Answer: A large Mean Absolute Deviation (MAD) tells you the values are spread out far from the mean.
In a set of data, we can add up all the values and divide by the total number of values. This will give you the mean. It tells you what the "center" of the data set is.
Now, you can take the difference of each individual value from the mean. If you take the absolute value of that number and find the mean, you will have the M.A.D. of the set of number.
The larger the MAD, the further away the values are from the mean.
Answer:
2
Step-by-step explanation:
add them aall up
Answer:
(x+1)²/6 - (y+3)²/12 = 1
Step-by-step explanation:
The standard form of writing the equation of an hyperbola is expressed as;
(x-h)²/a² - (y-b)²/b² = 1 where (h,k) is the centre of the hyperbola.
Given the equation:
6x²-3y²+12x-18y-3 = 0
We are to convert it to the standard form of writing the equation of a hyperbola.
Collecting the like terms will give;
(6x²+12x)-(3y²+18y)-3 = 0
Divide through by 3
(2x²+4x)-(y²+6y)-1 = 0
Completing the square of the equation in parenthesis and adding the constants to the other side of the equation:
(2x²+4x)-(y²+6y+(6/2)²)-1 = 0+(6/2)²
(2x²+4x)-(y²+6y+9)-1 = 9
2x²+4x -{(y+3)²} = 9+1
2x²+4x - (y+3)² = 10
Divide through by 2
x²+2x -(y+3)²/2 = 5
(x²+2x+(2/2)²)-(y+3)²/2 = 5+(2/2)²
(x²+2x+1) - (y+3)²/2 = 5+1
(x+1)²-(y+3)²/2 = 6
Divide through by 6
(x+1)²/6 - (y+3)²/12 = 1
The resulting equation is the required standard form of a hyperbola with centre at (-1, -3)
