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2 years ago

An expression containing the operator && is true either or both of its operands are true. True/False

Computers and Technology

1 answer:

Luba_88 [7]2 years ago

4 0

Answer:

True.

Explanation:

An expression containing the && and operator is only true if both of the operands present in the expression are true otherwise it will give the result as false if either one of them is false or both of them are false.In the question it states that the expression in true either or both of it's operand are true.

Hence the answer is true.

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A workgroup database is a(n) _____ database

Step2247 [10]

It is a shared database

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2 years ago

We have said that the average number of comparisons need to find a target value in an n-element list using sequential search is

bija089 [108]

Answer:

Part a: If the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b: If the list contains n elements (where n is even) the middle terms are at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c: The average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

Explanation:

Suppose the list is such that the starting index is 0.

Part a

If list has 15 elements, the middle item would be given at 7th index i.e.

there are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(8,9,10,11,12,13,14) above it. It will have to run 8 comparisons to find the middle term.

If list has 17 elements, the middle item would be given at 8th index i.e.

there are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(9,10,11,12,13,14,15,16) above it.It will have to run 9 comparisons to find the middle term.

If list has 21 elements, the middle item would be given at 10th index i.e.

there are 10 indices (0,1,2,3,4,5,6,7,8,9) below it and 10 indices (11,12,13,14,15,16,17,18,19,20) above it.It will have to run 11 comparisons to find the middle term.

Now this indicates that if the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b

If list has 16 elements, there are two middle terms as one at would be at 7th index and the one at 8th index .There are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(9,10,11,12,13,14,15) above it. It will have to run 9 comparisons to find the middle terms.

If list has 18 elements, there are two middle terms as one at would be at 8th index and the one at 9th index .There are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(10,11,12,13,14,15,16,17) above it. It will have to run 10 comparisons to find the middle terms.

If list has 20 elements, there are two middle terms as one at would be at 9th index and the one at 10th index .There are 9 indices(0,1,2,3,4,5,6,7,8) below it and 9 indices(11,12,13,14,15,16,17,18,19) above it. It will have to run 11 comparisons to find the middle terms.

Now this indicates that if the list contains n elements (where n is even) the middle terms are at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c

So the average number of comparisons is given as

((n+1)/2+(n+2)/2)/2=(2n+3)/4

So the average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

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2 years ago

Which of the following stakeholders makes policies for a green economy?

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Answer:

government is the answer

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2 years ago

Write two methods: encrypt and decrypt. encrypt should #take as input a string, and return an encrypted version #of it according

Harman [31]

Answer:

The code is given below with appropriate comments

Explanation:

CIPHER = (("D", "A", "V", "I", "O"),

("Y", "N", "E", "R", "B"),

("C", "F", "G", "H", "K"),

("L", "M", "P", "Q", "S"),

("T", "U", "W", "X", "Z"))

# Add your code here!

def encrypt(plaintext):

theList = []

for char in plaintext:

if char.isalpha():

char = char.upper()

if char == "J":

char = "I"

theList.append(char)

if len(theList) % 2 == 1:

theList.append("X")

for i in range(0, len(theList), 2):

if theList[i] == theList[i + 1]:

theList[i + 1] = "X"

findex = [-1, -1]

sindex = [-1, -1]

for j in range(len(CIPHER)):

for k in range(len(CIPHER)):

if theList[i] == CIPHER[j][k]:

findex = [j, k]

if theList[i + 1] == CIPHER[j][k]:

sindex = [j, k]

# same row

if (findex[0] == sindex[0]):

findex[1] += 1

sindex[1] += 1

if findex[1] == 5:

findex[1] = 0

if sindex[1] == 5:

sindex[1] = 0

theList[i] = CIPHER[findex[0]][findex[1]]

theList[i + 1] = CIPHER[sindex[0]][sindex[1]]

# same column

elif (findex[1] == sindex[1]):

findex[0] += 1

sindex[0] += 1

if findex[0] == 5:

findex[0] = 0

if sindex[0] == 5:

sindex[0] = 0

theList[i] = CIPHER[findex[0]][findex[1]]

theList[i + 1] = CIPHER[sindex[0]][sindex[1]]

else:

theList[i] = CIPHER[findex[0]][sindex[1]]

theList[i + 1] = CIPHER[sindex[0]][findex[1]]

return "".join(theList)

def decrypt(ciphertext):

theString = ""

findex = [-1, -1]

sindex = [-1, -1]

for i in range(0, len(ciphertext), 2):

for j in range(len(CIPHER)):

for k in range(len(CIPHER)):

if ciphertext[i] == CIPHER[j][k]:

findex = [j, k]

if ciphertext[i + 1] == CIPHER[j][k]:

sindex = [j, k]

if (findex[0] == sindex[0]):

findex[1] -= 1

sindex[1] -= 1

if findex[1] == -1:

findex[1] = 4

if sindex[1] == -1:

sindex[1] = 4

theString += CIPHER[findex[0]][findex[1]]

theString += CIPHER[sindex[0]][sindex[1]]

# same column

elif (findex[1] == sindex[1]):

findex[0] -= 1

sindex[0] -= 1

if findex[0] == -1:

findex[0] = 4

if sindex[0] == -1:

sindex[0] = 4

theString += CIPHER[findex[0]][findex[1]]

theString += CIPHER[sindex[0]][sindex[1]]

else:

theString += CIPHER[findex[0]][sindex[1]]

theString += CIPHER[sindex[0]][findex[1]]

return theString

# Below are some lines of code that will test your function.

# You can change the value of the variable(s) to test your

# function with different inputs.

# If your function works correctly, this will originally

# print: QLGRQTVZIBTYQZ, then PSHELXOWORLDSX

print(encrypt("PS. Hello, worlds"))

print(decrypt("QLGRQTVZIBTYQZ"))

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One factor affecting digital camera quality is the number of pixels, measured in ____, used to store the data for each image.

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One factor affecting digital camera quality is the number of pixels, measured in megapixels, use to store the data for each image.

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2 years ago

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