Answer:
30.128 days
Explanation:
Given that:
Mean = 25
Standard deviation = 4
Confidence interval = 90% = 0.9
Since the confidence interval should not exceed 90%
Then using z test table
P(z) = 0.9
For 0.8997 , we get = 1.28
For 0.9015, we get = 1.29
∴
By solving
Z = 1.282
Thus, the duration to be used so that it will not exceed 90% C.I is:
Z = (x - μ)/σ
1.282= ( x - 25)/4
1.282 * 4 = x - 25
(1.282*4)+25 = x
x = 30.128 days
Answer:
thoroughly scrutinizing, especially in a disconcerting way.
Explanation:
Answer:
a) 0.01
b) 150 cm^3/s
c) 300 cm^3/s
d) 25 cm^3/s
Explanation:
a) We know that :
Q=ΔP/R
R=8ηl/π*r^4
Givens:
r^2 = 0.1 r_1
Plugging known information to get :
Q=ΔP/R
=ΔP*π*r^4/8*η*l
Q_2/r_2^4 =Q_1/r_1^4
Q_2=Q_1/r_1^4*r_2^4
=Q_1/r_1^4*r*0.0001*r_1^4
Q_2 = 0.01
b) From the rate flow of the fluid we know that :
Q=ΔP/R (1)
F=η*Av/l (2)
R=8*ηl/π*r^4 (3)
<em>Where: </em>
ΔP is the change in the pressure .
r is the raduis of the tube .
l is the length of the tube .
η is the coefficient of the vescosity of the fluid .
R is the resistance of the fluid .
Givens: Q1 = 100 cm^3/s , ΔP= 1.5
Plugging known information into EQ.1 :
Q=ΔP/R
Q_2/ΔP2=Q_1/ΔP
Q_2=150 cm^3/s
c) we know that :
F = η*Av/l
can be written as :
ΔP = F/A = η*v/l
Givens: η_2 = 3η_1
Q=ΔP/R
Q=η*v/l*R
Q_2/η_2=Q_1/η_1
Q_2=300 cm^3/s
d) We know that :
Q=ΔP/R
R=8*ηl/π*r^4
Givens: l_2 = 4*l_1
Plugging known information to get :
Q=ΔP/R
Q=ΔP*π*r^4/8*ηl
Q_2/l_2=Q_1/l_1
Q_2 = 25 cm^3/s