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3 years ago

Easy question

Mathematics

2 answers:

bazaltina [42]3 years ago

6 0

If for $x$ cumulative frequency the marks are $y$ (as given on the graph), we have $x$ students with marks equal to or less than $y$

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.

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.

Anarel [89]3 years ago

5 0

Answer:

d)33

e)80

Step-by-step explanation:

I'm sorry I don't really know how to explain this.

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The coordinates of the endpoints of AB and CD are A(2,

Ratling [72]

Answer:

Option 1: CD is a perpendicular bisector of AB

Step-by-step explanation:

Let us find out the slopes of various line segments and the Distances and then we will draw the conclusions accordingly.

Formula to find slope

$m= \frac{y_2-y_1}{x_2-x_1}$

Formula to Find Distance between two points

$D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

mAB ( represents , Slope of AB )

1. $mAC= \frac{3-2}{2-5}=\frac{1}{-3}=-\frac{1}{3}$

2. $mBC=\frac{2-1}{5-8}=\frac{1}{-3}=-\frac{1}{3}$

3. $mCD=\frac{5-2}{6-5}=\frac{3}{1}=3$

4. $AC=\sqrt{(3-2)^2+(2-5)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

5. $BC=\sqrt{(2-1)^2+(5-8)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

mAC = mBC , and C is common point , hence these three are collinear points making a straight line whole slope is $-\frac{1}{3}$

$mAB=-\frac{1}{3}$

$mCD=3$

$mAB \times mCD = -\frac{1}{3} \times 3 = -1$

Hence CD ⊥ AB

Also

From Point 4 and point 5 above , we see that

AC = CB

Hence CD bisect AB at C, also CD ⊥ AB

There fore

CD is a perpendicular bisector of AB

Therefor option 1 is true

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Answer:

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Step-by-step explanation:

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