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2 years ago

Pls help, will give brainliest if your answer is right! Question in image.

Mathematics

1 answer:

Blababa [14]2 years ago

7 0

Step-by-step explanation:

Mean = Average : Sum of the data ÷ Number of data

So find the average number of lates per student

4 = Lates

30 (sum of frequencies) = Frequencies

So 4 ÷ 30 = 7.5

Therefore the answer is 7.5..

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2 years ago

These two scalene triangles are similar with a scale of 2:3

yarga [219]

Answer:

the answer is PQ/GH=3/2

Step-by-step explanation:

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3 years ago

What is 4 1/3 divided by 3

kipiarov [429]

Answer:

13/9 or 1 4/9

Step-by-step explanation:

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2 years ago

A family has two cars. The first car has a fuel efficiency of 40 miles per gallon of gas and the second has a fuel efficiency of

kozerog [31]

Answer:

Step-by-step explanation:

Let x = gallons consumed by car 1

Let y = gallons consumed by car 2

We set up our equations:

35x+40y = 1850 eq1

x+y = 50 eq2

Substituting eq 2 into eq 1,

35x+40(50-x) = 1850

35x+2000-40x = 1850

-5x+2000 = 1850

-5x = -150

x = 30

Substitute value of x into eq 2.

x+y = 50

30+y = 50

y = 20

Car 1 consumes 30 gallons of gas.

Car 2 consumes 20 gallons of gas.

6 0

3 years ago

The set of points (-3, 4), (-1, 1), (-3,-2), and (-5,1) identifies the vertices of a quadrilateral. Which is the most specific d

KIM [24]

Answer:

Option (4). Rhombus

Step-by-step explanation:

From the figure attached,

Distance AB = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

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

= $\sqrt{(-3)^2+(-2)^2}$

= $\sqrt{13}$

Distance BC = $\sqrt{(4-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance CD = $\sqrt{(-2-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance AD = $\sqrt{(1+2)^2+(-5+3)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Slope of AB ( $m_{1}$ ) = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

= $\frac{4-1}{-3+5}$

= $\frac{3}{2}$

Slope of BC ( $m_{2}$ ) = $\frac{4-1}{-3+1}$

= $-\frac{3}{2}$

If AB and BC are perpendicular then,

$m_{1}\times m_{2}=-1$

But it's not true.

[ $m_{1}\times m_{2}=(\frac{3}{2})(-\frac{3}{2})$ = - $\frac{9}{4}$ ]

It shows that the consecutive sides of the quadrilateral are not perpendicular.

Therefore, ABCD is neither square nor a rectangle.

Slope of diagonal BD = $\frac{4+2}{-3+3}$

= Not defined (parallel to y-axis)

Slope of diagonal AC = $\frac{1-1}{-1+5}$

= 0 [parallel to x-axis]

Therefore, both the diagonals AC and BD will be perpendicular.

And the quadrilateral formed by the given points will be a rhombus.

5 0

3 years ago