I don't know what you have to do but I'm trying to do my first answer srry
The wall area is the product of the room perimeter and the room height:
A₁ = (2*(12.5 ft + 10.5 ft))*(8.0 ft) = 368 ft²
The window and door area together is
A₂ = 2*((4 ft)*(3 ft)) + (7 ft)*(3 ft) = 45 ft²
The area of one roll of wallpaper is
A₃ = (2.5 ft)*(30 ft) = 75 ft²
Then the number of rolls of wallpaper required will be
1.1*(A₁ - A₂)/A₃ ≈ 4.74
5 rolls of wallpaper should be purchased.
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As a practical matter, not much of the window and door area can be saved. The rolls are 30 inches wide, but the openings are 36 inches wide. Some will likely have to be cut from two strips. The strips will have to be the full length of the wall, and the amount cut likely cannot be used elsewhere. If the window and door area cannot be salvaged, then likely ceiling(5.4) = 6 rolls will be needed (still allowing 10% for matching and waste).
65 kg i think or 143 pounds
Answer:
If we solve for k we can do this:
So then we have at last 75% of the data withitn two deviations from the mean so the limits are:
Step-by-step explanation:
We don't know the distribution for the scores. But we know the following properties:
For this case we can use the Chebysev theorem who states that "At least of the values lies between and "
And we need the boundaries on which we expect at least 75% of the scores. If we use the Chebysev rule we have this:
If we solve for k we can do this:
So then we have at last 75% of the data withitn two deviations from the mean so the limits are: