Answer:
C
Step-by-step explanation:
40x + 3
2p + 7p = 747
9p = 747 divded by 9 = 83
2 * 83 = 166 7 * 83 = 581
Hope it helps
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)
Answer:
solution is (3, - 1)
Step-by-step explanation:
given the 2 equations
x + y = 2 → (1)
x - y = 4 → (2)
Adding the 2 equations term by term will eliminate the y-term
(x + x) + (y - y) = (2 + 4)
2x = 6 ( divide both sides by 2 )
x = 3
substitute x = 3 in either (1) or (2) for y
(1) → 3 + y = 2 ( subtract 3 from both sides )
y = 2 - 3 = - 1
solution is (3, - 1)
1 is 1,530 if you need help with 2 I’ll help u with that also