Easy question
2 answers:
If for cumulative frequency the marks are $y$ (as given on the graph), we have $x$ students with marks equal to or less than $y$
d)
Total students are $140$. Hence three-fifths of that is $140\times \frac{3}{5}=84$. Find the marks at $84$ in the graph for the passing marks.
e)
From the graph we have $60$ students having less than or equal to $30$ marks. Total students are $140$.
Subtracting, $140-60=80$ we get the numbee of students achieving more than 30 marks.
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9514 1404 393
Explanation:
Use the identities ...
csc(x) = 1/sin(x)
cot(x) = cos(x)/sin(x)
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