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3 years ago

The box plots show the weights, in pounds, of the dogs in two different animal shelters.

Mathematics

2 answers:

Dmitriy789 [7]3 years ago

7 0

Answer:

The median weight for shelter A is greater than that for shelter B.

The data for shelter B are a symmetric data set.

The interquartile range of shelter A is greater than the interquartile range of shelter B.

Step-by-step explanation:

The median weight for shelter A is greater than that for shelter B.

The median of A = 21 and the median of B = 18 true

The median weight for shelter B is greater than that for shelter A.

The median of A = 21 and the median of B = 18 false

The data for shelter A are a symmetric data set.

False, looking at the box it is not symmetric

The data for shelter B are a symmetric data set.

true, looking at the box it is symmetric

The interquartile range of shelter A is greater than the interquartile range of shelter B.

IQR = 28 - 17 = 11 for A

IQR for B = 20 -16 = 4 True

ANTONII [103]3 years ago

7 0

Answer:

The median weight for shelter A is greater than that for shelter B.

The data for shelter B are a symmetric data set.

The interquartile range of shelter A is greater than the interquartile range of shelter B.

Step-by-step explanation:

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Is 1/3 greater than 5/6

Harman [31]

1/3 or 5/6

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Calculations

1/3
1 ÷ 3 = 0.3333....

5/6
5 ÷ 6 = 0.83333...

0.83333... is greater then 0.3333.....
Therefore 5/6 is greater then 1/3
1/3 is smaller then 5/6

5 0

3 years ago

20 points!

riadik2000 [5.3K]

Its Your option B.
i.e. 1.325 * 10 ^ 6

3 0

3 years ago

There is a bag filled with 5 blue and 6 red marbles. A marble is taken at random from the bag, the colour is noted and then it i

il63 [147K]

Answer: 36/121

Step-by-step explanation:

There are 6 red marbles out of total 11.

The probability of drswing two red marbles is 6/11 x 6x11 = 36/121

4 0

3 years ago

Find , , and if and terminates in quadrant .

ioda

sin2x =12/13

cos2x = 5/13

tan2x = 12/5

STEP - BY - STEP EXPLANATION

What to find?

• sin2x

• cos2x

• tan2x

Given:

tanx = 2/3 = opposite / adjacent

We need to first make a sketch of the given problem.

Let h be the hypotenuse.

We need to find sinx and cos x, but to find sinx and cosx, first determine the value of h.

Using the Pythagoras theorem;

hypotenuse² = opposite² + adjacent²

h² = 2² + 3²

h² = 4 + 9

h² =13

Take the square root of both-side of the equation.

h =√13

This implies that hypotenuse = √13

We can now proceed to find the values of ainx and cosx.

Using the trigonometric ratio;

$\sin x=\frac{opposite}{\text{hypotenuse}}=\frac{2}{\sqrt[]{13}}$ $\cos x=\frac{adjacent}{\text{hypotenuse}}=\frac{3}{\sqrt[]{13}}$

And we know that tanx =2/3

From the trigonometric identity;

sin 2x = 2sinxcosx

Substitute the value of sinx , cosx and then simplify.

$\sin 2x=2(\frac{2}{\sqrt[]{13}})(\frac{3}{\sqrt[]{13}})$ $=\frac{12}{13}$

Hence, sin2x = 12/13

cos2x = cos²x - sin²x

Substitute the value of cosx, sinx and simplify.

$\begin{gathered} \cos 2x=(\frac{3}{\sqrt[]{13}})^2-(\frac{2}{\sqrt[]{13}})^2 \\ \\ =\frac{9}{13}-\frac{4}{13} \\ =\frac{5}{13} \end{gathered}$

Hence, cos2x = 5/13

tan2x = 2tanx / 1- tan²x

$\tan 2x=\frac{2\tan x}{1-\tan ^2x}$ $=\frac{2(\frac{2}{3})}{1-(\frac{2}{3})^2}$ $=\frac{\frac{4}{3}}{1-\frac{4}{9}}$ $=\frac{\frac{4}{3}}{\frac{9-4}{9}}$ $=\frac{\frac{4}{3}}{\frac{5}{9}}$ $=\frac{4}{3}\times\frac{9}{5}=\frac{4}{1}\times\frac{3}{5}=\frac{12}{5}$