The suspense and reaction time of the car . So for example if a car pulled out suddenly and was about to hit you, you could make the car so it functionally knows how to react
Answer:
The WPA2 shared key is incorrect is the correct answer.
Explanation:
The WPA2 shared key is incorrect because when the technician installs a new wireless thermostat for the purpose to control the temperature of the meeting room. Then, admin determines that the thermostat is not connecting to the control system through the internet and the admin authenticate that its parameter of receiving thermostat is associated with the AP. So, that's why the following option is correct.
Answer:
dataFile << salary;
Explanation:
To write salary to a file (payroll.dat) using ofstream, you make use of the following instruction:
<em>ofstream dataFile;
</em>
<em>myfile.open ("payroll.dat");
</em>
<em>myfile <<salary;
</em>
<em>myfile.close();</em>
<em />
This line creates an instance of ofstream
<em>ofstream dataFile;
</em>
This line opens the file payroll.dat
<em>myfile.open ("payroll.dat");
</em>
This is where the exact instruction in the question is done. This writes the value of salary to payroll.dat
<em>myfile <<salary;
</em>
This closes the opened file
<em>myfile.close();</em>
<em />
<em />
Answer:
cyber-extortion
Explanation:
Ashley Baker has been the webmaster for Berryhill Finance only ten days when she received an e-mail that threatened to shut down Berryhill's website unless Ashley wired payment to an overseas account. Ashley was concerned that Berryhill Finance would suffer huge losses if its website went down, so she wired money to the appropriate account. The author of the e-mail successfully committed cyber-extortion.
Answer:
Each time you insert a new node, call the function to adjust the sum.
This method has to be called each time we insert new node to the tree since the sum at all the
parent nodes from the newly inserted node changes when we insert the node.
// toSumTree method will convert the tree into sum tree.
int toSumTree(struct node *node)
{
if(node == NULL)
return 0;
// Store the old value
int old_val = node->data;
// Recursively call for left and right subtrees and store the sum as new value of this node
node->data = toSumTree(node->left) + toSumTree(node->right);
// Return the sum of values of nodes in left and right subtrees and
// old_value of this node
return node->data + old_val;
}
This has the complexity of O(n).
Explanation:
