What two factors make up the skills of a good typist

Computers and Technology

1 answer:

Vikki [24]3 years ago

5 0

Answer:

I feel to be a good typist one would have to have a good grasp of reading, grammar, and the ability to make errors to a lesser degree than more so. I can type, myself, 120-150 wpm with an error rate of 0-2. I also have the ability of not having to look at the keyboard as I type, and have a good eye on catching errors.

Explanation:

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Assume that cell F5 to F10 In a spreadsheet contains numeric value of salary earned by some workers and cell F12 contains the va

snow_tiger [21]

Answer:

=F5*$F$12+F5

Explanation:

If we want to increment the salaries in the cell F5, we must multiply the cell F5 by cell F12, and then we must sum that result.

If we want to drag the formula from the cell F5 to F10, we must use the dollar symbol $ to apply the same percent in our formula.

For example:

F12 = 5% = 0.05

F5 = 10,000

=F5*$F$12+F5

=10,000×0.05+10,000 = 10,500

5 0

3 years ago

a password to a certain database consists of digits that cannot be repeated. if the password is known to consist of at least 8 d

BaLLatris [955]

Answer:

$\frac{10!}{2}mins$

Explanation:

12 seconds to try one combination will be equivalent to $\frac{1}{12}\times 60 = \frac{1}{5} \ mins$

Password contain at least 8 digit i.e. password can contain 8, 9, 10 digit.

Password cannot contain more than 10 digit because it will give room for repetition which it is clearly stated that digit cannot be repeated.

Possible digit that can be used: 9,8,7,6,5,4,3,2,1,0.

Total number of passwords combination possible for each position in 8 digit.

1st position = 10, 2nd position = 9, 3rd position = 8, 4th position = 7, 5th position = 6, 6th position = 5, 7th position = 4, 8th position = 3. Total number of passwords combination possible in 8 digit is equivalent to $\frac{10!}{2}$ .

Total number of passwords combination possible for each position in 9 digit.

1st position = 10, 2nd position = 9, 3rd position = 8, 4th position = 7, 5th position = 6, 6th position = 5, 7th position = 4, 8th position = 3, 9th position = 2. Total number of passwords combination possible in 9 digit is equivalent to $\frac{10!}{1}$ .

Total number of passwords combination possible for each position in 10 digit.

1st position = 10, 2nd position = 9, 3rd position = 8, 4th position = 7, 5th position = 6, 6th position = 5, 7th position = 4, 8th position = 3, 9th position = 2, 10th position = 1. Total number of passwords combination possible in 10 digit is equivalent to 10!.

Adding them up and multiplying by $\frac{1}{5} \ mins$ to get the total number of time needed to guarantee access to database = $[\frac{10!}{2}\times\ \frac{10!}{1} \times 10!] \frac{1}{5}\ mins = \frac{10!}{2}$