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RUDIKE [14]
3 years ago
6

Write a program that prompts the user to enter a series of numbers between 0 and 10 asintegers. The user will enter all numbers

on a single line, and scores will be separatedby spaces. You cannot assume the user will supply valid integers, nor can you assumethat they will enter positive numbers, so make sure that you validate your data. You donot need to re-prompt the user once they have entered a single line of data. Once youhave collected this data you should compute the following: • The largest value • Thesmallest value • The range of values • The mode (the value that occurs the most often) •A histogram that visualizes the frequency of each number Here’s a sample running of theprogram.
Computers and Technology
1 answer:
Anna71 [15]3 years ago
7 0

Answer:

The program in Python is as follows:

import collections

from collections import Counter

from itertools import groupby

import statistics

str_input = input("Enter a series of numbers separated by space: ")

num_input = str_input.split(" ")

numList = []

for num in num_input:

         if(num.lstrip('-').isdigit()):

                   if num[0] == "-" or int(num)>10:

                             print(num," is out of bound - rejecting")

                   else:

                             print(num," is valid - accepting")

                             numList.append(int(num))

         else:

                   print(num," is not a number")

print("Largest number: ",max(numList))

print("Smallest number: ",min(numList))

print("Range: ",(max(numList) - min(numList)))

print("Mode: ",end = " ")

freqs = groupby(Counter(numList).most_common(), lambda x:x[1])

print([val for val,count in next(freqs)[1]])

count_freq = {}

count_freq = collections.Counter(numList)

for item in count_freq:

         print(item, end = " ")

         lent= int(count_freq[item])

         for i in range(lent):

                   print("#",end=" ")

print()

Explanation:

See attachment for program source file where comments are used as explanation. (Lines that begin with # are comments)

The frequency polygon is printed using #'s to represent the frequency of each list element

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Dima020 [189]

Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

^nC_r = \frac{n!}{(n-r)!r!}

Where n = 15\ and\ r = 4

^{15}C_4 = \frac{15!}{(15-4)!4!}

^{15}C_4 = \frac{15!}{11!4!}

^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}

^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}

^{15}C_4 = \frac{32760}{24}

^{15}C_4 = 1365

<em>Hence, there are 1365 ways </em>

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

<em>From the given parameters; 3 out of 15 is detective;</em>

So;

p = 3/15

p = 0.2

q = 1 - p

q = 1 - 0.2

q = 0.8

Solving further using binomial;

(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

Probability =  ^nC_3pq^3

Substitute 4 for n, 0.2 for p and 0.8 for q

Probability =  ^4C_3 * 0.2 * 0.8^3

Probability =  \frac{4!}{3!1!} * 0.2 * 0.8^3

Probability = 4 * 0.2 * 0.8^3

Probability = 0.4096

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

<em>Probability that at least one is defective + Probability that at none is defective = 1</em>

Probability that none is defective is calculated as thus;

Probability =  q^n

Substitute 4 for n and 0.8 for q

Probability =  0.8^4

Probability = 0.4096

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

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Alisiya [41]

Answer:

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Explanation:

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-The new and delete operators do not return a null value on failure but the malloc/free functions do.

- The new/delete operator memory is allocated from free store while the malloc/free functions allocate from heap.

- The new/delete operators can add a new memory allocator to help with low memory but the malloc/free functions can't.

- The compiler calculates the size of the new/delete operator array while the malloc/free functions manually calculate array size as specified.

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Answer:

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Strike Level 2 is one of the two innately equal mapping and assurance strategies characterized in the Berkeley paper. It has not been broadly sent in the business to a great extent since it requires an extraordinary plate highlights. Since plate creation volumes decides the cost, it is increasingly practical to utilize standard circles for the RAID frameworks.  

Points of interest: "On the fly" it gives information blunder amendment. Amazingly high information move rates is possible.Higher the information move rate required,better the proportion of information circles to ECC plates. Moderately basic controller configuration contrasted with the other RAID levels 3,4 and 5.  

Impediments: Very high proportion of the ECC plates to information circles with littler word sizes - wasteful. Passage level expense are extremely high - and requires exceptionally high exchange rate prerequisite to legitimize. Exchange rate is equivalent to that of the single plate, best case scenario (with shaft synchronization). No business executions can't/not industrially suitable.

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3 years ago
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