A(w) = -w² + 100w
Step-by-step explanation: Initially, we need to express the length (l) of the rectangular area in terms of the width (w). Given that the total perimeter of the rectangle is 200 feet implies that 2(w + l) = 200, leading to l = 100 - w. Hence, the area A can be given as width multiplied by length: A = w * l = w * (100 - w) = -w² + 100w. Consequently, A(w) = -w² + 100w.
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.
Answer:
Step-by-step breakdown:
You may consider a structure like the following while keeping in mind your specific notation.
1. | ~( (~Q ->~R) v (R & ~Q) ) Assume
2. | | ~(~Q ->~R) Assume
3. | | | ~(R & ~Q) Assume
4. | | | ~R v Q 3, De Morgan
5. | | | ~Q -> ~R 4 Material implication
6. | | | # 2, 5 Results in a contradiction
7. | | R & ~Q 3-6 indirect proof outcome
8. | ~(~Q ->~R) -> (R & ~Q ) 2-7 Discharging the conditional proof
9. | (~Q ->~R) v (R & ~Q) 8 Material implication derived
10 | # 1,9 Results in contradiction
11. (~Q ->~R) v (R & ~Q) 1-10 Completion via indirect proof
