Answer:
100 cm
Explanation:
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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
The answer is the test is being tested towards the lungs the test is done by scanning your body the tools are called “the x rat visional lock space” and the rubber tool is called a deeldo it’s purple with a pencil looking shape perfect for the body.
Answer:
1
Created on Nov 3, 2018 @author: ASLand
7import atexit
#Read, nanes of both files
Rrintll"Enter tvo files to be compared below
userliamel input ("Enter the nome of the first file: ")
userliame2 input("Enter the name of the second file: ")
ROpen each file
f1 - open(userNamel, r')
@17 f2 = opan(useriame 2, )
tread all the lines into a list
d1 f1.readlines ()
d2 f2.readlines()
re equivalent, print "Yes" else pri
oiterate, and conpare
#11
the
y
if dl == d2:
print("Yes")
atexit
elif for i in range(@, min(len (d1), len(d2))):
if di[i]!=d2[i]:
PCint("No")
print(d1[i])
pcint(d2[])
Answer: True
Explanation:
Permanent molds do not collapse, unlike expendable molds so the mold must be opened before appreciable cooling contraction occurs in order to prevent cracks from developing in the casting.
The metal casting becomes solid inside the mold after it has been poured. But during the process of manufacture, before the would cools any further, they usually remove the metal cast in order to stop excess contractions of the solid metal casting in the mold. This is done to prevent prevent cracks from developing in the casting since permanent mold do not collapse.
