Answer:
5.0674
Explanation:
Given that :
Mean, μ = 5 pounds
Standard deviation, σ = 0.1
Given that weight are normally distributed ;
From the Z table, the Zscore or value for 75th percentile weight is :
P(Z < z) = 0.75
z = 0.674
Using the relation :
Zscore = (x - μ) / σ
x = weight
0.674 = (x - 5) / 0.1
0.674 * 0.1 = x - 5
0.0674 = x - 5
0.0674 + 5 = x - 5 + 5
5.0674 = x
The weight which corresponds to the 75th percentile is 5.0674
The following pseudocode describes how a widget company computes the price of an order from the total price and the number of the widgets that were ordered. Read the number of widgets. Multiple the number of widgets by the price per widget of 9.99. Compute the tax (5.5 percent of the total price). Compute the shipping charge (.40 per widget). The price of the order is the sum of the total widget price, the tax, and the shipping charge. Print the price of the order
Answer:
The program in Python is as follows:
widget = int(input("Widgets: "))
price = widget * 9.9
tax = price * 0.55
ship = price * 0.40
totalprice = price + tax + ship
print("Total Price: $",totalprice)
Explanation:
The question is incomplete, as what is required is not stated.
However, I will write convert the pseudocode to a programming language (in Python)
Get the number of widgets
widget = int(input("Widgets: "))
Calculate price
price = widget * 9.9
Calculate the tax
tax = price * 0.55
Calculate the shipping price
ship = price * 0.40
Calculate the total price
totalprice = price + tax + ship
Print the calculated total price
print("Total Price: $",totalprice)
Given two regular expressions r1 and r2, construct a decision procedure to determine whether the language of r1 is contained in the language r2; that is, the language of r1 is a subset of the language of r2.
Answer:
Test if L(M1-2) is empty.
Construct FA M2-1 from M1 and M2 which recognizes the language L(>M2) - L(>M1).
COMMENT: note we use the algorithm that is instrumental in proving that regular languages are closed with respect to the set difference operator.
Test if L(M2-1) is empty.
Answer yes if and only if both answers were yes.
Explanation:
An algorithm must be guaranteed to halt after a finite number of steps.
Each step of the algorithm must be well specified (deterministic rather than non-deterministic).
Three basic problems:
Given an FA M and an input x, does M accept x?
Is x in L(M)
Given an FA M, is there a string that it accepts?
Is L(M) the empty set?
Given an FA M, is L(M) finite?
Algorithm for determining if M accepts x.
Simply execute M on x.
Output yes if we end up at an accepting state.
This algorithm clearly halts after a finite number of steps, and it is well specified.
This algorithm is also clearly correct.
Testing if L(M) is empty.
Incorrect "Algorithm"
Simulate M on all strings x.
Output yes if and only if all strings are rejected.
The "algorithm" is well specified, and it is also clearly correct.
However, this is not an algorithm because there are an infinite number of strings to simulate M on, and thus it is not guaranteed to halt in a finite amount of time.
COMMENT: Note we use the algorithm for the first problem as a subroutine; you must think in this fashion to solve the problems we will ask.
Correct Algorithm
Simulate M on all strings of length between 0 and n-1 where M has n states.
Output no if and only if all strings are rejected.
Otherwise output yes.
This algorithm clearly halts after a finite number of steps, and it is well specified.
The correctness of the algorithm follows from the fact that if M accepts any strings, it must accept one of length at most n-1.
Suppose this is not true; that is, L(M) is not empty but the shortest string accepted by M has a length of at least n.
Let x be the shortest string accepted by M where |x| > n-1.
Using the Pumping Lemma, we know that there must be a "loop" in x which can be pumped 0 times to create a shorter string in L.
This is a contradiction and the result follows.
COMMENT: There are more efficient algorithms, but we won't get into that.
Testing if L(M) is finite
Incorrect "Algorithm"
Simulate M on all strings x.
Output yes if and only if there are a finite number of yes answers.
This "algorithm" is well specified and correct.
However, this is not an algorithm because there are an infinite number of strings to simulate M on, and thus it is not guaranteed to halt in a finite amount of time.
COMMENT: Note we again use the algorithm for the first problem as a subroutine.
Correct Algorithm
Simulate M on all strings of length between n and 2n-1 where M has n states.
Output yes if and only if no string is accepted.
Otherwise output no.
This algorithm clearly halts after a finite number of steps, and it is well specified.
The correctness of the algorithm follows from the fact that if M accepts an infinite number of strings, it must accept one of length between n and 2n-1.
This builds on the idea that if M accepts an infinite number of strings, there must be a "loop" that can be pumped.
This loop must have length at most n.
When we pump it 0 times, we have a string of length less than n.
When we pump it once, we increase the length of the string by at most n so we cannot exceed 2n-1. The problem is we might not exceed n-1 yet.
The key is we can keep pumping it and at some point, its length must exceed n-1, and in the step it does, it cannot jump past 2n-1 since the size of the loop is at most n.
This proof is not totally correct, but it captures the key idea.
COMMENT: There again are more efficient algorithms, but we won't get into that.
Other problems we can solve using these basic algorithms (and other algorithms we've seen earlier this chapter) as subroutines.
COMMENT: many of these algorithms depend on your understanding of basic set operations such as set complement, set difference, set union, etc.
Given a regular expression r, is Lr finite?
Convert r to an equivalent FA M.
COMMENT: note we use the two algorithms for converting a regular expression to an NFA and then an NFA to an FA.
Test if L(M) is finite.
Output the answer to the above test.
Given two FAs M1 and M2, is L(M1) = L(M2)?
Construct FA M1-2 from M1 and M2 which recognizes the language L(>M1) - L(>M2).
COMMENT: note we use the algorithm that is instrumental in proving that regular languages are closed with respect to the set difference operator.
Test if L(M1-2) is empty.
Construct FA M2-1 from M1 and M2 which recognizes the language L(>M2) - L(>M1).
COMMENT: note we use the algorithm that is instrumental in proving that regular languages are closed with respect to the set difference operator.
Test if L(M2-1) is empty.
Answer yes if and only if both answers were yes.
How are dates and times stored by Excel?
Answer:
Regardless of how you have formatted a cell to display a date or time, Excel always internally stores dates And times the same way. Excel stores dates and times as a number representing the number of days since 1900-Jan-0, plus a fractional portion of a 24 hour day: ddddd. tttttt
Explanation:
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computer is an ............. machine because once a task is intitated computer proceeds on its own t ill its completion.
Answer:
I think digital,versatile
computer is an electronic digital versatile machine because once a task is initiated computer proceeds on its own till its completation.
The _________ attack exploits the common use of a modular exponentiation algorithm in RSA encryption and decryption, but can be adapted to work with any implementation that does not run in fixed time.
A. mathematical.
B. timing.
C. chosen ciphertext.
D. brute-force.
Answer:
chosen ciphertext
Explanation:
Chosen ciphertext attack is a scenario in which the attacker has the ability to choose ciphertexts C i and to view their corresponding decryptions – plaintexts P i . It is essentially the same scenario as a chosen plaintext attack but applied to a decryption function, instead of the encryption function.
Cyber attack usually associated with obtaining decryption keys that do not run in fixed time is called the chosen ciphertext attack.
Theae kind of attack is usually performed through gathering of decryption codes or key which are associated to certain cipher texts The attacker would then use the gathered patterns and information to obtain the decryption key to the selected or chosen ciphertext.Hence, chosen ciphertext attempts the use of modular exponentiation.
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While saving a document to her hard drive, Connie's computer screen suddenly changed to display an error message on a blue background. The error code indicated that there was a problem with her computer's RAM. Connie's computer is affected by a(n) __________.
Answer:
The right answer is "Hardware crash".
Explanation:
According to the runtime error message, this same RAM on your machine was problematic. This excludes file interoperability or compliance problems as well as program error possibilities.Assuming implementation performance problems exist, the timeframe that would save the information would be typically longer, but there's still a lower possibility that the adequacy and effectiveness color will become blue but instead demonstrate warning would appear.Thus the above is the right solution.
Write a method that prints on the screen a message stating whether 2 circles touch each other, do not touch each other or intersect. The method accepts the coordinates of the center of the first circle and its radius, and the coordinates of the center of the second circle and its radius.
The header of the method is as follows:
public static void checkIntersection(double x1, double y1, double r1, double x2, double y2, double r2)
Hint:
Distance between centers C1 and C2 is calculated as follows:
d = Math.sqrt((x1 - x2)2 + (y1 - y2)2).
There are three conditions that arise:
1. If d == r1 + r2
Circles touch each other.
2. If d > r1 + r2
Circles do not touch each other.
3. If d < r1 + r2
Circles intersect.
Answer:
The method is as follows:
public static void checkIntersection(double x1, double y1, double r1, double x2, double y2, double r2){
double d = Math.sqrt(Math.pow((x1 - x2),2) + Math.pow((y1 - y2),2));
if(d == r1 + r2){
System.out.print("The circles touch each other"); }
else if(d > r1 + r2){
System.out.print("The circles do not touch each other"); }
else{
System.out.print("The circles intersect"); }
}
Explanation:
This defines the method
public static void checkIntersection(double x1, double y1, double r1, double x2, double y2, double r2){
This calculate the distance
double d = Math.sqrt(Math.pow((x1 - x2),2) + Math.pow((y1 - y2),2));
If the distance equals the sum of both radii, then the circles touch one another
if(d == r1 + r2){
System.out.print("The circles touch each other"); }
If the distance is greater than the sum of both radii, then the circles do not touch one another
else if(d > r1 + r2){
System.out.print("The circles do not touch each other"); }
If the distance is less than the sum of both radii, then the circles intersect
else{
System.out.print("The circles intersect"); }
}
