Answer:
Step-by-step explanation:
Given that a teacher gives a test to a large group of students. The results are closely approximated by a normal curve
mu =74 and sigma =8
A grade starts from 100-8 = 92nd percentile
Z score for 92nd percentile = 1.405
X score = 74+8(1.405) = 85.24
--------------------
B cut off is to next 16%
Hence C would start for scores below 100-(8+16) = 76%
76th percentile = 0.705*8+74 =79.64
Step 1:
<span>Calculate the effective thermal conductivity of the wall or ceiling:
</span>
K_eff = [ (13 ÷ 8)(0.12) + (16 - (13 ÷ 8)) × (0.04)] ÷ 16
K_eff =<span> [ 0.195 + 0.565] </span>÷<span> 16
</span>
K_eff = 0.76 ÷ 16
K_eff = 0.0475 W/ (m K)
Step 2:
Calculate <span>the interior ceiling area:
</span>Area of each of the interior side walls = <span>8.82 m x 8.64 m
= 76.2 m</span>²
Area of the interior ceiling = 8.64 m × <span>8.64 m
</span> = 74.6 m²
H = - k·A·(Δ - T) ÷ <span>(thickness)
</span>
H = - 0.0475 ÷ (379.45 × 20) ÷ 45/8
H = - ( - 0.95 × 379.45 ) ÷<span> 0.1429
</span>
H = <span>2.52 kW </span>