3^2w^9
Step by step solution :
Step 1 :
Equation at the end of step 1 :
0 - (9w • (0 - 3w8))
Step 2 :
Multiplying exponential expressions :
2.1 w1 multiplied by w8 = w(1 + 8) = w9
Multiplying exponents :
2.2 32 multiplied by 31 = 3(2 + 1) = 33
Equation at the end of step 2 :
0 - -33w9
Step 3 :
Final result :
33w9
the apple tree is 20 and the oak tree is 40
hopes this helps,
Xeno
![\mathbb P(X\le163)=\mathbb P\left(\dfrac{X-155}7\le\dfrac{163-155}7\right)\approx\mathbb P(Z\le1.1429)\approx0.8735](https://tex.z-dn.net/?f=%5Cmathbb%20P%28X%5Cle163%29%3D%5Cmathbb%20P%5Cleft%28%5Cdfrac%7BX-155%7D7%5Cle%5Cdfrac%7B163-155%7D7%5Cright%29%5Capprox%5Cmathbb%20P%28Z%5Cle1.1429%29%5Capprox0.8735)
which puts the student at about the 87th percentile.
first off, let's notice the arithmetic sequence is moving as
![\bf 11~~,~~\stackrel{11+4}{15}~~,~~\stackrel{15+4}{19}~~,~~\stackrel{19+4}{23}...71](https://tex.z-dn.net/?f=%20%5Cbf%2011~~%2C~~%5Cstackrel%7B11%2B4%7D%7B15%7D~~%2C~~%5Cstackrel%7B15%2B4%7D%7B19%7D~~%2C~~%5Cstackrel%7B19%2B4%7D%7B23%7D...71%20)
so is adding "4" to the current term in order to get the next term, meaning the "common difference" of d = 4, and the first term is of course 11.
so, hmmm what ordinal value is the last number of 71 anyway?
![\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ d=4\\ a_1=11 \end{cases} \\\\\\ 71=11+(n-1)4\implies 71=11+4n-4\implies 71=7+4n \\\\\\ 64=4n\implies \cfrac{64}{4}=n\implies 16=n](https://tex.z-dn.net/?f=%20%5Cbf%20n%5E%7Bth%7D%5Ctextit%7B%20term%20of%20an%20arithmetic%20sequence%7D%20%5C%5C%5C%5C%20a_n%3Da_1%2B%28n-1%29d%5Cqquad%20%20%5Cbegin%7Bcases%7D%20n%3Dn%5E%7Bth%7D%5C%20term%5C%5C%20a_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5C%20d%3D%5Ctextit%7Bcommon%20difference%7D%5C%5C%20----------%5C%5C%20d%3D4%5C%5C%20a_1%3D11%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%2071%3D11%2B%28n-1%294%5Cimplies%2071%3D11%2B4n-4%5Cimplies%2071%3D7%2B4n%20%5C%5C%5C%5C%5C%5C%2064%3D4n%5Cimplies%20%5Ccfrac%7B64%7D%7B4%7D%3Dn%5Cimplies%2016%3Dn%20)
low and behold, so 71 is the 16th term, well, let's get their sum then.
![\bf \textit{sum of a finite arithmetic sequence} \\\\ S_n=\cfrac{n(a_1+a_n)}{2}\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ ----------\\ n=16\\ a_1=11\\ a_n=71 \end{cases} \\\\\\ S_{16}=\cfrac{16(11+71)}{2}\implies S_{16}=8(82)\implies S_{16}=656](https://tex.z-dn.net/?f=%20%5Cbf%20%5Ctextit%7Bsum%20of%20a%20finite%20arithmetic%20sequence%7D%20%5C%5C%5C%5C%20S_n%3D%5Ccfrac%7Bn%28a_1%2Ba_n%29%7D%7B2%7D%5Cqquad%20%20%5Cbegin%7Bcases%7D%20n%3Dn%5E%7Bth%7D%5C%20term%5C%5C%20a_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5C%20----------%5C%5C%20n%3D16%5C%5C%20a_1%3D11%5C%5C%20a_n%3D71%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20S_%7B16%7D%3D%5Ccfrac%7B16%2811%2B71%29%7D%7B2%7D%5Cimplies%20S_%7B16%7D%3D8%2882%29%5Cimplies%20S_%7B16%7D%3D656%20)
There are three groups of people: 25 years old and younger, between 25 to 50, and 50 years old and older. Their fractions must equal to 1 because these three together form the whole.
1 = 1/3 + 2/7 + x
where x is the fraction for people ages 25 to 50 years old
x = 8/21
So, the actual number of people ages 25 to 50 years old is:
84(8/21) = 32 people