Answer:
y = 0.834X - 1.58015
Slope = 0.8340 ; Intercept = - 1.5802
y = 40.9539
19.93
0.9765
Explanation:
X: Rainfall volume
6
12
14
16
23
30
40
52
55
67
72
81
96
112
127
Y : Runoff
4
10
13
14
15
25
27
48
38
46
53
72
82
99
100
The scatterplot shows a reasonable linear trend between the Rainfall volume and run off.
The estimated regression equation obtained using a linear regression calculator is :
y = 0.834X - 1.58015
y = Runoff ; x = Rainfall volume
Slope = 0.8340 ; Intercept = - 1.5802
Point estimate for Runoff, when, x = 51
y = 0.834X - 1.58015
y = 0.834(51) - 1.58015
y = 40.95385
y = 40.9539
d.)
Point estimate for standard deviation :
s = 5.145
σ = s * √n
σ = √15 * 5.145
= 19.93
e.)
r² = Coefficient of determination gives the proportion of explained variance in Runoff due to the regression line. From the model output, the r² value = 0.9765. Which means That about 97.65% Runoff is due to Rainfall volume.
Answer: General purpose branch circuit
Explanation:
General purpose branch circuit are the type of circuits that are used mainly to supply light to two or more receptacle outlets for small appliances. This circuits are about 120v can be used either in residential, commercial and industrial buildings.
Answer:
a) 
b) 
c) 
d) 
Explanation:
Non horizontal pipe diameter, d = 25 cm = 0.25 m
Radius, r = 0.25/2 = 0.125 m
Entry temperature, T₁ = 304 + 273 = 577 K
Exit temperature, T₂ = 284 + 273 = 557 K
Ambient temperature, 
Pipe length, L = 10 m
Area, A = 2πrL
A = 2π * 0.125 * 10
A = 7.855 m²
Mass flow rate,
Rate of heat transfer,
a) To calculate the convection coefficient relationship for heat transfer by convection:
Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.
c) The surface temperature of the pipe:
Smear coefficient of the pipe, 
b) Heat loss from the pipe to the environment:
d) The required fan control power is 25.125 W as calculated earlier above
Answer: True
Explanation:
Engineering stress is the applied load divided by the original cross-sectional area of a material. It is also known as nominal stress. It can also be defined as the force per unit area of a material. Engineering Stress is usually in large numbers.
While Engineering strain is the amount that a material deforms per unit length in a tensile test. It can also be defined as extension per unit length. It has no unit as it is a ratio of lengths. Engineering Strain is in small numbers.
