Answer:
1.1m
Step-by-step explanation:
Let L denote longer pipe and S denote shorter part
given:
L:S = 9:2
or L/S = 9/2
2L = 9S
S = (2/9)L-----(eq 1)
Also given:
L - 1.65 = 3S
L - 3S = 1.65 ----- (eq 2)
By substitution, sub eq 1 into eq 2
L - 3(2/9)L = 1.65
L - (2/3)L = 1.65
L [ 1 - (2/3) ] = 1.65
L/3 = 1.65
L = 1.65 (3) = 4.95 (substitute into equation 1)
S = (2/9)(4.95) = 1.1m
The factors of 32 are 1, 2, 4, 8, 16, and 32.
The factors of 20 are 1, 2, 4, 5, 10, and 20.
The common factors of 32 and 20 are 1, 2, and 4.
The greatest one is 4.
32 = 4 x 8
20 = 4 x 5
32 + 20 = (4 x 8) + (4 x 5)
32 + 20 = <em>4 (8 + 5)</em>
Answer: Probability of trucks having both defects is 4%.
Step-by-step explanation:
Since we have given that
Probability of trucks having defective brake linings only = 6%
Probability of trucks having defective steering only = 3%
Probability of trucks having neither defect = 87%
Probability of trucks having either defect - 100 - 87 = 13%
Let the probability of trucks having both defects be 'x'.
P(defective brake lining) = P(B) = 0.06+x
P(defective steering) = P(S) = 0.03+x
So, Probability of the trucks having both defects is given by

So, Probability of trucks having both defects is 4%.
Answer:
For cobalt-60, which has a half-life of 5.27 years, 50% remains after 5.27 years (one half-life), 25% remains after 10.54 years (two half-lives), 12.5% remains after 15.81 years (three half-lives), and so on.
Step-by-step explanation:
You can figue out the rest.