A computer password consists of six characters.a. How many different passwords are possible if each character may be any lowerca
1 answer:
Answer:
a)
b) 0.27778
Step-by-step explanation:
a)
There are 26 lowercase letters and 10 digits.
The character that must be a digit can be generated in 10 different ways and the others, in 26+10=36 different ways.
Since the password is 6 characters long, by the Fundamental Rule of Counting there are
10*36*36*36*36*36 = 604,661,760
In scientific notation this number can be expressed as
b)
By the <em>Fundamental Rule of Counting</em>, there would be
different combinations of 6 characters, but only the number computed in a) are valid combinations, so the probability of generating a valid password is
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Note that 6% converted to a decimal number is 6/100=0.06. Also note that 6% of a certain quantity x is 0.06x.
Here is how much the worker earned each year:
In the year 1985 the worker earned <span>$10,500.
</span>In the year 1986 the worker earned $10,500 + 0.06($10,500). Factorizing $10,500, we can write this sum as:
$10,500(1+0.06).
In the year 1987 the worker earned
$10,500(1+0.06) + 0.06[$10,500(1+0.06)].
Now we can factorize $10,500(1+0.06) and write the earnings as:
$10,500(1+0.06) [1+0.06]=
.
Similarly we can check that in the year 1987 the worker earned
, which makes the pattern clear.
We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:
.
Factorizing, we have
![=10,500[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]=10,500\cdot[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]](https://tex.z-dn.net/?f=%3D10%2C500%5B1%2B1.06%2B%281.06%29%5E2%2B%281.06%29%5E3%2B...%2B%281.06%29%5E%7B26%7D%5D%3D10%2C500%5Ccdot%5B1%2B1.06%2B%281.06%29%5E2%2B%281.06%29%5E3%2B...%2B%281.06%29%5E%7B26%7D%5D)
We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula
(where a is the first term and r is the common ratio) we have:
.
Finally, multiplying 10,500 by 59.17 we have 621.285 ($).
The answer we found is very close to D. The difference can be explained by the accuracy of the values used in calculation, most important, in calculating
.
Answer: D
Here is how much the worker earned each year:
In the year 1985 the worker earned <span>$10,500.
</span>In the year 1986 the worker earned $10,500 + 0.06($10,500). Factorizing $10,500, we can write this sum as:
$10,500(1+0.06).
In the year 1987 the worker earned
$10,500(1+0.06) + 0.06[$10,500(1+0.06)].
Now we can factorize $10,500(1+0.06) and write the earnings as:
$10,500(1+0.06) [1+0.06]=
Similarly we can check that in the year 1987 the worker earned
We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:
Factorizing, we have
We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula
Finally, multiplying 10,500 by 59.17 we have 621.285 ($).
The answer we found is very close to D. The difference can be explained by the accuracy of the values used in calculation, most important, in calculating
Answer: D