How do I solve it. I am having trouble getting the answer
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1.) C - Final exams: midterm = Q3 92 - Q1 80 = 12 and finals = Q3 85 - Q1 78 = 7, so finals exam has the smaller IQR.
2.) B - exam median is much higher than the class median: refer to the attached image of class and exam box and whisker graph
3.) A - <span>IQR is a better measure of spread for movies than it is for basketball games: because the data for movies are quite wide or apart from each other than the data in the basketball games
4.) A - </span><span>mean for April is higher than October's mean: mean in April is 67 and mean in October is 60
5.) A - </span><span>Neither data set has suspected outliers: meaning there's no data that is further apart from the group or set
6.) B - </span>There is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical
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7.) C - </span>college spread is best described by the IQR. The high school spread is best described by the standard deviation: this is because of the wide range of data in the college than the data in the high school.
1.) C - Final exams: midterm = Q3 92 - Q1 80 = 12 and finals = Q3 85 - Q1 78 = 7, so finals exam has the smaller IQR.
2.) B - exam median is much higher than the class median: refer to the attached image of class and exam box and whisker graph
3.) A - <span>IQR is a better measure of spread for movies than it is for basketball games: because the data for movies are quite wide or apart from each other than the data in the basketball games
4.) A - </span><span>mean for April is higher than October's mean: mean in April is 67 and mean in October is 60
5.) A - </span><span>Neither data set has suspected outliers: meaning there's no data that is further apart from the group or set
6.) B - </span>There is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical
<span>
7.) C - </span>college spread is best described by the IQR. The high school spread is best described by the standard deviation: this is because of the wide range of data in the college than the data in the high school.
keeping in mind that the vertex is between the focus point and the directrix, in this cases we have the focus point above the directrix, meaning is a vertical parabola opening upwards, Check the picture below, which means the "x" is the squared variable.
now, the vertical distance from the focus point to the directrix is , which means the distance "p" is half that or 1/8, and is positive since it's opening upwards.
since the vertex is 1/8 above the directrix, that puts the vertex at , meaning the y-coordinate for the vertex is 2.
