Answer: the standard deviation STD of machine B is s (Lb) = 0.4557
Explanation:
from the given data, machine A and machine B produce half of the rods
Lt = 0.5La + 0.5Lb
so
s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)
but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)
so we substitute
s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)
0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)
0.64 = 0.25 + s²(Lb) + 0.4s(Lb)
s²(Lb) + 0.4s(Lb) - 0.39 = 0
s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2
s (Lb) = 0.4557
therefore the standard deviation STD of machine B is s (Lb) = 0.4557
Answer:
A pointer is spun on a fair wheel of chance having its periphery labeled Trom 0 to 100. (a) Whhat is the sample space for this experiment? (b)What is the probability that the pointer will stop between 20 and 35? (c) What is the probability that the wheel will stop on 58?
Explanation:
thats all you said
Answer:
λ^3 = 4.37
Explanation:
first let us to calculate the average density of the alloy
for simplicity of calculation assume a 100g alloy
80g --> Ag
20g --> Pd
ρ_avg= 100/(20/ρ_Pd+80/ρ_avg)
= 100*10^-3/(20/11.9*10^6+80/10.44*10^6)
= 10744.62 kg/m^3
now Ag forms FCC and Pd is the impurity in one unit cell there is 4 atoms of Ag since Pd is the impurity we can not how many atom of Pd in one unit cell let us calculate
total no of unit cell in 100g of allow = 80 g/4*107.87*1.66*10^-27
= 1.12*10^23 unit cells
mass of Pd in 1 unit cell = 20/1.12*10^23
Now,
ρ_avg= mass of unit cell/volume of unit cell
ρ_avg= (4*107.87*1.66*10^-27+20/1.12*10^23)/λ^3
λ^3 = 4.37
Answer:
T_warm = 47.22 C
Explanation:
Using energy balance for the system:
m_1*h_1 + m_2*h_2 = m_3*h3 ... Eq1
h_i = c_p. T_i ... Eq 2
m_1 + m_2 = m_3 ... steady flow system (Eq 3)
Substitute Eq 2 and Eq3 in Eq1
m_3 = 0.8 + 1 = 1.8 kg/s
(0.8)*(4.18)*( 348-273) + (1)*(4.18)*( 298-273) = 1.8 * 4.18 *T_3
T_3 = 355.3 / (1.8*4.18) = 47.22 C