What do u mean ....yhjdjennenenebbenhdjdjd
Answer: 0.55567
Step-by-step explanation:
Given : A certain brand of refrigerator has a useful life that is normally distributed with mean 10 years and standard deviation 3 years. The useful lives of these refrigerators are independent.
i.e.
![\mu=10\text{ years}\\\sigma=3\text{ years}](https://tex.z-dn.net/?f=%5Cmu%3D10%5Ctext%7B%20years%7D%5C%5C%5Csigma%3D3%5Ctext%7B%20years%7D)
Let
and
are two randomly selected refrigerator's life whose sum will exceed third selected refrigerator
.
So that, ![X =X_1+X_2-1.9\times X_3](https://tex.z-dn.net/?f=X%20%3DX_1%2BX_2-1.9%5Ctimes%20X_3)
Mean ![=E(X_1)+E(X_2)-E(X_3)=10+10-1.9\times10 =1](https://tex.z-dn.net/?f=%3DE%28X_1%29%2BE%28X_2%29-E%28X_3%29%3D10%2B10-1.9%5Ctimes10%20%3D1)
Standard deviation =![\sqrt{(Var(X_1)+Var(X_1)+Var(X_1))}](https://tex.z-dn.net/?f=%5Csqrt%7B%28Var%28X_1%29%2BVar%28X_1%29%2BVar%28X_1%29%29%7D)
![=\sqrt{(3^2+3^2+(1.9\times3)^2)}](https://tex.z-dn.net/?f=%3D%5Csqrt%7B%283%5E2%2B3%5E2%2B%281.9%5Ctimes3%29%5E2%29%7D)
![=7.1056315694](https://tex.z-dn.net/?f=%3D7.1056315694)
Z-score : ![z=\dfrac{X-E[x]}{\sqrt{Var[x]}}=\dfrac{0-1}{7.0156315694}\approx-0.14](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7BX-E%5Bx%5D%7D%7B%5Csqrt%7BVar%5Bx%5D%7D%7D%3D%5Cdfrac%7B0-1%7D%7B7.0156315694%7D%5Capprox-0.14)
Now , The probability that the total useful life of two(i..e n=2) randomly selected refrigerators will exceed 1.9 times the useful life of a third randomly selected refrigerator would be :-
![P(Z>-0.14)=P(Z](https://tex.z-dn.net/?f=P%28Z%3E-0.14%29%3DP%28Z%3C0.14%29%3D0.55567)
Hence, the required probability is 0.55567.
3^2= 9. And 9 times 0.2 is equal to C. 1.8
Answer:
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Step-by-step explanation:
Answer:
6(1-5m) = 6
−
30
m
3(4+3r) = 12
+
9
r
3(6r+8) = 18
r
+
24
4(811+ 1+2) = 3256
-(-2-n) 7) = 14
+
7
n
-6(7k+11) = −
42
k
−
66
-3(71+1) = −
216
-6(1 +116) = −
702
-10(a - 5) = −
10
a
+
50
-3(1 + 2v) = −
3
−
6
v
-4(3r+2) = −
12
r
−
8
(3 - 76)-2 = −
75
(-2018x+20) = −
2018
x
+
20
(7 + 190)-15 = 182
(x + 1)14 = 14
x
+
14