Answer:
A) while (num >= 0)
Explanation:
To understand why we need to focus on the module and division operation inside the loop. num % 10 divide the number by ten and take its remainder to then add this remainder to sum, the important here is that we are adding up the number in reverse order and wee need to repeat this process until we get the first number (1%10 = 1), therefore, num need to be one to compute the last operation.
A) this is the correct option because num = 1 > 0 and the last operation will be performed, and after the last operation, num = 1 will be divided by 10 resulting in 0 and 0 is not greater than 0, therefore, the cycle end and the result will be printed.
B) This can not be the option because this way the program will never ends -> 0%10 = 0 and num = 0/10 = 0
C) This can not be the option because num = 1 > 1 will produce an early end of the loop printing an incomplete result
D) The same problem than C
E) There is a point, before the operations finish, where sum > num, this will produce an early end of the loop, printing an incomplete result
Answer:
1. General purpose file server.
2. Virtual Desktop Infrastructure Server.
3. Virtualized Backup Server.
All except : Database server
Explanation:
As a simple definition, we can tell, data deduplication is an elimination of redundant data in data set and storing only one copy of the same data. It is done by identifying double byte patterns through data analysis, removing double data and replacing it with reference pointed to stored, single piece of data.
Complete Question
The complete question is shown on the first uploaded image
Answer:
a) The elongation of the composite bar is given as δ = 0.072 in
b) The axial stress induced in each material is = 5485.7 psi
Explanation:
The explanation to the answer above is shown on the second uploaded image
Answer:
a) 25000 pcs/yr
b) cost per part produced is $ 1.84 per pc
c) Cpc = $ 1.365 per pcs
Explanation:
a)
the relation to calculate the number of parts produced annually by manual process is;
Q = Hw / Tc
Hw is the hourly rate ( 2000hr/yr) and Tc is the cycle time ( 4.8 min)
so we substitute
Q = (2000 × 60) / 4.8
= 120000 / 4.8
= 25000 pcs/yr
b)
cost per part produced
the relation to calculate the cost per part produced is expressed as;
Cpc = $23(Hw) / Q
Cpc is the cost per part produced
so we substitute
Cpc = $23(2000) / 25000
= 46000 / 25000
= $ 1.8 per pc
therefore cost per part produced is $ 1.84 per pc
c)
for the robot cell, at a service life of 4 years and a 10% rate of return , the factor is expressed as;
f = [r(1 + r)^t] / [((1 + r)^t ) - 1 ]
our rate r = 10% = 0.1 and our t = 4
so we substitute
f = [0.1 (1 + 0.1)^4] / [((1 + 0.1)^4 ) - 1 ]
f = 0.14641 / 0.4641
f = 0.3155
now we find the total cost
TC = 120000(0.3155) + 2000(0.3) + 2500
TC = $ 40,960
next we find the parts produced annually
Q = (2000 × 60) / 4
Q = 120,000 / 4
Q = 30000 pcs/yr
finally we find the cost per part produced;
Cpc = TC / Q
we substitute
Cpc = 40,960 / 30000
Cpc = $ 1.365 per pcs
