All categories

2 years ago

What is mean median mode range?

Mathematics

1 answer:

adoni [48]2 years ago

3 0

Answer:

Mean- the average of data, meaning you add up all the numbers, and then divide the sum of the numbers by how many numbers there are. For example:

2, 2, 3, 4, 6, 6, 9.

7 numbers total.

2 + 2 + 3 + 4 + 6 + 6 + 9 = 32

32 ÷ 7 = 4.57

The mean would be 4.57.

Median- the middle point of the numbers. You cross off each of the numbers until you have one last number standing. If you have an even number of numbers, you find the middle of the last two numbers you crossed off. For example:

2, 2, 3, 4, 6, 6, 9

2, 3, 4, 6, 6

3, 4, 6

4 would be the median.

Mode- the largest number on the data.

Range- subtract the mode (largest number) from the smallest number to get range. For example:

2, 2, 3, 4, 6, 6, 9

9 - 2 = 7

7 would be the range.

Step-by-step explanation:

You might be interested in

Who knows the area of this ? #struggleatmath

photoshop1234 [79]

I think it’s 6. There are 4 squares and 4 triangles. 2 triangles put together turn into a square, so that would make 2 extra squares. 4+2=6.

8 0

3 years ago

Thanks to whoever helped me out last time but I have another question.

yulyashka [42]

Answer:

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Find a polynomial P(x) with real coefficients having a degree 4, leading coefficient 4, and zeros 3-i and 4i.

Alisiya [41]

now, this polynomial has roots of 3-i and 4i, namely 3 - i and 0 + 4i.

let's bear in mind that a complex root never comes all by her lonesome, her sibling is always with her, the conjugate, so if 3 - i is there, 3 + i is also coming along, likewise if 0 + 4i is there, her sibling 0 - 4i is also there.

$\bf \begin{cases} x=3-i\implies &x-3+i=0\\ x=3+i\implies &x-3-i=0\\ x=4i\implies &x-4i=0\\ x=-4i\implies &x+4i=0 \end{cases}\\\\[-0.35em] ~\dotfill\\\\ (x-3+i)(x-3-i)(x-4i)(x+4i)=\stackrel{y}{0} \\[2em] \underset{\textit{difference of squares}}{[(x-3)+i][(x-3)-i]}\underset{\textit{difference of squares}}{[x-4i][x+4i]}=0$

$\bf [(x-3)^2-i^2][x^2-(4i)^2]=y\implies [(x-3)^2-(-1)][x^2-(4^2i^2)]=0 \\[2em] [(x-3)^2-(-1)][x^2-(16(-1))]=0\implies [(x-3)^2+1][x^2+16]=0 \\[2em] [(x^2-6x+9)+1][x^2+16]=y\implies (x^2-6x+10)(x^2+16)=0 \\\\\\ x^4-6x^3+10x^2+16x^2-96x+160=0 \\\\\\ x^4-6x^3+26x^2-96x+160=0 \\\\\\ \stackrel{\textit{multiplying both sides by 4}}{4(x^4-6x^3+26x^2-96x+160)=4(0)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 4x^4-24x^3+104x^2-384x+640=y~\hfill$

3 0

3 years ago

Write the slope intercept form of the equation of the line through the given points x intercept of -5 and y intercept of -1. Pls

soldier1979 [14.2K]

bearing in mind that an x-intercept is when the graph touches the x-axis and when that happens y = 0, and a y-intercept is when the graph touches the y-axis and when that happens x = 0.

$\bf \underset{x-intercept}{(\stackrel{x_1}{-5}~,~\stackrel{y_1}{0})}\qquad \underset{y-intercept}{(\stackrel{x_2}{0}~,~\stackrel{y_2}{-1})} ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-1}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{(-5)}}}\implies \cfrac{-1}{0+5}\implies -\cfrac{1}{5}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{-\cfrac{1}{5}}[x-\stackrel{x_1}{(-5)}] \\\\\\ y=-\cfrac{1}{5}(x+5)\implies y = -\cfrac{1}{5}x-1$

4 0

3 years ago

What is the area of the parallelogram shown below?

aivan3 [116]

Answer:135

Step-by-step explanation:

Area of parallelogram is base x height.

4 0

3 years ago