There are 3 methods to solve this. elimination substitution and graphing but i am going to use the elimination method.
x+2y=17
<u> x-y =2
</u> 0+3y =15 ( subtracted down to eliminate the x)( x-x=0, 2y-(-y)=3y, and 17-2=15)
3y=15 (divide both sides by 3 to solve for y)
y=5
<u>substitute the y=5 in any of the above equations and solve for x
ie... ( </u>meaning where you find y in the equation, u replace it with a 5)
it will be easier to solve for x in (x-y=2) so i will use that one.
x-(5)=2 ( add 5 on both sides to solve for x)
x=7
You can convert them to decimals and multiply. Remember when using decimals, it is over 10 or 100 <span />
If the co-vertices are (0, 3) and (0, -3) where x is 0 and y has a value, then y is the minor axis. That means that the x axis is the major axis. Because of what the co-vertices are, the center of the ellipse is at the origin. The formula for an ellipse that has a horizontal major axis is
. The a value will always be larger than the b value, therefore, the a value goes under the coordinate that is the major axis. Here, its the x-axis. a is the distance that the outer edge of the ellipse is from the center. It's 8 units away from the center along the x axis and 3 units along the y axis from the center. So a = 8 and a^2 = 64; b = 3 and b^2 = 9. Our formula then is
The Rome data center is best described by the mean. The New York data center is best described by the median. The third option C is correct.
<h3>The Mean and Median:</h3>
The mean of a data set is the average of all the terms in the data set. The median of a data set is the value of the midpoint term in the frequency distribution.
From the given information, the table can be better expressed as:
High Low Q1 Q3 IQR Median Mean σ
Rome 18 1 3 7 4 6.5 6.4 4.3
NY 14 1 4.5 8.5 4 5.5 6.1 3.2
- From the data sets in the table, the distribution for Rome is not largely diverse, and there isn't much departure from the mean value. It indicates that in the data set of Rome families, no outliers have occurred.
- In New York, the data indicate a distinct outlier for New York families in Q3. This is due to the fact that the gap is so large, the mean may not be a good choice for determining the measure of the central tendency.
Therefore, we can conclude that, the Rome data center is best described by the mean and the median will be utilized to determine the central tendency in New York.
Learn more about mean and median here:
brainly.com/question/14532771
