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.

where are the choices?

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00