Answer:
c
Step-by-step explanation:
x+x=90
2x=90 ,x=45 ,∆BAD=45
Answer:
2(x+5)*(x-9)
Step-by-step explanation:
Answer:
The weight of the water in the pool is approximately 60,000 lb·f
Step-by-step explanation:
The details of the swimming pool are;
The dimensions of the rectangular cross-section of the swimming pool = 10 feet × 20 feet
The depth of the pool = 5 feet
The density of the water in the pool = 60 pounds per cubic foot
From the question, we have;
The weight of the water in Pound force = W = The volume of water in the pool given in ft.³ × The density of water in the pool given in lb/ft.³ × Acceleration due to gravity, g
The volume of water in the pool = Cross-sectional area × Depth
∴ The volume of water in the pool = 10 ft. × 20 ft. × 5 ft. = 1,000 ft.³
Acceleration due to gravity, g ≈ 32.09 ft./s²
∴ W = 1,000 ft.³ × 60 lb/ft.³ × 32.09 ft./s² = 266,196.089 N
266,196.089 N ≈ 60,000 lb·f
The weight of the water in the pool ≈ 60,000 lb·f
Answer:
1191.4 ; 34.5
Step-by-step explanation:
Given the data:
29; 37; 38; 40; 58; 67; 68; 69; 76; 86; 87; 95; 96; 96; 99; 106; 112; 127; 145; 150
The sample variance and standard deviation can be obtained thus :
Σ(X - m)² / (n - 1)
Where, m = mean of the sample
n = sample size
The standard deviation equals, sqrt(variance )
Using a calculator:
The variance, σ² ;
Mean = Σx / n = 1681 / 20 = 84.05
(x -m)^2
[(29-84.05)^2 + (37-84.05)^2 + (38-84.05)^2 + (40-84.05)^2 + (58-84.05)^2 + (67-84.05)^2 + (68-84.05)^2 + (69-84.05)^2 + (76-84.05)^2 + (86-84.05)^2 + (87-84.05)^2 + (95-84.05)^2 + (96-84.05)^2 + (96-84.05)^2 + (99-84.05)^2 + (106-84.05)^2 + (112-84.05)^2 + (127-84.05)^2 + (145-84.05)^2 + (150-84.05)^2] / 19
22636.95 / 19
= 1191.4184 = 1191.42
Standard deviation = sqrt( Variance)
Standard deviation = sqrt(1191.4184)
Standard deviation = 34.516929 = 34.52
