Answer:
The probability that a product is defective is 0.2733.
Step-by-step explanation:
A product consists of 3 parts. If any one of the part is defective the whole product is considered as defective.
The probability of the 3 parts being defective are:
P (Part 1 is defective) = 0.05
P (part 2 is defective) = 0.10 P (part 3 is defective) = 0.15
Compute the probability that a product is defective as follows:
P (Defective product) = 1 - P (non-defective product)
= 1 - P (None of the 3 parts are defective)
= 1 - P (Part 1 not defective) × P (Part 2 not defective) × P (Part 1 not defective)
![=1-[(1-0.05)\times(1-0.10)\times (1-0.15)]\\=1-[0.95\times0.90\times0.85]\\=1-0.72675\\=0.27325\\\approx0.2733](https://tex.z-dn.net/?f=%3D1-%5B%281-0.05%29%5Ctimes%281-0.10%29%5Ctimes%20%281-0.15%29%5D%5C%5C%3D1-%5B0.95%5Ctimes0.90%5Ctimes0.85%5D%5C%5C%3D1-0.72675%5C%5C%3D0.27325%5C%5C%5Capprox0.2733)
Thus, the probability that a product is defective is 0.2733.
Answer:
221
Step-by-step explanation:
wdad
Answer:
12 muffins for $9.60
Step-by-step explanation:
Answer:
82.79MPa
Step-by-step explanation:
Minimum yield strength for AISI 1040 cold drawn steel as obtained from literature, Sᵧ = 490 MPa
Given, outer radius, r₀ = 25mm = 0.025m, thickness = 6mm = 0.006m, internal radius, rᵢ = 19mm = 0.019m,
Largest allowable stress = 0.8(-490) = -392 MPa (minus sign because of compressive nature of the stress)
The tangential stress, σₜ = - ((r₀²p₀)/(r₀² - rᵢ²))(1 + (rᵢ²/r²))
But the maximum tangential stress will occur on the internal diameter of the tube, where r = rᵢ
σₜₘₐₓ = -2(r₀²p₀)/(r₀² - rᵢ²)
p₀ = - σₜₘₐₓ(r₀² - rᵢ²)/2(r₀²) = -392(0.025² - 0.019²)/2(0.025²) = 82.79 MPa.
Hope this helps!!
