Answer
given,
thickness of a flange on an aircraft component is uniformly distributed between 0.95 and 1.05 millimeters.
X = U[0.95,1.05] 0.95≤ x ≤ 1.05
the cumulative distribution function of flange
F(x) = P{X≤ x}=
=
b) P(X>1.02)= 1 - P(X≤1.02)
= 
= 0.3
c) The thickness greater than 0.96 exceeded by 90% of the flanges.
d) mean = 
= 1
variance = 
= 0.000833
Answer:
The answer is 0.97
Step-by-step explanation:
Hi, we need to use the log properties taking into account that Ln(2) is aprox. 0.69 and Ln(3) is aprox. 1.10. I think you can understand better as we solve it, so here it goes.

Since:


Then

Now, we have everything where we want it, in terms of Ln(2) and Ln(3), now it is easy to solve.


Best of luck.
Answer:
10x+20y= 350
Step-by-step explanation:
Add the decimal. 3.50
Answer:
Step-by-step explanation:
L.C.M. of 2 and 3=6
Raise to the power 6