Mathematics High School


Calculate measures of central tendency and
dispersion for the following:

Find the mean, median, midrange, mode, range, standard deviation and Quartile graph

Data: 54, 92, 101, 83, 45, 102, 99,76, 79, 85, 90, 100,
98, 71, 65, 63, 90, 88, 70, 60, 90, 58,81

Answers

Answer 1

Answer:

The mean is 80, the median is 83, and the mode is 90

Step-by-step explanation:

Related Questions

find the area of the parallelogram with the given information:​

Answers

Answer:

There is a area of triangle and there is two trapezium so 8×17=136

How many solutions does this equation have. 3x + 99 = 3 (x + 33)

Answers

Answer:

Infinitely many solutions

Step-by-step explanation:

3x + 99 = 3 (x+33)

distribute 3 to x and 33

3 * x = 3x

3 * 33 = 99

3x + 99 = 3x + 99

If the expressions on both sides of the equal sign are the same, then solving for them would result in 0 = 0 meaning that the equation has infinitely many solutions

Could somebody help n explain pls thx

Answers

Answer:

lines are perpendicular

Step-by-step explanation:

• Parallel lines have equal slopes

• The product of the slopes of perpendicular lines = - 1

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Rearrange the given equations into this form and extract the slope

3x - 2y = - 6 ( subtract 3x from both sides )

- 2y = - 3x - 6 ( divide through by - 2 )

y = [tex]\frac{3}{2}[/tex] x + 3 ← in slope- intercept form

with slope m = [tex]\frac{3}{2}[/tex]

----------------------------------

4x + 6y = 2 ( subtract 4x from both sides )

6y = - 4x + 2 ( divide through by 6 )

y = - [tex]\frac{4}{6}[/tex] x + [tex]\frac{2}{6}[/tex] , that is

y = - [tex]\frac{2}{3}[/tex] x + [tex]\frac{1}{3}[/tex] ← in slope- intercept form

with slope m = - [tex]\frac{2}{3}[/tex]

--------------------------------------

[tex]\frac{3}{2}[/tex] ≠ - [tex]\frac{2}{3}[/tex] , then lines are not parallel

[tex]\frac{3}{2}[/tex] × - [tex]\frac{2}{3}[/tex] = - 1

Thus the lines are perpendicular to each other

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