Answer:
Each portable fire extinguisher should only be used for its specific type of fire. Class A extinguishers could cause a Class B fire to spread or electrocution in a Class C fire. A Class B extinguisher could fail to completely extinguish a Class A fire, causing the flame to re-ignite later.
Answer:
some connecting rods have spit holes
Answer:
Δr=20.45 %
Explanation:
Given that
Rake angle α = 15°
coefficient of friction ,μ = 0.15
The friction angle β
tanβ = μ
tanβ = 0.15
β=8.83°
2φ + β - α = 90°
φ=Shear angle
2φ + 8.833° - 15° = 90°
φ = 48.08°
Chip thickness r given as


r=0.88
New coefficient of friction ,μ' = 0.3
tanβ' = μ'
tanβ' = 0.3
β'=16.69°
2φ' + β' - α = 90°
φ'=Shear angle
2φ' + 16.69° - 25° = 90°
φ' = 49.15°
Chip thickness r' given as


r'=0.70
Percentage change


Δr=20.45 %
Answer:
a) Sample size = 1691
b) 95% Confidence Interval = (0.3696, 0.4304)
Explanation:
(a) How large a sample n should they take to estimate p with 2% margin of error and 90% confidence?
The margin of error is given by

Where z is the corresponding z-score for 90% confidence level
z = 1.645 (from z-table)
for p = 0.50 and 2% margin of error, the required sample size would be

(b) The advocacy group took a random sample of 1000 consumers who recently purchased this mobile phone and found that 400 were happy with their purchase. Find a 95% confidence interval for p.
The sample proportion is
p = 400/1000
p = 0.40
z = 1.96 (from z-table)
n = 1000
The confidence interval is given by

Therefore, we are 95% confident that the proportion of consumers who bought the newest generation of mobile phone were happy with their purchase is within the range of (0.3696, 0.4304)
What is Confidence Interval?
The confidence interval represents an interval that we can guarantee that the target variable will be within this interval for a given confidence level.
The total evaporation loss of water is 87.873 ×
lbm /day.
<u>Explanation:</u>
Assume A is the water and B is air.
A is diffusing to non diffusing B.

By the table 6.2 - 1 at 25°C, the diffusivity of air and water system is
.
Total pressure P = 1 atm = 101.325 KPa
= 23.76 mm Hg
=
= 0.03126 atm
= 3.167 K Pa
When air surrounded is dry air, then
= 0 mm Hg
R = 8.314 
= 

= 101.325 - 3.167
= 98.158 K Pa

= 101.325 - 0
= 101.325 K Pa

= 99.733 K Pa
Z = 1 ft = 0.3048 m
T = 298 K
= 
= 0.11077 ×
mol/
.s
= 0.11077 ×
× 18 × (60×60×24)
= 0.1723
/
.day

Area of individual pipe is

A = 0.00051 
= 0.1723 × 0.00051
= 0.000087873 lbm/day
In 1000 ft length of ditch,there will be a 10 pipes. The amount of evaporation water is
= 10 × 0.000087873 = 0.00087873 lbm /day
The total evaporation loss of water is 87.873 ×
lbm /day.