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wariber [46]
3 years ago
14

5.20 LAB: Output values below an amount Write a program that first gets a list of integers from input. The input begins with an

integer indicating the number of integers that follow. Then, get the last value from the input, which indicates a threshold. Output all integers less than or equal to that last threshold value. Ex: If the input is: 5 50 60 140 200 75 100 the output is: 50,60,75,
Computers and Technology
1 answer:
Sveta_85 [38]3 years ago
6 0

Answer:

Following are the code to the given question:

x= input()#defining a variable x that inputs value from the user-end

l = x.split(" ")#defining variable l that splits the x variable value with a space

t= int(l[len(l)- 1])#defining a variable t that decrease the l length and convert the value into integer

for i in range(1, (len(l) - 1)):#use for loop to calculate the range of the list

      if int(l[i]) < t:#defining if block that checks the list value is lessthan t

          print(l[i])#print list values

Output:

Please find the attached file.

Explanation:

In the above-given code, an "x" variable is declared that inputs the value from the user-end, and use the "l" variable that splits all the value and defined a "t" variable that adds value into the list and convert the value into an integer.

In the next step, for loop is declared that counts the lists and use a conditional statement that checks list value is less than t and use the print the method that prints the list values.  

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Answer:

num = int(input("Enter a number: "))

if (num % 2) == 0:

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3 years ago
Sketch f(x) = 5x2 - 20 labelling any intercepts.​
Norma-Jean [14]

Answer:

  • The graph of the function is attached below.
  • The x-intercepts will be: (2, 0), (-2, 0)
  • The y-intercept will be: (-20, 0)

Explanation:

Given the function

f\left(x\right)\:=\:5x^2-\:20

As we know that the x-intercept(s) can be obtained by setting the value y=0

so

y=\:5x^2-\:20

switching sides

5x^2-20=0

Add 20 to both sides

5x^2-20+20=0+20

5x^2=20

Dividing both sides by 5

\frac{5x^2}{5}=\frac{20}{5}

x^2=4

\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=\sqrt{4},\:x=-\sqrt{4}

x=2,\:x=-2

so the x-intercepts will be: (2, 0), (-2, 0)

we also know that the y-intercept(s) can obtained by setting the value x=0

so

y=\:5(0)^2-\:20

y=0-20

y=-20

so the y-intercept will be: (-20, 0)

From the attached figure, all the intercepts are labeled.

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2 years ago
A digital certificate system: Group of answer choices uses digital signatures to validate a user's identity. is used primarily b
kompoz [17]

Answer:

uses third party CAs to validate a user's identity

Explanation:

The Digital Certificate is the only means that technically and legally guarantees the identity of a person on the Internet. This is an essential requirement for institutions to offer secure services over the Internet. Further:

The digital certificate allows the electronic signature of documents The recipient of a signed document can be sure that it is the original and has not been tampered with and the author of the electronic signature cannot deny the authorship of this signature.

The digital certificate allows encryption of communications. Only the recipient of the information will be able to access its content.

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3 years ago
Modify an array's elements Write a for loop that iterates from 1 to numberSamples to double any element's value in dataSamples t
ValentinkaMS [17]

Answer:

See explaination for program code

Explanation:

%save as AdjustMinValue.m

%Matlab function that takes three arguments

%called number of samples count, user samples

%and min vlaue and then returns a data samples

%that contains values which are double of usersamples

%that less than min vlaue

%

function dataSamples=AdjustMinValue(numberSamples, userSamples, minValue)

dataSamples=userSamples;

%for loop

for i=1:numberSamples

%checking if dataSamples value at index,i

%is less than minValue

if dataSamples(i)<minValue

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dataSamples(i)= 2*dataSamples(i);

end

end

end

Sample output:

Note:

File name and calling method name must be same

--> AdjustMinValue(4, [2,12,9,20],10)

ans =

4 12 18 20

3 0
3 years ago
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