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katen-ka-za [31]
3 years ago
6

The top 3 most popular male names of 2017 are Oliver, Declan, and Henry according to babynames. Write a program that modifies th

e male_names set by removing a name and adding a different name. Sample output with inputs: 'Oliver' 'Atlas' { 'Atlas', 'Declan', 'Henry' } NOTE: Because sets are unordered, the order in which the names in male_names appear may differ from above.
Engineering
1 answer:
Flura [38]3 years ago
7 0

Answer:

The code is given below in Python with appropriate comments

Explanation:

# convert list to set

male_names = set(['Oliver','Declan','Henry'])

# get remove and add name from user

remove_name = input("Enter remove name: ")

add_name = input("Enter add name: ")

# remove name from set

male_names.remove(remove_name)

# add new name ij set

male_names.add(add_name)

# sort the set

a = sorted(male_names)

# print the set

print(a)

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Use Routh's stability criterion to determine how many roots with positive real parts the following equations have:
Pavlova-9 [17]

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a) no roots not in LHP

b) 2 roots not in LHP

c) 2 roots not in the LHP

d) 2 roots not in the LHP

e) 2 roots not in LHP

Explanation:

a) s^4 + 8s^3 + 32s^2 + 80s + 100 = 0\\\\s^4:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:32\:\:\:\:\:\:100\\s^3:\:\:\:8\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:80\\s^2:\:\:\:22\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:100\\s^1:\:\:\:80-\frac{800}{22} =43.6\\s^0:\:\:\:100

No roots not in the LHP

b) s^5 + 10s^4 + 30s^3 + 80s^2+344s + 480 =0 \\\\s^5:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:30\:\:\:\:\:\:344\\s^4:\:\:\:10\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:80\:\:\:\:\:\:480\\s^3:\:\:\:22\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:296\\s^2:\:\:\:-545\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:480\\s^1:\:\:\:490\\s^0:\:\:\:480

2 roots not in the LHP

c) s^4 + 2s^3 + 7s^2 -2s + 8 = 0 \\\\s^4:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:7\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:8\\s^3:\:\:\:2\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-2\\s^2:\:\:\:8\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:8\\s^1:\:\:\:-4\\s^0:\:\:\:8

There are roots in the RHP (not all coefficients are greater than 0).

2 roots not in the LHP

d) s^4 + 2s^3 + 7s^2 -2s + 8 = 0 \\\\s^3:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:20\\s^2:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:78\\s^1:\:\:\:-58\\s^0:\:\:\:78

There are two sign changes in the first column of the Routh array.

2 roots not in the LHP

e) s^4 + 2s^3 + 7s^2 -2s + 8 = 0 \\\\s^4:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:6\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:25\\s^3:\:\:\:4\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:12\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: new \:\:row \\s^2:\:\:\:3\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:25\\s^1:\:\:\:12-\frac{100}{3}=-21.3 \\s^0:\:\:\:25

2 roots not in LHP

check:

 a (s) = 0  ⇒

 s^2 = -3 \limits^+_- 4j = 5e^{j(\pi \limits^+_- 0.92)}\\\\s = \sqrt5 e^{j( \frac{\pi}{2} \limits^+_-  0.46)+n\pi j},\:\:\:\:\: n= 0, 1\\

3 0
3 years ago
The mean of hours that the average person watches television each day is 4.18 hours with a standard deviation of 1.19 hours. Fin
luda_lava [24]

Answer:

z = \frac{3-4.18}{1.19}=-0.992

z = \frac{5-4.18}{1.19}=0.689

And we can find this probability with this difference:

P(-0.992

And then we can conclude that the probability that someone watches between 3 and 5 hours a day is approximately 0.591 using a normal distribution

Explanation:

For this case we can define the random variable X as "hours that a person watches television". For this case we don't have the distribution for X but we have the following parameters:

\mu = 4.18,\sigma =1.19

We can assume that the distribution for X is normal

X \sim N(\mu = 4.18 , \sigma =1.19)

And we want to find this probability:

P(3

And we can use the z score formula given by:

z=\frac[X- \mu}{\sigma}

And we can find the z score for each limit and we got:

z = \frac{3-4.18}{1.19}=-0.992

z = \frac{5-4.18}{1.19}=0.689

And we can find this probability with this difference:

P(-0.992

And then we can conclude that the probability that someone watches between 3 and 5 hours a day is approximately 0.591 using a normal distribution

6 0
3 years ago
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