From the box plot, it can be seen that for grade 7 students,
The least value is 72 and the highest value is 91. The lower and the upper quartiles are 78 and 88 respectively while the median is 84.
Thus, interquatile range of <span>the resting pulse rate of grade 7 students is upper quatile - lower quartle = 88 - 78 = 10
</span>Similarly, from the box plot, it can be seen that for grade 8 students,
The
least value is 76 and the highest value is 97. The lower and the upper
quartiles are 85 and 94 respectively while the median is 89.
Thus, interquatile range of the resting pulse rate of grade 8 students is upper quatile - lower quartle = 94 - 85 = 9
The difference of the medians <span>of the resting pulse rate of grade 7 students and grade 8 students is 89 - 84 = 5
Therefore, t</span><span>he difference of the medians is about half of the interquartile range of either data set.</span>
Step-by-step explanation:
1 pound = 16 ounces
5 pound = 16 x 5 = 80
Cost per ounce = 15
Cost of 80 ounces = 15 x 80 = 1200
Option D is the correct answer
This problem can be solved directly by neglecting the
initial velocity given. All we have to do is to plug in t = 4 in the given
equation. That is:
h = -16^2 + 123 * 4
h = 236 ft
Given:
Q = 0.0764 kJ = 764 J, heat input
W = -830 J, useful work done
From the 1st Law of Thermodynamics, the change in internal energy is
δE = Q - W
= 764 - (-830)
= 1594 J = 1.594 kJ
Answer: 1.59 kJ (two sig. fig.)
