Answer:
NO
Step-by-step explanation:
To find out which observation to classify as an outlier, whether the largest or not, a very good approach or way to do this is to apply the 1.5(IQR) rule.
According to the rule, for finding the largest observation in the data that can be classified as an outlier, we would use the formula = Q3 + 1.5(IQR).
Q3 = 120
IQR = Q3 - Q1 = 120 - 95 = 25
Lets's plug these values into Q3 + 1.5(IQR)
We have,
120 + 1.5(25)
= 157.5
Since our max in the observation is given as 155, the largest observation in the data set cannot be set as an outlier because 157.5 which we got from our calculation is higher than the max value we have in the data set.
Our answer is NO.
However, the smallest observation should be set as outlier because:
Q1 - 1.5(IQR) = 95 - (1.5*25) = 57.5, which gives us an outlier that falls within our data range.
Answer:
the best predicted pulse rate of a woman who is 66 in tall = 79.78 beats per minute
Step-by-step explanation:
We are given that;
Regression line: y^ = 18.4 + 0.93x
Now we want to find out the best predicted pulse rate of a woman who is 66 in tall.
Thus;
If x = 66, then the predicted height is:y^ = 18.4 + 0.93(66)
y^ = 79.78 beats per minute
Answer:
.
Step-by-step explanation:
Let 
, 
, and 
be constants, and let 
. The equation 
represents a parabola in a plane with vertex at 
.
For example, for 
, 
, 
, and 
.
A parabola is entirely above the 
-axis only if this parabola opens upwards, with the vertex above the 
-axis.
The parabola opens upwards if and only if the leading coefficient is positive: 
.
For the vertex to be above the -axis, the 
-coordinate of that point, , must be strictly positive. Thus, 
.
Among the choices:
does not meet the requirements. Since , this parabola would open downwards, not upwards as required.
does not meet the requirements. Since 
and is negative, the vertex of this parabola would be below the -axis.- meet both requirements:
and . 
(for which 
) would touch the -axis at its vertex.
