Answer:
7.7 kN
Explanation:
The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.
It can be expressed by using the formula:
where;
fracture toughness K = 137 MPa
geometry factor Y = 1
applied stress 
= ???
crack length a = 2mm = 0.002
∴
Now, the tensile impact obtained is:
P = A × σ
P = 1728.289 × 4.5
P = 7777.30 N
P = 7.7 kN
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average speed (in km/h) of a car stuck in traffic that drives 12 kilometers in 2 hours.
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
This statement is b which is true: hope this helped
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