Answer:
Step-by-step explanation:
The graph can take on three forms as displayed in the figure.
(a) having no intersections
(b) having a single point of intersection (tangency)
(c) having two points of intersection.
Consequently, the maximum number of intersections that these graphs can yield is 2, as illustrated in figure (c).
Response:
2/9 = 0.22
Clarification:
There are two ways to select the first number that is odd and less than 5: 1 and 3.
For each of these, the second number drawn can be any of the values from 1 to 9, giving us a total of 18 options.
Out of these, the only pairs that result in a sum less than 5 are (1,1), (1,2), (1,3), and (3,1). Thus we have 4 combinations from the total of 18:
4/18 = 2/9 = 0.22
Answer:
I believe it's 2.
Step-by-step explanation:
10, and -10
1.4×5=7
0.8×10=8
1.4×10=14
1×15=15
15+14+8+7=44
44÷4=11
LQ of 44=11
LQ=10 minutes
11×3=33
UQ= 29 minutes
The Range is 19 minutes
Detailed breakdown:
Commence with the individual boxes. For determining the number of students in each category, calculate Frequency density × The difference in the category. (if it's 5-15, the difference is 10)
This results in the counts of students in each range.
Next, determine the LQ of 44, which is 11.
Then locate the 11th student's score; in this instance, it resides in the 5-15 range. 7 students have already surpassed it, with 8 in the 5-15 range. Hence, the 11th lies within the bounds of 5-15, making the middle 10.
Repeat this process for the UQ.
The interquartile range is calculated as UQ-LQ, yielding 29-10=19 minutes.
I hope this helps, though I'm not entirely sure if my explanation is coherent and I'm unclear on the terminology I've used for these categories.
