bcases%7D%0AA%3D%5Ctextit%7Baccumulated%20amount%7D%5Cto%20%26%5C%24100%5C%5C%0AP%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cto%20%26%5C%2450%5C%5C%0Ar%3Drate%5Cto%208%5C%25%5Cto%20%5Cfrac%7B8%7D%7B100%7D%5Cto%20%260.08%5C%5C%0An%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%0A%5Ctextit%7Bannnually%2C%20thus%20once%7D%0A%5Cend%7Barray%7D%5Cto%20%261%5C%5C%0At%3Dyears%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A100%3D50%5Cleft%281%2B%5Cfrac%7B0.08%7D%7B1%7D%5Cright%29%5E%7B1%5Ccdot%20t%7D%5Cimplies%20100%3D50%281.08%29%5Et%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B100%7D%7B50%7D%3D1.08%5Et%5Cimplies%202%3D1.08%5Et%5Cimplies%20log%282%29%3Dlog%281.08%5Et%29%0A%5C%5C%5C%5C%5C%5C%0A)
![\bf log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t\\\\ -------------------------------\\\\ ](https://tex.z-dn.net/?f=%5Cbf%20log%282%29%3Dt%5Ccdot%20log%281.08%29%5Cimplies%20%5Ccfrac%7Blog%282%29%7D%7Blog%281.08%29%7D%3Dt%5Cimplies%209.0065%5Capprox%20t%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A)
now, for the second amount, if the Principal is 500, the accumulated amount is 1000 when doubled,
![\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\to &\$1000\\ P=\textit{original amount deposited}\to &\$500\\ r=rate\to 8\%\to \frac{8}{100}\to &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annnually, thus once} \end{array}\to &1\\ t=years \end{cases} \\\\\\ 1000=500\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 1000=500(1.08)^t \\\\\\ ](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Ctextit%7BCompound%20Interest%20Earned%20Amount%7D%0A%5C%5C%5C%5C%0AA%3DP%5Cleft%281%2B%5Cfrac%7Br%7D%7Bn%7D%5Cright%29%5E%7Bnt%7D%0A%5Cquad%20%0A%5Cbegin%7Bcases%7D%0AA%3D%5Ctextit%7Baccumulated%20amount%7D%5Cto%20%26%5C%241000%5C%5C%0AP%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cto%20%26%5C%24500%5C%5C%0Ar%3Drate%5Cto%208%5C%25%5Cto%20%5Cfrac%7B8%7D%7B100%7D%5Cto%20%260.08%5C%5C%0An%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%0A%5Ctextit%7Bannnually%2C%20thus%20once%7D%0A%5Cend%7Barray%7D%5Cto%20%261%5C%5C%0At%3Dyears%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A1000%3D500%5Cleft%281%2B%5Cfrac%7B0.08%7D%7B1%7D%5Cright%29%5E%7B1%5Ccdot%20t%7D%5Cimplies%201000%3D500%281.08%29%5Et%0A%5C%5C%5C%5C%5C%5C%0A)
![\bf \cfrac{1000}{500}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t) \\\\\\ log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B1000%7D%7B500%7D%3D1.08%5Et%5Cimplies%202%3D1.08%5Et%5Cimplies%20log%282%29%3Dlog%281.08%5Et%29%0A%5C%5C%5C%5C%5C%5C%0Alog%282%29%3Dt%5Ccdot%20log%281.08%29%5Cimplies%20%5Ccfrac%7Blog%282%29%7D%7Blog%281.08%29%7D%3Dt%5Cimplies%209.0065%5Capprox%20t%5C%5C%5C%5C%0A-------------------------------)
now, for the last, Principal is 1700, amount is then 3400,
![\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\to &\$3400\\ P=\textit{original amount deposited}\to &\$1700\\ r=rate\to 8\%\to \frac{8}{100}\to &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annnually, thus once} \end{array}\to &1\\ t=years \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Ctextit%7BCompound%20Interest%20Earned%20Amount%7D%0A%5C%5C%5C%5C%0AA%3DP%5Cleft%281%2B%5Cfrac%7Br%7D%7Bn%7D%5Cright%29%5E%7Bnt%7D%0A%5Cquad%20%0A%5Cbegin%7Bcases%7D%0AA%3D%5Ctextit%7Baccumulated%20amount%7D%5Cto%20%26%5C%243400%5C%5C%0AP%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cto%20%26%5C%241700%5C%5C%0Ar%3Drate%5Cto%208%5C%25%5Cto%20%5Cfrac%7B8%7D%7B100%7D%5Cto%20%260.08%5C%5C%0An%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%0A%5Ctextit%7Bannnually%2C%20thus%20once%7D%0A%5Cend%7Barray%7D%5Cto%20%261%5C%5C%0At%3Dyears%0A%5Cend%7Bcases%7D)
8c = 14 - 5c - 1
13c = 13
c = 1