Question

Help With homework questions please:
I tried all I can,these I have seem very hard.

Answer 1

Questions:

SELECT ALL THAT APPLY: Which of the following phrases can be used within this doStuff method?
class BankAccount {
private double balance;
static double dollar;
public static void doStuff (double given) {
double interest;
...
}
 A. this.balance
 B. this.dollar
C. this.given
 D. this.interest
E. none of the above

Complete this coding to compute the balance at the end of 10 years when you put $1000 in a
savings account at the end of each year and earn 5% interest annually. Use only one arithmetic
operator in your answer.
int k = 1;
double balance = 1000;
while (k <= 10)
{ k = k + 1;
balance = ________ _________ ________ + 1000;
}
System.out.println ("saving $1000 per year gives $" + balance + " after 10 years.");

The Test class is given below. Assume that ann and sue are variables of this Test class, where y is a sta
class Test
{ private int x = 5;
private static int y = 5;
public void think ()
{ x++;
y++;
}
public void print ()
{ System.out.println (x * y);
}
}

A. 36
B. 42
C. 49
D. none of the above


Complete this coding in such a way that it prints out all the values read in that are between 100 and
999 inclusive. Assume that read() is a method that obtains a number that the user enters.
for (double x = read(); x != 0; x = read())
if (x <= _________ _________ _______ )
System.out.println (x);

my thought please check (1000 && x >= 100)

Complete this coding to compute the gross pay that a person receives when she gets time-and-ahalf
for all hours over 40 worked in a week.
if (hoursWorked <= 40)
grossPay = hoursWorked * ratePerHour;
else
grossPay = 40 * ratePerHour + ________ _________ ________ * ratePerHour * 1.5;

my answer used to be hoursWorked.

The first number that this coding prints is _____ _________ ___________ .
for (x = 1; x < 100; x *= 2);
System.out.println (x);
System.out.println (x);

printed number = 128.

Complete this statement to round a given positive double value to the nearest integer value.
int rounded = _______ _________ _________ (given + 0.5);

Answer 2

What do you think? Any guesses?

Answer 3

@Lyrae ,For the first question, I'm just putting what I think. What if we use this.given, will that work with the method?

Answer 4

I think we can use this.interest as well, as we have interest as a variable.

Answer 5

A good guess! but sadly this one is a little bit tricky.

Lets try and see what we have available
```
class BankAccount {
private double balance; <- Instance variable
static double dollar; <- Static variable

public static void doStuff (double given) { <- Static method and declaration of
double interest; <- Another local variable a local variabe
...
}
```
the answers
```
A. this.balance <- (non-static) instance variable
B. this.dollar <- (non-static) instance variable
C. this.given <- (non-static) instance variable
D. this.interest <- (non-static) instance variable
E. none of the above
```

Can you figure out what I'm trying to show you?
Do you know of the differences between static and non-static methods and variables?

Answer 6

I know difference between static and non static methods. Not the variables.

Answer 7

The major difference is that a static variable can be used in both static and non-static methods. An instance variable (non-static) can however only be accessed from non-static methods.

Answer 8

Furthermore the "this" keyword can only be used to access instance variables (i.e. non static variables).

Answer 9

According to your explanation, it looks like we can use this.given and this.interest in the method.

Answer 10

Sorry, I confused you a bit there :P

given and interest are local variables, "this" keyword can only be used to access instance variables.

Answer 11

It's my fault for writing non-static in the parentheses.

Answer 12

okay, but am I correct in my answers, or did I do something wrong?

Answer 13

Can you use "this" keyword to access given and interest?
What did I write in my two previous posts?

Answer 14

Looks like I can't use "this" to access given and interest. So, it looks like I'm wrong.

Answer 15

Yes! Any clue what the answer might be now?

Answer 16

the answer is most likely using this.balance with this.dollar as this question an have up to two answers.

Answer 17

dollar is a static variable, can you use "this" to access static variables?
balance is an instance variable, can you access instance variables in static methods?

What did i write in my previous posts?

Answer 18

No,
Then the only answer remains is this.balance

Answer 19

balance is an instance variable, can you access instance variables in static methods?

Answer 20

Yes,

Answer 21

"... An instance variable (non-static) can however only be accessed from non-static methods."

Answer 22

My method is static.... ,
it looks like none of the answers are correct but I can't leave this question blank.

Answer 23

Also, I have figured out question number 3 while the site was glitching. The answer to number 3 is 42.

Answer 24

How about "E. none of the above"?

As for the third I think a part of the question is missing?

Answer 25

I figured it out, but the first one has none of the above. Can you check my latest answer for question number 4?
my latest answer is : if (x<= 999 && >= 100);

Answer 26

4 Is almost correct, there's a x missing.

Answer 27

5 is almost correct. Remember that hoursWorked contains all hours in the week including the 40 you've already calculated with 40*ratePerHour

Answer 28

@Lyrae

Answer 29

I still can't figure thatone out. Ican say hoursWorked/2for half time

Answer 30

whatdo you think of my answer so far? for 5

Answer 31

Nope, you are not there quite yet.

Answer 32

I also thought on this, and 5 % can be put as 0.05

hoursWorked + 0.05 and then

Answer 33

can this work with gross pay calculation?

Answer 34

how would you start calculating gross pay using what the question gives?

Answer 35

hoursWorked contains all hours worked in the week including the initial 40.
in the first expression hoursWorked * ratePerHour you calculate the pay for the first 40 hours.

What you are supposed to enter is how many excess hours the person worked.
So if a person works 48 hours the excess time is 8 hours.
or if another person works 43 hours the excess time is three hours.'

How do you calculate the excess time if the total time is in hoursWorked?

Answer 36

I don't know but the best thing to do is to do 40/5

Answer 37

48 - 40 = 8
43 - 40 = 3
total time - 40 = excess time

Answer 38

so, hoursWorked -40

Answer 39

yup

Answer 40

This one is hard too...

Complete this statement to round a given positive double value to the nearest integer value.
int rounded = _______ _________ _________ (given + 0.5);

my answer was 1. But I need help on understanding what I have to do on this one.

Answer 41

1. ?

Answer 42

I thought of the math rule 5 or more raise it high. so I thought 0.5 would round to 1

Answer 43

The correct answer have to do with casting. When a double is casted to an integer all decimals will be lost.

For example
```java
double a = 4.2,
b = 2.9;

int c = (int) a;
int d = (int) b;

System.out.println("c="+c+", d="+ d);
```
prints
```
c=4, d=2
```
Java don't round anything automatically or mathematically.

Answer 44

so, int rounded = (int) given +0.5

I also found a few other things that we can try:
Math.ceil (x)
(int) round (float a)

Answer 45

Yah, all of those should work.

Answer 46

can you check my answer please?
I wantedto put int rounded = (int) given + 0.5.Whatwould you put?

Answer 47

That method is valid. You just might want to put some other paren in there:

int rounded = (int) (given + 0.5)

See, that makes sure it is the sum that gets cast to an int.

Answer 48

okay :). I get it.
Iwanted to ask if you know anything about this:

Complete this coding to compute the balance at the end of 10 years when you put $1000 in a
savings account at the end of each year and earn 5% interest annually. Use only one arithmetic
operator in your answer.

Answer 49

I tried something and I ended up doing (0.05 * 10). How do I calculate the balance? thequestion is confusing

Answer 50

Well, the arithmetic operators ar /*+-. If you can only use one of them, you need to find out which one. However, algebra can help here. It is the same idea as:

5x+x = 6x

If I show that the long way:

5x+x factor out the x:
x(5+1)
x(6)
6x

Well, you can do the same thing with:
interest = ballance * .08
ballance + interest = new balance

And loop it 10 times.

So
interest = ballance * .08
balance += interest

But that still has two operators....

However, if I sub in for interest:

balance += ballance * .08

Still two, but now it looks a lot more like the x(5+1) and can be simplified more...

Answer 51

so, in my case, Ican just do this: balance = (balance *0.05) + 1000 ?

Answer 52

the 0.05 represents 5%

Answer 53

the x(5+1) can be x(6)

Answer 54

Well, 1000 is the original ballance. This wil compound every year, so:

balance = (balance *0.05) + balance

Answer 55

Do you see how to get rid of the + now?

Answer 56

Yes, I see. I can say that (1000 * 0.05) could be the answer, and we don't need the balance as our original is 1000

Answer 57

1000 * 0.05 = 50

Answer 58

Um, not quite. Lets just take this:

balance *0.05 + balance

What do I get if I factor out ballance?

Answer 59

if you do
balance *0.05 + balance
-balance - balance
_________________________
0.05

if the balance is factored out, we have just plain 0.05

Answer 60

No. To factor is not to subtract, but to divide:

\(\dfrac{x}{x}=1\)

I will just use b for ballance:

\(b*.05+b\)

\(b\left(\dfrac{b*.05}{b}+\dfrac{b}{b}\right)\)

Do you see how that would cancel, leaving a lot of 1s?

Answer 61

yes, I see it.
I have b (0.05)
It seems like the ( means to multiply .

Answer 62

No, you dropped part still...

It would simplify to:

b(1*.05+1)

Now, the multiply by 1 is simple, that is just itself:

b(.05+1)

But what happens to the .05+1?

Answer 63

it becomes 1.05

b (1.05)

Answer 64

=) Yah.

so in code:

ballance = ballance * 1.05

Or:
ballance *= 1.05

Answer 65

And that is one operator...

Answer 66

okay :)
Thank you so much for the help.
Will you be online later today?

Answer 67

I have a homework piece with fill in the blank coding. If I need any help, I will be online.

Answer 68

Well, I may or may not. Never know.

Answer 69

okay:)
see you next time! :)

Answer 70

A quick note on:
"so, int rounded = (int) given +0.5

I also found a few other things that we can try:
Math.ceil (x)
(int) round (float a)"

Math.ceil(double x) works if you subtract five (half down rounding). To get a similar effect of casting to int you have to use Math.floor() (half up rounding). This won't however help you as the java floor, ceil and round functions returns doubels and not ints.

If you try to substitute
```
int rounded = (int) given +0.5
```
with
```
int rounded = Math.floor(given +0.5);
```
the compiler will most likely say something like
```
Main.java:6: error: possible loss of precision
int rounded = Math.floor(given+0.5);
^
required: int
found: double
```

The compiler is essentially hinting that assigning a double to int will result in loss of decimals. To fix this you have cast the result to an int before assigning it.
```
int rounded = (int) Math.floor(given +0.5);
```
However since the result is same as if you didn't use Math.floor(), I don't see any reason to to use it.

The best option (if possible) is to use the built in Math.round(double x)
```
int rounded = (int) Math.round(given);
```
it deals with some special cases, not covered by the casting trick.

Answer 71

Thank you, :)
Thank you Lyrae and emccormick :)