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grandymaker [24]
3 years ago
13

How many 5-letter passwords can be created if no letter can be used twice within the password?

Mathematics
2 answers:
serious [3.7K]3 years ago
7 0
<span>If there are 5 letters used only, then 5! or 5*4*3*2 = 120
If all 26 letters of the alphabet may be used for the code, then:
26*25*24*23*22
</span><span>If you mean letters of the alphabet, and not case sensitive, it's</span><span>26*25*24*23*22
= 7893600</span>
bonufazy [111]3 years ago
5 0
26*25*24*23*22=7893600

or you can use

26!/(26-5)!=7893600
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A student's course grade is based on one midterm that counts as 15% of his final grade, one class project that counts as 15% of
kow [346]

Answer:

Overall final score = 77.75% ; Grade = C.

Step-by-step explanation:

The approach to solve this question is to realize that the marks have to be converted into the respective percentages of the whole course. This means that the marks of all the components have to be normalized according to the grading breakdown.

Project Marks = 97/100. Weightage = 15%. So 97*15/100 = 14.55/15.

This means that the student received 14.55 marks in the project out of 15.

Similarly for other components:

Mid-Term Marks = 83/100. Weightage = 15%. So 83*15/100 = 12.45/15.

Homework Marks = 82/100. Weightage = 35%. So 82*35/100 = 28.7/35.

Finals Marks = 63/100. Weightage = 35%. So 63*35/100 = 22.05/35.

After the conversion process, add up the normalized marks, which are now acting as the percentages earned in all the components.

Aggregate Percentage = 14.55 + 12.45 + 28.7 + 22.05 = 77.75%.

According to the grade scale, the student receives a C because 70 is less than 77.75 and 77.75 is less than 80.

Summarizing, the student receives a C at 77.75%!!!

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