Answer:
def typeHistogram(it,n):
d = dict()
for i in it:
n -=1
if n>=0:
if str(type(i).__name__) not in d.keys():
d.setdefault(type(i).__name__,1)
else:
d[str(type(i).__name__)] += 1
else:
break
return list(d.items())
it = iter([1,2,'a','b','c',4,5])
print(typeHistogram(it,7))
Explanation:
- Create a typeHistogram function that has 2 parameters namely "it" and "n" where "it" is an iterator used to represent a sequence of values of different types while "n" is the total number of elements in the sequence.
- Initialize an empty dictionary and loop through the iterator "it".
- Check if n is greater than 0 and current string is not present in the dictionary, then set default type as 1 otherwise increment by 1.
- At the end return the list of items.
- Finally initialize the iterator and display the histogram by calling the typeHistogram.
Answer:
876100
019343
Explanation:
10s complement of a decimal number is obtained by the following process:
- Obtain 9s complement ( Subtract each digit by 9)
- Add 1 to the result
1) 123900
9s complement => (9-1)(9-2)(9-3)(9-9)(9-0)(9-0)
= 876099
Adding 1 , 10s complement of 123900 = 876100
2) 980657
9s complement = (9-9)(9-8)(9-0)(9-6)(9-5)(9-7)
= 019342
Adding 1 , 10s complement of 980657 = 019343
Answer:
a. Power cycle the printer.
Explanation:
Power Cycle: To unplug the printer and restart it, is called power cycling. The peripheral devices often tend to stop working and the cause of this is not always easily figured out. So the first and easiest way that can be done to fix this issue is to run a power cycle. For this, you have to turn off the printer and unplug it. Then wait for a few seconds (at least 30) and plug in the printer again. Turn the printer on. Power cycle helps to resolve many basic issues. It is the easiest step to take before checking the printer cable, reinstalling printer or resetting the print spooler.
False. you can make different slides have different transitions
Answer:
#include<iostream>
#include<iomanip>
using namespacestd;
int main ()
{
int x1[3][3]={1,2,3,4,5,6,7,8,9};
int x2[3][3];
int i,j;
for(i=0;i<3;i++)
for(j=0;j<3;j++)
x2[i][j] = x1[i][j];
cout<<"copy from x1 to x2 , x2 is :";
for(i=0;i<3;i++)
for(j=0;j<3;j++)
cout<<x2[i][j]<<" ";
cout<<endl;
system("pause");
return 0;
}
/* Sample output
copy from x1 to x2 , x2 is :1 2 3 4 5 6 7 8 9
Press any key to continue . . .
*/
Explanation:
