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inessss [21]
2 years ago
6

Carl knows that water moves through different kinds of soil at different rates. How easily water moves through a soil is known a

s permeability. Carl decides to compare the permeabilities of different soil types. To do this, Carl takes five identical flower pots with holes in the bottom and fills each one with different soil: rocky, gravely dirt from the side of the road potting soil from the store clay soil from a nearby creek sand from a store dirt from his own backyard Carl pours one liter of water into each pot and measures how much water flows out of the bottom in one minute.
Computers and Technology
1 answer:
Serggg [28]2 years ago
6 0

Answer:

To get the same same results from all pots "amount of water should be same" for all pots.

Explanation:

As Carl want to measure and compare the amount of water that flows in pot in one minute from all all pots. He should keep the amount of water constant for all pots to get the desired results.

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A laptop gets replaced if there's a hardware issue. Which stage of the hardware lifecycle does this scenario belong to?
Ilya [14]

Answer: Maintenance. This is the stage where software is updated and hardware issues are fixed if and when they occur. Retirement. In this final stage, hardware becomes unusable or no longer needed and it needs to be properly removed from the fleet

Explanation:

3 0
2 years ago
In which job role would a course in 3D modeling help with professional career prospects?
ale4655 [162]

Answer:

An eye for detail and good visualization skills.

Explanation:

In which job role would a course in 3D modeling help with professional career prospects? Computer Programmer. Multimedia Artist. Technical Support Specialist.

4 0
2 years ago
Write a function string middle(string str) that returns a string containing the middle character in str if the length of str is
Katen [24]

Answer:

function getMiddle(s) {

return s.length % 2 ? s.substr(s.length / 2, 1) : s.substr((s.length / 2) - 1, 2);

}

// I/O stuff

document.getElementById("submit").addEventListener("click", function() {

input = document.getElementById("input").value;

document.getElementById("output").innerHTML = getMiddle(input);

});

Explanation:

// >>> is an unsigned right shift bitwise operator. It's equivalent to division by 2, with truncation, as long as the length of the string does not exceed the size of an integer in Javascript.

// About the ~ operator, let's rather start with the expression n & 1. This will tell you whether an integer n is odd (it's similar to a logical and, but comparing all of the bits of two numbers). The expression returns 1 if an integer is odd. It returns 0 if an integer is even.

// If n & 1 is even, the expression returns 0.

// If n & 1 is odd, the expression returns 1.

// ~n & 1 inverts those two results, providing 0 if the length of the string is odd, and 1 if the length of the sting is even. The ~ operator inverts all of the bits in an integer, so 0 would become -1, 1 would become -2, and so on (the leading bit is always the sign).

// Then you add one, and you get 0+1 (1) characters if the length of the string is odd, or 1+1 (2) characters if the length of the string is even.

6 0
2 years ago
Write the recursive function max_depth; it is passed a binary (any binary tree, not necessarily a binary search tree) and a valu
OleMash [197]

Answer:

A python code was used in writing of the recursive function max_depth; and passed a binary (any binary tree, not necessarily a binary search tree) and a value as arguments.

Explanation:

Solution

To solve this given question we make use of Python programming language.

Thus

THE CODE:

class BinaryTreeNode:

def __init__(self, value, left=None, right=None):

self.value = value

self.left = left

self.right = right

# recursive max_depth function implementation

def max_depth(tree, value):

if tree == None:

return -1

# recursive calls to this function

leftDepth = max_depth(tree.left, value)

rightDepth = max_depth(tree.right, value)

# return the maximum depth

if tree.value == value or leftDepth > -1 or rightDepth > -1:

return 1 + max(leftDepth, rightDepth)

else:

return max(leftDepth, rightDepth)

btn1 = BinaryTreeNode(2)

btn2 = BinaryTreeNode(3, None, btn1)

btn3 = BinaryTreeNode(3)

btn4 = BinaryTreeNode(3)

btn5 = BinaryTreeNode(2, btn2, btn3)

btn6 = BinaryTreeNode(1, btn4)

tree = BinaryTreeNode(3, btn5, btn6)

print('max_depth(tree, 1):',max_depth(tree, 1))

print('max_depth(tree, 2):',max_depth(tree, 2))

print('max_depth(tree, 3):',max_depth(tree, 3))

Note: Kindly find an attached copy of the Implementation of max_depth() function, Code to test the max_depth() function and the sample output below.

8 0
2 years ago
In the simulation, player 2 will always play according to the same strategy. The number of coins player 2 spends is based on wha
baherus [9]

The simulation, player 2 will always play according to the same strategy.

Method getPlayer2Move below is completed by assigning the correct value to result to be returned.

Explanation:

  • You will write method getPlayer2Move, which returns the number of coins that player 2 will spend in a given round of the game. In the first round of the game, the parameter round has the value 1, in the second round of the game, it has the value 2, and so on.

#include <bits/stdc++.h>  

using namespace std;

bool getplayer2move(int x, int y, int n)  

{

   int dp[n + 1];  

   dp[0] = false;  

   dp[1] = true;  

   for (int i = 2; i <= n; i++) {  

       if (i - 1 >= 0 and !dp[i - 1])  

           dp[i] = true;  

       else if (i - x >= 0 and !dp[i - x])  

           dp[i] = true;  

       else if (i - y >= 0 and !dp[i - y])  

           dp[i] = true;  

       else

           dp[i] = false;  

   }  

   return dp[n];  

}  

int main()  

{  

   int x = 3, y = 4, n = 5;  

   if (findWinner(x, y, n))  

       cout << 'A';  

   else

       cout << 'B';  

   return 0;  

}

8 0
2 years ago
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