Answer:
rhombus
Step-by-step explanation:
A quadrilateral with equal-length sides is a <em>rhombus</em>.
_____
A square is a special case of a rhombus.
Answer:
Explain the circumstances for which the interquartile range is the preferred measure of dispersion
Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.
What is an advantage that the standard deviation has over the interquartile range?
The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.
Step-by-step explanation:
Previous concepts
The interquartile range is defined as the difference between the upper quartile and the first quartile and is a measure of dispersion for a dataset.

The standard deviation is a measure of dispersion obatined from the sample variance and is given by:

Solution to the problem
Explain the circumstances for which the interquartile range is the preferred measure of dispersion
Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.
What is an advantage that the standard deviation has over the interquartile range?
The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.
Answer:
9/25 = 36%
Step-by-step explanation:
First add up all of the students (150)
Then divide 54 by 150, 0.36
0.36 = 36% = 9/25
To find the 'maximum height' we will need to take the derivative of h(t) = –16t² + 32t + 6 then set it equal to zero, then solve for t. this t will be the time at which the ball reaches it's maximum height.
<h3>Given</h3>
A(-3, 1), B(4, 5)
<h3>Find</h3>
coordinates of P on AB such that AP/PB = 5/2
<h3>Solution</h3>
AP/PB = 5/2 . . . . . desired result
2AP = 5PB . . . . . . multiply by 2PB
2(P-A) = 5(B-P) . . . meaning of the above
2P -2A = 5B -5P . . eliminate parentheses
7P = 2A +5B . . . . . collect P terms
P = (2A +5B)/7 . . . .divide by the coefficient of P
P = (2(-3, 1) +5(4, 5))/7 . . . . substitute the given points
P = (-6+20, 2+25)/7 . . . . . . simplify
P = (2, 3 6/7)