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.
Answer:
d)33
e)80
Step-by-step explanation:
I'm sorry I don't really know how to explain this.
You might be interested in
Answer:
1,2
-2 2
Step-by-step explanation:
Answer:
Step-by-step explanation:
a₂ = 9
a₃ = a₂-1.5
a₄ = a₃-1.5
a₅ = a₄-1.5
etc.
The terms have a common difference of -1.5, so the sequence is arithmetic.
Answer:
x=10, y=-1
Step-by-step explanation:
x-y=11
2x+y=19
3x=30
x=10
10-y=11
-y=1
y=-1
USED ELIMINATION BC THERE WAS ALREADY A Y AND -Y
5/2 = 2.5
4/16 = 0.25
2.5/0.25 = 10
Your answer would be 10
hope it helps!
Answer:
4 and 7
Step-by-step explanation:
X is a variable
any number that can be multiplied is a factor
