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3 years ago

10

A data set is made up of these values. 4, 6, 7, 8, 9, 12, 15 Find the interquartile range. O A. 6 - 4 = 2 O B. 12- 6 = 6 O C. 15

- 4 = 11 O D. 15 - 8 = 7

1 answer:

galina1969 [7]3 years ago

3 0

Answer:

The answer is b

Step-by-step explanation:

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ILL MARK BRAINLIEST IF YOU HELP ME PLSS

-BARSIC- [3]

Answer:

Angle 5 = 35 degrees

Step-by-step explanation:

A straight line is equal to 180 degrees, meaning 1 + 2 + 5 = 180 degrees because they make up a straight line.

Angle 2 = 55 degrees

Angle 1 = 90 degrees (because it is a right angle)

Angle 5 = ?

Set up an equation:

90 + 55 + angle 5 = 180

145 + angle 5 = 180

Angle 5 = 35

4 0

3 years ago

write an equation in point-slope form for the perpendicular bisector of the segment with endpoints at A(-2,2) and B(5,4)

dimaraw [331]

The equation in point-slope form for the perpendicular bisector of the segment with endpoints at A(-2,2) and B(5,4) is $y - 3 = \frac{-7x}{2}+ \frac{21}{4}$

<h3><u>Solution:</u></h3>

Given that we have to write equation in point-slope form for the perpendicular bisector of the segment with endpoints at A(-2,2) and B(5,4)

Let us first find the slope of given line AB

<em><u>The slope "m" of the line is given as:</u></em>

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Here the given points are A(-2,2) and B(5,4)

$\text {Here } x_{1}=-2 ; y_{1}=2 ; x_{2}=5 ; y_{2}=4$

$m=\frac{4-2}{5-(-2)}=\frac{2}{7}$

Thus the slope of line with given points is $\frac{2}{7}$

We know that product of slopes of given line and slope of line perpendicular to given line is always -1

$\begin{array}{l}{\text {slope of given line } \times \text { slope of perpendicular bisector }=-1} \\\\ {\frac{2}{7} \times \text { slope of perpendicular bisector }=-1} \\ \\{\text {slope of perpendicular bisector }=\frac{-7}{2}}\end{array}$

The perpendicular bisector will run through the midpoint of the given points

So let us find the midpoint of A(-2,2) and B(5,4)

<em><u>The midpoint formula for given two points is given as:</u></em>

$\text {For two points }\left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right), \text { midpoint } \mathrm{m}(x, y) \text { is given as }$

$m(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

Substituting the given points A(-2,2) and B(5,4)

$m(x, y)=\left(\frac{-2+5}{2}, \frac{2+4}{2}\right)=\left(\frac{3}{2}, 3\right)$

Now let us find the equation of perpendicular bisector in point slope form

The perpendicular bisector passes through points (3/2, 3) and slope -7/2

<em><u>The point slope form is given as:</u></em>

$y - y_1 = m(x - x_1)$

$\text { Substitute } \mathrm{m}=\frac{-7}{2} \text { and }\left(x_{1}, y_{1}\right)=\left(\frac{3}{2}, 3\right)$

$y - 3 = \frac{-7}{2}(x - \frac{3}{2})\\\\y - 3 = \frac{-7x}{2}+ \frac{21}{4}$

Thus the equation in point-slope form for the perpendicular bisector of the segment with endpoints at A(-2,2) and B(5,4) is found out

7 0

4 years ago

Sam is a manager in an office

masya89 [10]

Answer:

Sorry but this is not really a question so we can't answer it

4 0

3 years ago

What is the minimum value for g(x)=x^2-10x+16? Enter your answer in the box.

Anna71 [15]

Answer:

- 9

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

To obtain this form use the method of completing the square

add/subtract (half the coefficient of the x- term)² to x² - 10x

g(x) = x² - 10x + (- 5)² - (- 5)² + 16

= x² - 10x + 25 - 25 + 16

= (x - 5)² - 9 ← in vertex form

vertex = (5, - 9), hence

minimum value = - 9

6 0

3 years ago

Solve the equation below for x. 100 = 25e3 In4 O A. 4 B. 3 e O C. X=In4 - 3 c In100 3. In25 O D.

Ray Of Light [21]

Dividing the given equation by 25 we get:

$\begin{gathered} \frac{100}{25}=\frac{25e^{3x}}{25}, \\ 4=e^{3x}\text{.} \end{gathered}$

Applying the natural logarithm to the above equation we get:

$\ln (4)=\ln (e^{3x})=3x\ln (e)=3x\text{.}$

Finally, solving for x we get:

$x=\frac{\ln (4)}{3}\text{.}$

Answer: Option A.

8 0

1 year ago