Mathematics, 21.12.2019 09:31 DeanWhinchester8970

How do you find mean meadian and range?

Answers: 3

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Median in the middle number in an ordered sequence of numbers
Ex. 2,4,6,9,10,11,14

The median is 9 because it is in the middle of the ordered series of numbers

Range in simply the largest number minus the smallest number in an ordered sequence of numbers
Ex. 14 - 2 = 12

Answer from: dbih2014

To find the Mean add up all the numbers then divide by how many numbers there are. to find the Range take the highest number in the set and the lowest and subtract them and then the Meadian Put all the numbers in numerical order if theres an odd number mark off till you get to the middle if theres an even number of the set the two middle one you will add them both then subtract by 2.

Answer from: Quest

last option: rhombus or kite

step-by-step explanation:

a kite and rhombus have the same shape. as we don't have any specific formula for the calculation of area of kite. the diagonals are used to calculate the area of a kite or rhombus.

the formula for area of rhombus is given by:

$a = \frac{1}{2}d_{1} d_{2}$

here, d_1 is the first diagonal and d_2 is the second diagonal.

so,

last option is the correct answer ..

Answer for 15 points (look at the image)

Answer from: Quest

geometric series converge the common ratio r has magnitude less than one.

$|r| < 1$

which is equivalent to

$-1 < r < 1$

for the example we need

|2-x| < 1

-1 < 2-x and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1 diverges

1b. r=50/100=1/2 converges

when these converge they converge to

$s = \dfrac{a}{1 -r}$

where a is the first term.

$s = \dfrac{100}{1 - 1/2} = 200$

answer: 200

1c. r=-1 so it's not the case that |r| < 1. diverges

2a. x is the common ratio so we converge when |x|< 1

2b. r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c. r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

$|x| < \frac 1 2 \sqrt[3]{2}$

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