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kakasveta [241]
2 years ago
8

Aii so is anyone pushing P? if u are u a g

Computers and Technology
2 answers:
Verdich [7]2 years ago
5 0

Answer:

keep it real

Explanation:

Pushing P essentially means to 'keep it real,' and is generally a positive term.

Vitek1552 [10]2 years ago
4 0
Nahh it’s all about ABOWWW
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Complete each sentence to describe different elements of an Excel worksheet.
barxatty [35]

Answer:

Answer is - cell, - column, - string value

Explanation:

  • <em>Cell - this is often referred to as the intersection of a single row and column.  </em>
  • <em>Column - this is a group of cells which are represented vertically.  </em>
  • <em>String - these are values that are inside the cell which are represented through texts or group of letters including acceptable symbols and characters.</em>
7 0
2 years ago
Read 2 more answers
For the past five days, the Howard family has traveled these miles: 389;126;419;93; and 394. Find the range and how. Show all yo
Fofino [41]
The answer is 326.

Range is found by subtracting the smallest number in the data set from the largest number.

Highest number: 419
Lowest number: 93

419-93
=326

(Next time make sure to post this under the Mathematics section.)
5 0
2 years ago
You are given an unsorted array x of elements that implement the Comparable interface. There are no duplicates in this array. Yo
Andrei [34K]

Answer:

boolean isEven = false;

if (x.length % 2 == 0)

isEven = true;

Comparable currentMax;

int currentMaxIndex;

for (int i = x.length - 1; i >= 1; i--)

{

currentMax = x[i];

currentMaxIndex = i;

for (int j = i - 1; j >= 0; j--)

{

if (((Comparable)currentMax).compareTo(x[j]) < 0)

{

currentMax = x[j];

currentMaxIndex = j;

}

}

x[currentMaxIndex] = x[i];

x[i] = currentMax;

}

Comparable a = null;

Comparable b = null;

if (isEven == true)

{

a = x[x.length/2];

b = x[(x.length/2) - 1];

if ((a).compareTo(b) > 0)

m = a;

else

m = b;

}

else

m = x[x.length/2];

8 0
3 years ago
Q1) Convert the decimal number 67 to binary?​
Mashutka [201]

Answer:

1000011

Explanation:

4 0
2 years ago
A certain list, L, contains a total of n numbers, not necessarily distinct, that are arranged in increasing order. If L1 is the
Free_Kalibri [48]

In this question, we are given ,

  • A certain list, L, contains a total of n numbers, not necessarily distinct, that are arranged in increasing order.
  • L1 is the list consisting of the first n1 numbers in L.
  • L2 is the list consisting of the last n2 numbers in L.

Explanation:

As per the information given in statement 1, 17 is a mode for L1 and 17 is a mode for L2.

Therefore, we can infer that ,

  • 17 must occur in L1, either same or a greater number of times as any other number in L1.
  • 17 must occur in L1, either same or a greater number of times as any other number in L2.

As all elements in L are in ascending order, we can also conclude that

  • Each number between last occurrence of 17 in L1 and the first occurrence of 17 in L2 must be equal to 17 only.
  • Therefore, 17 occurs either same or greater number of times as any other number in L.
  • Thus, 17 is a mode for L.

However, from this statement, we cannot conclude anything about the mode of L1, L2, or L.

Hence, statement 2 is not sufficient to answer the question.

Therefore, 17 is a mode for L1 and 17 is a mode for L2.

3 0
2 years ago
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