Answer:
The couples are not all on one axis or plane for that matter but if the A and B connector had to be specified it would go by the yz axis diagonal to the x axis with a magnitude of about 15. The direction of the axis would be pointed up to the second quadrant. Hope this was helpful
Explanation:
Answer:
maybe it's twang because of the blade tension
Answer: B, repetitive practice! hope this helps. :)
Explanation:
(a) If a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.
(b) For a kitten to be at 90th percentile, the minimum weight is 146.45 g.
<h3>
Weight distribution of the kitten</h3>
In a normal distribution curve;
- 2 standard deviation (2d) below the mean (M), (M - 2d) is at 2%
- 1 standard deviation (d) below the mean (M), (M - d) is at 16 %
- 1 standard deviation (d) above the mean (M), (M + d) is at 84%
- 2 standard deviation (2d) above the mean (M), (M + 2d) is at 98%
M - 2d = 125 g - 2(15g) = 95 g
M - d = 125 g - 15 g = 110 g
95 g is at 2% and 110 g is at 16%
(16% - 2%) = 14%
(110 - 95) = 15 g
14% / 15g = 0.93%/g
From 95 g to 99 g:
99 g - 95 g = 4 g
4g x 0.93%/g = 3.72%
99 g will be at:
(2% + 3.72%) = 5.72%
Thus, if a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.
<h3>Weight of the kitten in the 90th percentile</h3>
M + d = 125 + 15 = 140 g (at 84%)
M + 2d = 125 + 2(15) = 155 g ( at 98%)
155 g - 140 g = 15 g
14% / 15g = 0.93%/g
84% + x(0.93%/g) = 90%
84 + 0.93x = 90
0.93x = 6
x = 6.45 g
weight of a kitten in 90th percentile = 140 g + 6.45 g = 146.45 g
Thus, for a kitten to be at 90th percentile, the approximate weight is 146.45 g
Learn more about standard deviation here: brainly.com/question/475676
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Answer:
(a) We are asked to compute the Brinell hardness for the given indentation. for HB, where P= 1000 kg, d= 2.3 mm, and D= 10 mm.
Thus, the Brinell hardness is computed as
![=2*1000hg/\pi (10mm)[10mm-\sqrt{(1000^2-(2.3mm)^2} ]](https://tex.z-dn.net/?f=%3D2%2A1000hg%2F%5Cpi%20%2810mm%29%5B10mm-%5Csqrt%7B%281000%5E2-%282.3mm%29%5E2%7D%20%5D)
(b) This part of the problem calls for us to determine the indentation diameter d which will yield a 270 HB when P= 500 kg.
![d=\sqrt{D^2-[D-\frac{2P}{(HB)\pi D} } ]^2\\=\sqrt{(10mm)^2-[10mm-\frac{2*500}{450( \pi10mm)} } ]^2](https://tex.z-dn.net/?f=d%3D%5Csqrt%7BD%5E2-%5BD-%5Cfrac%7B2P%7D%7B%28HB%29%5Cpi%20D%7D%20%7D%20%5D%5E2%5C%5C%3D%5Csqrt%7B%2810mm%29%5E2-%5B10mm-%5Cfrac%7B2%2A500%7D%7B450%28%20%5Cpi10mm%29%7D%20%7D%20%5D%5E2)
