Answer:
the answer is 5.
given, 2c-7c+8=-17
or, -5c=-17-5
or, c=-25/-5
therefore the value of c is 5...
Let
be the random variable for the number of marks a given student receives on the exam.
10% of students obtain more than 75 marks, so

where
follows a standard normal distribution. The critical value for an upper-tail probability of 10% is

where
denotes the CDF of
, and
denotes the inverse CDF. We have

Similarly, because 20% of students obtain less than 40 marks, we have

so that

Then
are such that


and we find

19 percent of the people polled who preferred burgers were teachers.
Answer:
in my opinion anush has the better performance
The answer to your question is 7.4.