Answer:
Step-by-step explanation:
hello :
2x²+12x+2y²−16y−150=0
divide by 2 : (x²+6x)+(y²-8y)-75 =0
(x²+6x+9)-9+(y²-8y+16)-16-75 =0
(x+3)²+(y-4)² =10²
The coordinates of the center are (−3,4), and the length of the radius is 10 units.
Answer:
below
Step-by-step explanation:
add -36 to both sides
divide both sides by 2
This is a quadratic equation
let's solve it by graphing it
The solution are the x-coordinates of the intersection points
The x-coordinates of the intersection points are approximatively -1.6 and 5.6
The intersection of two planes is a line. :) Hope it works.
Step-by-step explanation:
q=24 m
cause its congurent
8. The mean of the grades (74) is substantially below the median (88), so the distribution is not symmetrical. Of course, half the grades are below the median, just as half are above the median. However, the mean being 88-74 = 14 points below the median means that the lower half of the grades will have an average at least twice that much, or 28 points, below the median. The distribution of grades must extend quite a bit further to the left of the median than it does to the right. Hence ...
B. The distribution is skewed left.
9. It seems likely the distribution has a number of low grades pulling the average down. There certainly exists the possibility that at least one of them qualifies as an outlier by being more than 1.5 times the IQR below the first quartile. That rule, however, only applies when the distribution is relatively symmetrical, which this one is not. There does not appear to be any recommended way to describe an outlier when the distribution is skewed and has a long tail.*
TRUE: The distribution likely has an outlier.
_____
* Math instruction these days rarely recognizes such subtleties. Since it is probable that at least one value is well below the bottom quartile, I've shown my guess at the expected answer as TRUE. In a real set of grades, I expect the tail of the distribution to have enough low grades that they cannot be considered to be outliers.
