Answer: ![\bold{(1)\ \dfrac{19,683}{64}\qquad (2)\ 16}](https://tex.z-dn.net/?f=%5Cbold%7B%281%29%5C%20%5Cdfrac%7B19%2C683%7D%7B64%7D%5Cqquad%20%282%29%5C%2016%7D)
<u>Step-by-step explanation:</u>
(1) (12, 18, 27, ...)
The common ratio is:
![r=\dfrac{a_{n+1}}{a_n}\quad r =\dfrac{18}{12}=\boxed{\dfrac{3}{2}}\quad \rightarrow \quad r=\dfrac{27}{18}=\boxed{\dfrac{3}{2}}](https://tex.z-dn.net/?f=r%3D%5Cdfrac%7Ba_%7Bn%2B1%7D%7D%7Ba_n%7D%5Cquad%20r%20%3D%5Cdfrac%7B18%7D%7B12%7D%3D%5Cboxed%7B%5Cdfrac%7B3%7D%7B2%7D%7D%5Cquad%20%5Crightarrow%20%5Cquad%20r%3D%5Cdfrac%7B27%7D%7B18%7D%3D%5Cboxed%7B%5Cdfrac%7B3%7D%7B2%7D%7D)
The equation is:
![a_n=a_o(r)^{n-1}\\\\Given:a_o=12,\ r=\dfrac{3}{2}\\\\\\Equation:\\a_n =12\bigg(\dfrac{3}{2}\bigg)^{n-1}\\\\\\\\9th\ term:\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{9-1}\\\\\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{8}\\\\\\.\quad =\large\boxed{\dfrac{19643}{64}}](https://tex.z-dn.net/?f=a_n%3Da_o%28r%29%5E%7Bn-1%7D%5C%5C%5C%5CGiven%3Aa_o%3D12%2C%5C%20%20r%3D%5Cdfrac%7B3%7D%7B2%7D%5C%5C%5C%5C%5C%5CEquation%3A%5C%5Ca_n%20%3D12%5Cbigg%28%5Cdfrac%7B3%7D%7B2%7D%5Cbigg%29%5E%7Bn-1%7D%5C%5C%5C%5C%5C%5C%5C%5C9th%5C%20term%3A%5C%5Ca_9%3D12%5Cbigg%28%5Cdfrac%7B3%7D%7B2%7D%5Cbigg%29%5E%7B9-1%7D%5C%5C%5C%5C%5C%5Ca_9%3D12%5Cbigg%28%5Cdfrac%7B3%7D%7B2%7D%5Cbigg%29%5E%7B8%7D%5C%5C%5C%5C%5C%5C.%5Cquad%20%3D%5Clarge%5Cboxed%7B%5Cdfrac%7B19643%7D%7B64%7D%7D)
![(2)\qquad \bigg(\dfrac{1}{16},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{2}\bigg)\\\\\\\text{The common ratio is}:\\\\r=\dfrac{a_{n+1}}{a_n}\quad r=\dfrac{\frac{1}{8}}{\frac{1}{16}}=\boxed{2}\quad \rightarrow \quad r=\dfrac{\frac{1}{4}}{\frac{1}{8}}=\boxed{2}](https://tex.z-dn.net/?f=%282%29%5Cqquad%20%5Cbigg%28%5Cdfrac%7B1%7D%7B16%7D%2C%5Cdfrac%7B1%7D%7B8%7D%2C%5Cdfrac%7B1%7D%7B4%7D%2C%5Cdfrac%7B1%7D%7B2%7D%5Cbigg%29%5C%5C%5C%5C%5C%5C%5Ctext%7BThe%20common%20ratio%20is%7D%3A%5C%5C%5C%5Cr%3D%5Cdfrac%7Ba_%7Bn%2B1%7D%7D%7Ba_n%7D%5Cquad%20%20r%3D%5Cdfrac%7B%5Cfrac%7B1%7D%7B8%7D%7D%7B%5Cfrac%7B1%7D%7B16%7D%7D%3D%5Cboxed%7B2%7D%5Cquad%20%5Crightarrow%20%5Cquad%20r%3D%5Cdfrac%7B%5Cfrac%7B1%7D%7B4%7D%7D%7B%5Cfrac%7B1%7D%7B8%7D%7D%3D%5Cboxed%7B2%7D)
The equation is:
![a_n=a_o(r)^{n-1}\\\\Given:a_o=\dfrac{1}{16},\ r=2\\\\\\Equation:\\a_n =\dfrac{1}{16}(2)^{n-1}\\\\\\\\9th\ term:\\a_9=\dfrac{1}{16}(2)^{9-1}\\\\\\a_9=\dfrac{1}{16}(2)^{8}\\\\\\.\quad =\large\boxed{16}](https://tex.z-dn.net/?f=a_n%3Da_o%28r%29%5E%7Bn-1%7D%5C%5C%5C%5CGiven%3Aa_o%3D%5Cdfrac%7B1%7D%7B16%7D%2C%5C%20%20r%3D2%5C%5C%5C%5C%5C%5CEquation%3A%5C%5Ca_n%20%3D%5Cdfrac%7B1%7D%7B16%7D%282%29%5E%7Bn-1%7D%5C%5C%5C%5C%5C%5C%5C%5C9th%5C%20term%3A%5C%5Ca_9%3D%5Cdfrac%7B1%7D%7B16%7D%282%29%5E%7B9-1%7D%5C%5C%5C%5C%5C%5Ca_9%3D%5Cdfrac%7B1%7D%7B16%7D%282%29%5E%7B8%7D%5C%5C%5C%5C%5C%5C.%5Cquad%20%3D%5Clarge%5Cboxed%7B16%7D)
Answer: m = 5/8
Step-by-step explanation:
Answer:
import pandas as pd
vec = pd.Series([7.12,24,4,18,12,9])
vec.plot(kind = 'hist')
Step-by-step explanation:
You can use python for that.
By doing
import pandas as pd
vec = pd.Series([7.12,24,4,18,12,9])
vec.plot(kind = 'hist')
And this is the result you get
Answer:
-2
Step-by-step explanation:
rise over run
![\frac{rise (-2)}{run 1}](https://tex.z-dn.net/?f=%5Cfrac%7Brise%20%28-2%29%7D%7Brun%201%7D)
If you are given a unit rate (ex: $0.59 per ounce) then you can multiple this by the weight of the cereal to find the cost of the entire box. Do this for both boxes, and then compare the costs.
Hope this helps. I'm not sure that I'm on the right path, but I can help you further if you provide more details.
Good luck!