Easy question I need the answers only for (d) and (e) parts.
plz help
will give the brainliest
Need explanation.
Please answer correctly.

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.

start with a line segment pq that we will copy. mark a point r that will be one endpoint of the new line segment. set the compasses' point on the point p of the line segment to be copied. adjust the compasses' width to the point q. the compasses' width is now equal to the length of the line segment pq.

step-by-step explanation:

answer: 2530

hope this