It takes 5.22 hours to install carpet in a room. Step-by-step explanation: To start, let's explore the relationship among yards, feet, and inches. 1. 1 yard = 3 feet. 2. 1 foot = 12 inches. 3. Therefore, 1 yard = 36 inches. Now, let's convert all measurements to inches. ∵ 1 yard = 36 inches ∴ 1 yard² = (36)² inches². - Fine Floors can outfit 15 square yards of carpeting in 4 hours and 30 minutes. ∵ 15 yard² equals 15 × (36)² inches², so 15 yards² = 19440 inches². ∵ 1 hour equals 60 minutes, thus 30 minutes = 0.5 hours. Consequently, 4 hours and 30 minutes = 4.5 hours, implying Fine Floors covers 19440 inches² in 4.5 hours. The dimensions of the room are 11 feet 9 inches by 13 feet 4 inches. ∵ 1 foot = 12 inches, therefore, 11 feet 9 inches = 11 × 12 + 9 = 141 inches, and 13 feet 4 inches = 13 × 12 + 4 = 160 inches. Hence, the room’s area is 141 × 160 = 22560 inches². Using the ratio method, we have area of carpeting (in²): time of carpeting (hr) hence, 19440: 4.5 = 22560: h. By cross-multiplying, 19440h = 4.5 × 22560 yields 19440h = 101520. Dividing both sides by 19440 results in h = 5.22 hrs.
Answer:
Step-by-step explanation:
It has been established that the count of drivers traveling between a specific origin and destination in a certain time frame follows a Poisson distribution with a mean μ = 20 (as indicated in the article "Dynamic Ride Sharing: Theory and Practice"†).
a) 
b) 
c) 
d) 2 standard deviations = 2(20) = 40
Thus, this means the range for 2 standard deviations is
20-40, 20+40
which equates to (0,60)
The result is 3.6y. By multiplying 0.3 by 12, we arrive at 3.6, and we include the variable y.
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.
