Answer:
Option C) 16(0.5x - 0.75y + 2)
Step-by-step explanation:
16(0.5x - 0.75y + 2) = 16*0.5x - 0.75y*16 + 2 *16
= 8.0x - 12.0y + 32
= 8x - 12y + 32
13x - 21x - 7 = 63
- 8x = 70
x = - 8.75
Answer:
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Answer:
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Step-by-step explanation:
Answer:
the probability that system’s failure is due to the radio = ![\dfrac{1}{3}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B3%7D)
Step-by-step explanation:
From the question given;
Let the mean lifetime of the radio
and the mean lifetime of the speaker ![\dfrac{1}{\lambda_2} = 500](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5Clambda_2%7D%20%3D%20500)
we can re-write both expressions as:
and ![\lambda_2= \dfrac{1}{500}](https://tex.z-dn.net/?f=%5Clambda_2%3D%20%5Cdfrac%7B1%7D%7B500%7D)
Let consider
to be the variables which are independent to the exponentially distributed mean of ![\dfrac{1}{\lambda _1} \ and \ \dfrac{1}{\lambda _2}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5Clambda%20_1%7D%20%5C%20and%20%5C%20%5Cdfrac%7B1%7D%7B%5Clambda%20_2%7D)
∴
![P(X_1< X_2) = \int ^{\infty}_{0} \ P (X_1](https://tex.z-dn.net/?f=P%28X_1%3C%20X_2%29%20%3D%20%5Cint%20%5E%7B%5Cinfty%7D_%7B0%7D%20%5C%20P%20%28X_1%3CX_2%7CX_1%20%3Dx%29%20%5Clambda_1%20e%5E%7B-%5Clambda_1%20%5C%20x%7D%20%5C%20dx)
![P(X_1< X_2) = \int ^{\infty}_{0} \ P (X_1](https://tex.z-dn.net/?f=P%28X_1%3C%20X_2%29%20%3D%20%5Cint%20%5E%7B%5Cinfty%7D_%7B0%7D%20%5C%20P%20%28X_1%3CX_2%29%20%5Clambda_1%20e%5E%7B-%5Clambda_1%20%5C%20x%7D%20%5C%20dx)
![P(X_1< X_2) = \int ^{\infty}_{0} \ e^{-\lambda_2 \ x} \lambda _1 \ e^{-\lambda_1 \ x} \ dx](https://tex.z-dn.net/?f=P%28X_1%3C%20X_2%29%20%3D%20%5Cint%20%5E%7B%5Cinfty%7D_%7B0%7D%20%5C%20e%5E%7B-%5Clambda_2%20%5C%20x%7D%20%5Clambda%20_1%20%5C%20e%5E%7B-%5Clambda_1%20%5C%20x%7D%20%20%5C%20dx)
![P(X_1< X_2) = \int ^{\infty}_{0} \lambda _1 \ e^{(-\lambda_1 +\lambda_2)} \ dx](https://tex.z-dn.net/?f=P%28X_1%3C%20X_2%29%20%3D%20%5Cint%20%5E%7B%5Cinfty%7D_%7B0%7D%20%5Clambda%20_1%20%5C%20e%5E%7B%28-%5Clambda_1%20%2B%5Clambda_2%29%7D%20%20%5C%20dx)
![P(X_1< X_2) = \dfrac{\lambda_1}{\lambda_1 + \lambda_2}](https://tex.z-dn.net/?f=P%28X_1%3C%20X_2%29%20%3D%20%5Cdfrac%7B%5Clambda_1%7D%7B%5Clambda_1%20%2B%20%5Clambda_2%7D)
replace the values now; we have:
![P(X_1< X_2) = \dfrac{\dfrac{1}{1000} }{\dfrac{1}{1000} + \dfrac{1}{500}}](https://tex.z-dn.net/?f=P%28X_1%3C%20X_2%29%20%3D%20%5Cdfrac%7B%5Cdfrac%7B1%7D%7B1000%7D%20%7D%7B%5Cdfrac%7B1%7D%7B1000%7D%20%2B%20%5Cdfrac%7B1%7D%7B500%7D%7D)
![P(X_1< X_2) = \dfrac{\dfrac{1}{1000} }{\dfrac{1+2}{1000} }](https://tex.z-dn.net/?f=P%28X_1%3C%20X_2%29%20%3D%20%5Cdfrac%7B%5Cdfrac%7B1%7D%7B1000%7D%20%7D%7B%5Cdfrac%7B1%2B2%7D%7B1000%7D%20%7D)
![P(X_1< X_2) = \dfrac{\dfrac{1}{1000} }{\dfrac{3}{1000} }](https://tex.z-dn.net/?f=P%28X_1%3C%20X_2%29%20%3D%20%5Cdfrac%7B%5Cdfrac%7B1%7D%7B1000%7D%20%7D%7B%5Cdfrac%7B3%7D%7B1000%7D%20%7D)
![P(X_1< X_2) = {\dfrac{1}{1000} } \times {\dfrac{1000}{3}](https://tex.z-dn.net/?f=P%28X_1%3C%20X_2%29%20%3D%20%7B%5Cdfrac%7B1%7D%7B1000%7D%20%7D%20%5Ctimes%20%7B%5Cdfrac%7B1000%7D%7B3%7D)
![P(X_1< X_2) = {\dfrac{1}{3} }](https://tex.z-dn.net/?f=P%28X_1%3C%20X_2%29%20%3D%20%7B%5Cdfrac%7B1%7D%7B3%7D%20%7D)
Thus, the probability that system’s failure is due to the radio = ![\dfrac{1}{3}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B3%7D)