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3 years ago

Help giving brainliest

Mathematics

1 answer:

wolverine [178]3 years ago

8 0

Answer:

Q1: 12.5

Q3: 16

IQR: 3.5

Step-by-step explanation:

Mean : 14.25

Median : 13.5

Mode : 13, 12

Range: 6

Minimum :12

Maximum: 18

Count n: 8

Sum: 114

Quartiles Quartiles:

Q1 --> 12.5

Q2 --> 13.5

Q3 --> 16

Interquartile

Range IQR: 3.5

Outliers: none

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LCD of (2/x^2) - (7/4x^2+12x)?

svlad2 [7]

Answer:

4x³ + 12x²

Step-by-step explanation:

I assume the denominators are x² and 4x although that is not what you wrote.

x² = x² (already factored)

4x² + 12x = 4x(x + 3)

LCD = 4 × x² × (x + 3)

LCD = 4x³ + 12x²

8 0

2 years ago

Whats the difference between range and interquartile range?

miv72 [106K]

Answer:

Step-by-step explanation:

The range is the difference between the highest and the lowest values in a set of data. The interquartile range, the IQR, is what's "inside" the box in a box plot, which consists of the difference between the central measures, which are the first and the third quartiles.

3 0

3 years ago

30 POINTS PLUS BRAINLIEST PLEASE ANSWER THESE MATH QUESTIONS.ALL TROLL ANSWERS WILL BE REPORTED MUST HE DONE IN 1HR OR.LESS.

ipn [44]

Answer:

4) wall = 114 ft^2

$118.56 to paint

5) 77 cm^2

Step-by-step explanation:

4) (15 x 8)-(2 x 3)=120-6=114 ft^2

114 x $1.04 = $118.56

5) (14 x 3.5) + (8 x 3.5) = 49 + 28 = 77 cm^2

8 0

3 years ago

Read 2 more answers

PROBLEMA MATEMÁTICO:

Tema [17]

Answer:

Step-by-step explanation:

divide los números entre cuántos hermanos y cuántos euros

7 0

3 years ago

A rectangular parking lot has an area of 15,000 feet squared, the length is 20 feet more than the width. Find the dimensions

faust18 [17]

Dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet

<h3>Solution:</h3>

Given that

Area of rectangular parking lot = 15000 square feet

Length is 20 feet more than the width.

Need to find the dimensions of rectangular parking lot.

Let assume width of the rectangular parking lot in feet be represented by variable "x"

As Length is 20 feet more than the width,

so length of rectangular parking plot = 20 + width of the rectangular parking plot

=> length of rectangular parking plot = 20 + x = x + 20

The area of rectangle is given as:

$\text {Area of rectangle }=length \times width$

Area of rectangular parking lot = length of rectangular parking plot $\times$ width of the rectangular parking

$\begin{array}{l}{=(x+20) \times (x)} \\\\ {\Rightarrow \text { Area of rectangular parking lot }=x^{2}+20 x}\end{array}$

But it is given that Area of rectangular parking lot = 15000 square feet

$\begin{array}{l}{=>x^{2}+20 x=15000} \\\\ {=>x^{2}+20 x-15000=0}\end{array}$

Solving the above quadratic equation using quadratic formula

General form of quadratic equation is

${ax^{2}+\mathrm{b} x+\mathrm{c}=0$

And quadratic formula for getting roots of quadratic equation is

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

In our case b = 20, a = 1 and c = -15000

Calculating roots of the equation we get

$\begin{array}{l}{x=\frac{-(20) \pm \sqrt{(20)^{2}-4(1)(-15000)}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{400+60000}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{60400}}{2}} \\\\ {x=\frac{-(20) \pm 245.764}{2 \times 1}}\end{array}$

$\begin{array}{l}{=>x=\frac{-(20)+245.764}{2 \times 1} \text { or } x=\frac{-(20)-245.764}{2 \times 1}} \\\\ {=>x=\frac{225.764}{2} \text { or } x=\frac{-265.764}{2}} \\\\ {=>x=112.882 \text { or } x=-132.882}\end{array}$

As variable x represents width of the rectangular parking lot, it cannot be negative.

=> Width of the rectangular parking lot "x" = 112.882 feet

=> Length of the rectangular parking lot = x + 20 = 112.882 + 20 = 132.882

Hence can conclude that dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet.

3 0

3 years ago