Answer:

Explanation:
Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:
![Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]](https://tex.z-dn.net/?f=Q_%7Bwater%7D%20%3D%20%281.4%5C%2CL%29%5Ccdot%28%5Cfrac%7B1%5C%2Cm%5E%7B3%7D%7D%7B1000%5C%2CL%7D%20%29%5Ccdot%20%281000%5C%2C%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%7D%20%29%5Ccdot%20%5B%284.187%5C%2C%5Cfrac%7BkJ%7D%7Bkg%5Ccdot%20%5E%7B%5Ctextdegree%7DC%7D%20%29%5Ccdot%20%28100%5E%7B%5Ctextdegree%7DC-25%5E%7B%5Ctextdegree%7DC%29%2B2257%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D%5D)

The heat liberated by the LP gas is:


A kilogram of LP gas has a minimum combustion power of
. Then, the required mass is:


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