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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Answer:
(- 4, - 12 ) , (4, 12 )
Step-by-step explanation:
Given the 2 equations
y = 3x → (1)
y = x² + 3x - 16 → (2)
Substitute y = x² + 3x - 16 into (1)
x² + 3x - 16 = 3x ( subtract 3x from both sides )
x² - 16 = 0 ( add 16 to both sides )
x² = 16 ( take the square root of both sides )
x = ± = ± 4
Substitute these values into (1) for corresponding values of y
x = - 4 : y = 3 × - 4 = - 12 ⇒ (- 4, - 12 )
x = 4 : y = 3 × 4 = 12 ⇒ (4, 12 )