(2,7)
(6,12)
(10,17)
Some moderator deleted my response for some reason. I did it mentally.. but I guess I'll show proof.
1) (2,7)
-4y = -5x - 18
-4(7) = -5(2) - 18
-28 = -10 -18
-28 = -28
2) (6,12)
<span>-4y=-5x-18
</span> -4(12) = -5(6) - 18
-48 = -30 - 18
-48 = -48
3) (10,17)
4y=-5x-18
-4(17) = -5(10) - 18
-68 = -50 - 18
-68 = -68
See the pattern? 2,6,10 and 7,12,17?
Done with substitution and trial and error :3333
The given situation represents a random sample because, in statistics, a simple random sampling refers to the subset of individuals chosen from a larger set called population, where the sample is chosen randomly.
In this case, the dialed number are a random event, so it can be called a simple random sample.
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)
The exponent expression ∛(56x⁷y⁵) is equivalent to the expression 2x²y∛(7xy²).
<h3>What is an exponent?</h3>
Exponential notation is the form of mathematical shorthand which allows us to write complicated expressions more succinctly. An exponent is a number or letter is called the base. It indicates that the base is to raise to a certain power. X is the base and n is the power.
The expression will be given as
![\sqrt[3]{56x^7y^5}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B56x%5E7y%5E5%7D)
The expression can be written as
![\rightarrow \sqrt[3]{2*2*2*7 \ x^3 \ x^3 \ x \ y^3 \ y^2 }\\\\\rightarrow 2*x*x*y * \sqrt[3]{7xy^2} \\\\\rightarrow 2x^2 y \sqrt[3]{7xy^2 }](https://tex.z-dn.net/?f=%5Crightarrow%20%5Csqrt%5B3%5D%7B2%2A2%2A2%2A7%20%5C%20x%5E3%20%5C%20x%5E3%20%5C%20x%20%5C%20y%5E3%20%20%5C%20y%5E2%20%7D%5C%5C%5C%5C%5Crightarrow%202%2Ax%2Ax%2Ay%20%2A%20%5Csqrt%5B3%5D%7B7xy%5E2%7D%20%5C%5C%5C%5C%5Crightarrow%202x%5E2%20y%20%5Csqrt%5B3%5D%7B7xy%5E2%20%7D)
More about the exponent link is given below.
brainly.com/question/5497425