Answer:
VER.
Explanation:
The diagnostic test that involves watching a computer monitor with alternating checkerboard patterns while an ecg is performed is known as a VER.
Answer:
Following are the statement which is given below:
if (gpa > 3.5) // check the condition if gpa exceeds 3.5.
{
deansList++; // increment of 1 in the deansList variable
cout << studentName; // display the studentName,
}
Explanation:
The description of the statement
- In the if statement we check the condition of "gpa" .if "gpa" is greater then 3.5 then it executed the statement inside the if block otherwise nothing will be executed.
- if control moves to the if part it adds to 1 to the deansList.
- Finally, display the value of studentName.
Answer:
3. (9 points) The goal of this problem is to show that our formal definitions of P and NP imply that P = NP cannot have a non-constructive proof. Throughout this problem we assume a consistent, binary, encoding of both Turing machines and 3-SAT instances. Note that the complexities are intentionally not tight to address potential overheads from scan- ning the tape linearly in order to look for certain symbols. (a) (3 points) Show that for any constant cı, there is a constant c2 such that if Mi is a Turing machine encoded by at most bits that correctly decides 3-SAT in O(n10) time (aka. 3-SAT E TIME(n10), then there is a Turing machine M, encoded by at most most c2 bits that runs in O(n100) time such that if the 3-SAT instance has a satisfying assignment, M, accepts and writes a satisfying assignment at the end of the tape. (b) (3 points) Let M ...Mbe Turing machines show that the language {(k, M... Mk,t,w) | w is an instance of 3-SAT and one of M, writes a satisfying assignment to w in at most t steps can be decided by a Turing machine in (k2107 10) time. (c) (3 points) Under the assumption that there exists an unknown Turing machine encodable in c bits that decides 3-SAT is in O(n10) time, give implementation details for a Turing Machine M that decides 3-SAT in O(n10000) time. Hint: the constant factor in the big-O is able to hide any function related ci and c2, which independent of input sizes, and thus constants.
Explanation:
3. (9 points) The goal of this problem is to show that our formal definitions of P and NP imply that P = NP cannot have a non-constructive proof. Throughout this problem we assume a consistent, binary, encoding of both Turing machines and 3-SAT instances. Note that the complexities are intentionally not tight to address potential overheads from scan- ning the tape linearly in order to look for certain symbols. (a) (3 points) Show that for any constant cı, there is a constant c2 such that if Mi is a Turing machine encoded by at most bits that correctly decides 3-SAT in O(n10) time (aka. 3-SAT E TIME(n10), then there is a Turing machine M, encoded by at most most c2 bits that runs in O(n100) time such that if the 3-SAT instance has a satisfying assignment, M, accepts and writes a satisfying assignment at the end of the tape. (b) (3 points) Let M ...Mbe Turing machines show that the language {(k, M... Mk,t,w) | w is an instance of 3-SAT and one of M, writes a satisfying assignment to w in at most t steps can be decided by a Turing machine in (k2107 10) time. (c) (3 points) Under the assumption that there exists an unknown Turing machine encodable in c bits that decides 3-SAT is in O(n10) time, give implementation details for a Turing Machine M that decides 3-SAT in O(n10000) time. Hint: the constant factor in the big-O is able to hide any function related ci and c2, which independent of input sizes, and thus constants.
Answer: Integrity loss
Explanation:
A set of data is said to have experienced a Loss of Integrity if some event has caused it to be incorrectly altered
The property that stores the text a user types into a Textbox is called the: control's text property.
<h3>The Textbox Control</h3>
The Textbox Control is employed in display a text or accept the text as an input from a user. Also, single line of text can be accepted as input at runtime from a VB.NET Windows.
The text that displays on the Textbox can also be set. The control's text property therefore stores the text a user types into Textbox.
In conclusion, the property that stores the text a user types into a Textbox is called the: control's text property.
Learn more about Textbox Control on:
brainly.com/question/5677552
