There's definitely something wrong with your for loop. Try this:
for x in range(n):
You might want to check if all your print statements are correct. You might be missing just a period and it will be counted wrong.
The device that provides ports to allow a laptop to easily connect to a full-sized monitor, keyboard, ac power adapter, and other peripheral devices is Docking Station.
Answer:
see explaination
Explanation:
a)
customerRecord.lastName
b)
customerPtr->lastName or (*customerPtr).lastName
c)
customerRecord.firstName
d)
customerPtr->firstName or (*customerPtr).firstName
e)
customerRecord.customerNumber
f)
customerPtr->customerNumber or (*customerPtr).customerNumber
g)
customerRecord.personal.phoneNumber
h)
customerPtr->personal.phoneNumber or (*customerPtr).personal.phoneNumber
i)
customerRecord.personal.address
j)
customerPtr->personal.address or (*customerPtr).personal.address
k)
customerRecord.personal.city
l)
customerPtr->personal.city or (*customerPtr).personal.city
m)
customerRecord.personal.state
n)
customerPtr->personal.state or (*customerPtr).personal.state
o)
customerRecord.personal.zipCode
p)
customerPtr->personal.zipCode or (*customerPtr).personal.zipCode
Answer:
Pay attention in your class.
Explanation:
If you paid attention in your class, you would have at least a basic understanding of the material.
Answer:
Replace the comment with:
(c) double hours = Math.abs (marker1 - marker2) / 60.0;
Explanation:
See attachment for right presentation of question
Analyzing the options:
(a): May return a negative result
This option will return a negative value if marker1 is less than marker2 because it subtracts the absolute value of marker2 from the absolute value marker1.
This literally is not different from marker1 - marker2
(b): Incorrect expression
This divides only marker2 by 60, then subtracts the quotient from marker1. This is not the expected expression.
(c) This option is correct
This correctly calculate the positive difference between marker1 and marker2 and the result is divided by 60.0 (note 60.0 not 60)
(d) & (e) Integer division
When a variable declared as double is divided by an integer variable or value, the result of the computation is not always accurate due to approximation
