Answer:
1. Scatter plots are used to observe relationships between variables. The position of each dot on the horizontal and vertical axis indicates values for an individual data point.
2. A Scatter Analysis is used when you need to compare two data sets against each other to see if there is a relationship. Scatter plots are a way of visualizing the relationship; by plotting the data points you get a scattering of points on a graph
Step-by-step explanation:
I am leaving the last two for someone else to have a chance to answer.
I hope my answers helped.
Answer:
4523.89m
So for a cylinder it is radius times two. Then multiply your answer by pie 3.14 then times high.
the parabola has maximum at 9, meaning is a vertical parabola and it opens downwards.
it has a symmetry at x = -5, namely its vertex's x-coordinate is -5.
check the picture below.
so then, we can pretty much tell its vertex is at (-5 , 9), and we also know it passes through (-7, 1)
![\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} y=a(x- h)^2+ k\qquad \leftarrow \textit{using this one}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=-5\\ k=9 \end{cases}\implies y=a[x-(-5)]^2+9\implies y=a(x+5)^2+9](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Bparabola%20vertex%20form%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20y%3Da%28x-%20h%29%5E2%2B%20k%5Cqquad%20%5Cleftarrow%20%5Ctextit%7Busing%20this%20one%7D%5C%5C%5C%5C%20x%3Da%28y-%20k%29%5E2%2B%20h%20%5Cend%7Barray%7D%20%5Cqquad%5Cqquad%20vertex~~%28%5Cstackrel%7B%7D%7B%20h%7D%2C%5Cstackrel%7B%7D%7B%20k%7D%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20h%3D-5%5C%5C%20k%3D9%20%5Cend%7Bcases%7D%5Cimplies%20y%3Da%5Bx-%28-5%29%5D%5E2%2B9%5Cimplies%20y%3Da%28x%2B5%29%5E2%2B9)
![\bf \textit{we also know that } \begin{cases} x=-7\\ y=1 \end{cases}\implies 1=a(-7+5)^2+9 \\\\\\ -8=a(-2)^2\implies -8=4a\implies \cfrac{-8}{4}=a\implies -2=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y=-2(x+5)^2+9~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bwe%20also%20know%20that%20%7D%20%5Cbegin%7Bcases%7D%20x%3D-7%5C%5C%20y%3D1%20%5Cend%7Bcases%7D%5Cimplies%201%3Da%28-7%2B5%29%5E2%2B9%20%5C%5C%5C%5C%5C%5C%20-8%3Da%28-2%29%5E2%5Cimplies%20-8%3D4a%5Cimplies%20%5Ccfrac%7B-8%7D%7B4%7D%3Da%5Cimplies%20-2%3Da%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20y%3D-2%28x%2B5%29%5E2%2B9~%5Chfill)
The second one is the function
The difference between 1/2 and 1/6 is 2/6. So to write an expression we could do x divided by 2/6=2/3. So instead of a division expression, we could make that into a multiplication problem 2/6 times 2/3=x. 2/6 times 2/3= 4/18 so your answer is 4/18 or 2/9
