Answer:
Bubble sort, sometimes referred to as sinking sort, is a simple sorting algorithm that repeatedly steps through the list, compares adjacent elements and swaps them if they are in the wrong order. The pass through the list is repeated until the list is sorted.
Explanation:
Look at the photo
Good luck
Answer:
0.19s
Explanation:
Queueing delay is the time a job waits in a queue before it can be executed. it is the difference in time betwen when the packet data reaches it destination and the time when it was executed.
Queueing delay =(N-1) L /2R
where N = no of packet =93
L = size of packet = 4MB
R = bandwidth = 1.4Gbps = 1×10⁹ bps
4 MB = 4194304 Bytes
(93 - 1)4194304 / 2× 10⁹
queueing delay =192937984 ×10⁻⁹
=0.19s
(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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Explanation:
1) Work done = force x distance x cos(θ)
= 0.15 x 6 x cos(30)
= 0.779
2) Ek = ½mv²
v = acceleration due to gravity so 9.81
Ek = ½(2)(9.81)²
Ek = 96.2361
3) v = (√(2em)) / m
= (√(2(96.2361)(2)) / 2
= 9.81 so especially with no time given, I can only assume the acceleration due to gravity but take it with a pinch of salt.
