What fraction times 24=12

Which statement best compares the spread of the data sets?

MrRa [10]

Answer:

Choice B is correct

Step-by-step explanation:

The Interquartile Range (IQR) for Florida, 11, is greater than the IQR for Australia, 4.

The spread of a data set is a measure of the dispersion or variability of the data set. The spread can be measured by various statistical quantities depending on the nature of the data (skewed or symmetric);

The IQR

Variance

Standard Deviation

Range

A box plot is a graphical representation of the five number summary;

The minimum, first quartile, median, third quartile, and the maximum value in that order.

The IQR is defined as;

third quartile - first quartile

With this definition, the Interquartile Range (IQR) for Florida is;

28 - 17 = 11

while the Interquartile Range (IQR) for Australia is;

14 - 10 = 4

Therefore, the Interquartile Range (IQR) for Florida, 11, is greater than the IQR for Australia, 4.

5 0

3 years ago

A ship sails 250km due North qnd then 150km on a bearing of 075°.1)How far North is the ship now? 2)How far East is the ship now

olga_2 [115]

Answer:

1) 288.8 km due North

2) 144.9 km due East

3) 323.1 km

4) 207°

Step-by-step explanation:

Bearing: The angle (in degrees) measured clockwise from north.

Trigonometric ratios

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Cosine rule

$c^2=a^2+b^2-2ab \cos C$

where a, b and c are the sides and C is the angle opposite side c

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

Draw a diagram using the given information (see attached).

Create a right triangle (blue on attached diagram).

This right triangle can be used to calculate the additional vertical and horizontal distance the ship sailed after sailing north for 250 km.

Question 1

To find how far North the ship is now, find the measure of the short leg of the right triangle (labelled y on the attached diagram):

$\implies \sf \cos(75^{\circ})=\dfrac{y}{150}$

$\implies \sf y=150\cos(75^{\circ})$

$\implies \sf y=38.92285677$

Then add it to the first portion of the journey:

⇒ 250 + 38.92285677... = 288.8 km

Therefore, the ship is now 288.8 km due North.

Question 2

To find how far East the ship is now, find the measure of the long leg of the right triangle (labelled x on the attached diagram):

$\implies \sf \sin(75^{\circ})=\dfrac{x}{150}$

$\implies \sf x=150\sin(75^{\circ})$

$\implies \sf x=144.8888739$

Therefore, the ship is now 144.9 km due East.

Question 3

To find how far the ship is from its starting point (labelled in red as d on the attached diagram), use the cosine rule:

$\sf \implies d^2=250^2+150^2-2(250)(150) \cos (180-75)$

$\implies \sf d=\sqrt{250^2+150^2-2(250)(150) \cos (180-75)}$

$\implies \sf d=323.1275729$

Therefore, the ship is 323.1 km from its starting point.

Question 4

To find the bearing that the ship is now from its original position, find the angle labelled green on the attached diagram.

Use the answers from part 1 and 2 to find the angle that needs to be added to 180°:

$\implies \sf Bearing=180^{\circ}+\tan^{-1}\left(\dfrac{Total\:Eastern\:distance}{Total\:Northern\:distance}\right)$

$\implies \sf Bearing=180^{\circ}+\tan^{-1}\left(\dfrac{150\sin(75^{\circ})}{250+150\cos(75^{\circ})}\right)$

$\implies \sf Bearing=180^{\circ}+26.64077...^{\circ}$

$\implies \sf Bearing=207^{\circ}$

Therefore, as bearings are usually given as a three-figure bearings, the bearing of the ship from its original position is 207°

8 0

2 years ago

Read 2 more answers

A road has a 10% grade, meaning increasing 1 unit of rise to every 10 units of run.

finlep [7]

Answer:

Part a) The elevation of the road is $6\°$

Part b) The rise is $0.2\ km$

Step-by-step explanation:

Part a) What is the elevation of the road to the nearest degree?

Let

y-----> the rise of the road ( vertical distance)

x ----> the run of the road (horizontal distance)

we have

y/x=1/10

we know that

The ratio y/x is equal to the tangent of the angle of the elevation of the road

Let

$\theta$ ----> angle of the elevation of the road

$tan(\theta)=y/x$

$tan(\theta)=1/10$

$\theta=arctan(1/10)=6\°$

Part b) If the road is two km long, how much does it rise?

using proportion

$1/10=y/2$

$y=2/10=0.2\ km$

3 0

3 years ago

Who wants to tal/k I'm bor/ed

MrMuchimi

Answer:

I talk

Step-by-step explanation:

I can talk anytime

5 0

3 years ago

Read 2 more answers

Need help with this please

Nastasia [14]

Answer:

158 correct

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers