Room 20 ft. long; ceiling 15 ft (cathedral), regular height 8 feet; triangle 7 feet high; window 15 feet x 3 feet; half moon window 10 foot diameter.

 The area of the walls and ceiling totals 898.53 ft²  If assumptions about the width being calculated from the positioning of the half moon window are correct.

PS This room will not pass inspection: there is no exit!. A standard door is required. (approximately 22 ft²)

Step-by-step explanation:

If the width of the wall at the ends of the room is 15 ft, (the base of the triangle)  sides of the triangle supporting the cathedral ceiling will be 10.25 ft

If the width of the wall supporting the cathedral ceiling is 12 ft, the sides supporting the ceiling will be about 9.22 ft, and is a better fit for the half moon window. (about 4 inches of clearance at nearest point)

The area of those triangular ends totals 84 ft² Subtract the area of the half-moon window,  (π5²/2 ≈ 39.270  leaving 44.73 ft²

The ceiling will be two sections 20× 9.22 , 184.4 ×2 = 386.8 ft²

End walls below triangles: 12×8×2 = 192 ft²

Long walls: 20×8×2 = 320 ft² Subtract the  15×3 window 320-45= 275 ft²

answer:

1. 0.04

2. 0.15

step-by-step explanation:

i just finished that assignment for properties of probability distributions.

isi

step-by-step explanation:

use that smart brain