You just have to combine like terms here
6+7n^3 is the final answer
first off, let's notice the parabola is a vertical one, therefore the squared variable is the x, and the parabola is opening upwards, meaning the coefficient of x² is positive.
let's notice the vertex, or U-turn, is at (-2, 2)
![\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \boxed{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{-2}{ h},\stackrel{2}{ k}) \\\\\\ y=+1[x-(-2)]^2+2\implies y=(x+2)^2+2](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~%5Ctextit%7Bparabola%20vertex%20form%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5Cboxed%7By%3Da%28x-%20h%29%5E2%2B%20k%7D%5C%5C%5C%5C%20x%3Da%28y-%20k%29%5E2%2B%20h%20%5Cend%7Barray%7D%20%5Cqquad%5Cqquad%20vertex~~%28%5Cstackrel%7B-2%7D%7B%20h%7D%2C%5Cstackrel%7B2%7D%7B%20k%7D%29%20%5C%5C%5C%5C%5C%5C%20y%3D%2B1%5Bx-%28-2%29%5D%5E2%2B2%5Cimplies%20y%3D%28x%2B2%29%5E2%2B2%20)
Answer: The correct answer is 18ft
Step-by-step explanation:
Answer:
a. Chi-square test of independence
Step-by-step explanation:
The chi square statistics is also used to test the hypothesis about the independence of two variables each of which is classified into a number of categories or attributes.
In the given problem the Equal , more or less are the attributes.
The goodness of fit test is applicable when the cell probabilities depend upon the unknown parameters provided that the unknown parameters are replaced with their estimates and provided that one degree of freedom is deducted for each parameter estimated.
can you send me the actual problem so that i can solve it
