All categories

3 years ago

Determine the interquartile range for the given data 11, 13, 16, 17, 8, 10, 14, 14, 17, 14

Mathematics

1 answer:

Elanso [62]3 years ago

7 0

Answer:

(IQR) interquartile range: 4.5!

Step-by-step explanation:

8, 10, 11, 13, 14, 14, 14, 16, 17

Median: 14

Lower quartile: 10.5

Upper quartile: 15

Interquartile range: 15 - 10.5 = 4.5

You might be interested in

A factory wants to redesign a box that will hold twice as many unit cubes add a box that measures 2 by 3 by 4 unit cubes.What ar

iragen [17]

finding the volume of cube is equal to 2 × 3 × 4

v of cubes is 24

now

twice of it is 2 ×24

8 0

3 years ago

FIND THE INDICATED PROBABILITY FOR THE FOLLOWING:

luda_lava [24]

The value of the probability P(A) is 0.40

<h3>How to determine the probability?</h3>

The given parameters about the probability are

P(A or B) = 0.6

P(B) = 0.3

P(A and B) = 0.1

To calculate the probability P(A), we use the following formula

P(A and B) = P(A) + P(B) - P(A or B)

Substitute the known values in the above equation

0.1 = 0.3 + P(A) - 0.6

Collect the like terms

P(A)= 0.1 - 0.3 + 0.6

Evaluate the expression

P(A)= 0.4

Hence, the value of the probability P(A) is 0.40

Read more about probability at

brainly.com/question/25870256

#SPJ1

3 0

2 years ago

Fine length of BC on the following photo.

MrMuchimi

Answer:

$BC=4\sqrt{5}\ units$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

step 1

In the right triangle ACD

Find the length side AC

Applying the Pythagorean Theorem

$AC^2=AD^2+DC^2$

substitute the given values

$AC^2=16^2+8^2$

$AC^2=320$

$AC=\sqrt{320}\ units$

simplify

$AC=8\sqrt{5}\ units$

step 2

In the right triangle ACD

Find the cosine of angle CAD

$cos(\angle CAD)=\frac{AD}{AC}$

substitute the given values

$cos(\angle CAD)=\frac{16}{8\sqrt{5}}$

$cos(\angle CAD)=\frac{2}{\sqrt{5}}$ ----> equation A

step 3

In the right triangle ABC

Find the cosine of angle BAC

$cos(\angle BAC)=\frac{AC}{AB}$

substitute the given values

$cos(\angle BAC)=\frac{8\sqrt{5}}{16+x}$ ----> equation B

step 4

Find the value of x

In this problem

$\angle CAD=\angle BAC$ ----> is the same angle

equate equation A and equation B

$\frac{8\sqrt{5}}{16+x}=\frac{2}{\sqrt{5}}$

solve for x

Multiply in cross

$(8\sqrt{5})(\sqrt{5})=(16+x)(2)\\\\40=32+2x\\\\2x=40-32\\\\2x=8\\\\x=4\ units$

$DB=4\ units$

step 5

Find the length of BC

In the right triangle BCD

Applying the Pythagorean Theorem

$BC^2=DC^2+DB^2$

substitute the given values

$BC^2=8^2+4^2$

$BC^2=80$

$BC=\sqrt{80}\ units$

simplify

$BC=4\sqrt{5}\ units$

7 0

3 years ago

Give all relationships between angle 1 and angle 2 select all that apply

kolbaska11 [484]

Answer:

vertical maybe if try it and see

6 0

2 years ago

A block of ice has a square top and bottom and rectangular sides. At a certain point in

Lilit [14]

Answer:

$\frac{dV}{dt} = 360 \ cm^3/h$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Geometry

Volume of a Rectangular Prism: V = lwh

Calculus

Derivatives

Derivative Notation

Differentiating with respect to time

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

Step 1: Define

$l = 30 \ cm\\w = l\\\frac{dl}{dt} = 2 \ cm/h\\h = 20 \ cm\\\frac{dh}{dt} = 3 \ cm/h$

Step 2: Differentiate

Rewrite [VRP]: $V = l^2h$
Differentiate [Basic Power Rule]: $\frac{dV}{dt} = 2l\frac{dl}{dt} \frac{dh}{dt}$

Step 3: Solve for Rate

Substitute: $\frac{dV}{dt} = 2(30 \ cm)(2 \ cm/h)(3 \ cm/h)$
Multiply: $\frac{dV}{dt} = 360 \ cm^3/h$

Here this tells us that our volume is decreasing (ice melting) at a rate of 360 cm³ per hour.

3 0

3 years ago