<span>the distribution is 43
</span>
The two numbers have 1 as a common factor and nothing else. Hence 1 is the HCF. This proves that the HCF of any two consecutive numbers is always a one.
Two integers are relatively prime (or coprime) if there is no integer greater than one that divides them both (that is, their greatest common divisor is one). For example, 12 and 13 are relatively prime, but 12 and 14 are not.
Plug x=3 into the equation:
5(3)+2
And simplify:
15+2
17
Answer:

Step-by-step explanation:
Given
--- rate of hard disk drives failure
--- number of hard disk drives
<em>See comment for complete question</em>
Required

First, calculate the probability that the none of the 4 selected is working;



Using the complement rule, the probability that at least 1 is working is:

This gives:


So you want to set up an equation for a weighted average. You know the final is 30% of the grade, so everything else is 70%. This gives you:
(Final)(.30) + (other grades)(.70) = course grade
The best grade the student can get would be if they get a hundred on the final, since that’s the best score you can make on the final. Then,
(100)(.30) + (82)(.70) = course grade
30 + 57.4 = course grade = 87.4 Which, If you round, the student would get an 87.
For the last part, we use the same equation, just filling in different parts.
(Final)(.30) + (other grades)(.70) = course grade
This time, we don’t know the grade for the final, but we know the course grade.
(Final)(.30) + (82)(.70) = 75
(Final)(.30) + 57.4 = 75
(Final)(.30) = 17.6
Final = (17.6)/(.30)
Final = 58.667 Which is approx a 59.