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zavuch27 [327]
3 years ago
9

The following program includes fictional sets of the top 10 male and female baby names for the current year. Write a program tha

t creates: A set all_names that contains all of the top 10 male and all of the top 10 female names. A set neutral_names that contains only names found in both male_names and female_names. A set specific_names that contains only gender specific names. Sample output for all_names: {'Michael', 'Henry', 'Jayden', 'Bailey', 'Lucas', 'Chuck', 'Aiden', 'Khloe', 'Elizabeth', 'Maria', 'Veronica', 'Meghan', 'John', 'Samuel', 'Britney', 'Charlie', 'Kim'}

Engineering
2 answers:
yuradex [85]3 years ago
4 0

Answer:

Please see attachment

Explanation:

Please see attachment

otez555 [7]3 years ago
3 0

Answer:

Define Variables and Use List methods to do the following

Explanation:

#<em>Conjoins two lists together</em>

all_names = male_names.union(female_names)

#<em>Finds the names that appear in both lists, just returns those</em>

neutral_names = male_names.intersection(female_names)

#<em>Returns names that are NOT in both lists</em>

specific_names = male_names.symmetric_difference(female_names)

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To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105
GenaCL600 [577]

Answer:

62.14\ \text{miles}

6213727.37\ \text{miles}

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = 10^5\ \text{mm}^{-2}

Volume of the metal = 1000\ \text{mm}^3

10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}

1\ \text{mile}=1609.34\ \text{m}

\dfrac{10^5}{1609.34}=62.14\ \text{miles}

The chain would extend 62.14\ \text{miles}

Dislocation density = 10^{10}\ \text{mm}^{-2}

Volume of the metal = 1000\ \text{mm}^3

10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}

\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}

The chain would extend 6213727.37\ \text{miles}

3 0
3 years ago
Base course aggregate has a target dry density of 119.7 Ib/cu ft in place. It will be laid down and compacted in a rectangular s
djyliett [7]

Answer:

total weight of aggregate =  5627528 lbs = 2814 tons  

Explanation:

we get  here volume of space to be filled with aggregate that is

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now filling space with aggregate of the density that is

density = 0.95 × 119.7

density = 113.72 lb/ft³

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dry weight = 48000 × 113.72

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we consider here percent moisture is by weigh

so weight of moisture in aggregate will be

weight of moisture = 0.031 × 5458320

weight of moisture = 169208 lbs

so here total weight of aggregate is

total weight of aggregate = 5458320 + 169208

total weight of aggregate =  5627528 lbs = 2814 tons  

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Lera25 [3.4K]

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D

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