Answer:
Average :
UCL = 4.15
LCL = 2.65
Range :
UCL = 2.75
LCL = 0
Explanation:
Given :
Sample size, n = 5
Average, X = 3.4
Range, R = 1.3
A2 for n = 5 ; equals 0.577 ( X chart table)
For the average :
Upper Control Limit (UCL) :
X + A2*R
3.4 + 0.577(1.3) = 4.1501
Lower Control Limit (LCL) :
X - A2*R
3.4 - 0.577(1.3) = 2.6499
FOR the range :
Upper Control Limit (UCL) :
UCL = D4*R
D4 for n = 5 ; equals = 2.114
UCL = 2.114*1.3 = 2.7482
Lower Control Limit (LCL) :
LCL = D3*R
D3 for n = 5 ; equals = 0
LCL = 0 * 1.3 = 0
Answer:
False
Explanation:
The formula of force that exists between two charges is expressed as;
F = kq1q2/r²
If two charges separated by one meter exert a 9 N force on each other, the;
9 = kq1q2/1²
9 = kq1q2 ..... 1
If the charges are pushed to a 3 meter separation, then;
F = kq1q2/3²
F = kq1q2/9 .... 2
Divide both equations;
9/F = (kq1q2)/ kq1q2/9
9/F = kq1q2 * 9/ kq1q2
9/F = 9
F = 9/9
F = 1N
Hence if the charges are pushed to a 3 meter separation, then the force on EACH charge will be 1N. Hence the answer is False
Answer:
the final angular velocity of the platform with its load is 1.0356 rad/s
Explanation:
Given that;
mass of circular platform m = 97.1 kg
Initial angular velocity of platform ω₀ = 1.63 rad/s
mass of banana 
= 8.97 kg
at distance r = 4/5 { radius of platform }
mass of monkey 
= 22.1 kg
at edge = R
R = 1.73 m
now since there is No external Torque
Angular momentum will be conserved, so;
mR²/2 × ω₀ = [ mR²/2 + (
R)² + R² ]w
m/2 × ω₀ = [ m/2 + ( )² + ]w
we substitute
w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1
w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )
w = 48.55 × [ 1.63 / ( 76.3908 ) ]
w = 48.55 × 0.02133
w = 1.0356 rad/s
Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s
this is evolutionary psychology
The number 3252.6 has 5 significant figures
