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Bezzdna [24]
3 years ago
11

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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Write y=1/6x+7 in standard form using intergers.​
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Answer:

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Find the tangent plane to the given surface of f(x,y)=6- 6/5 x-y at the point (5, -1, 1). Make sure tat your final answer for th
HACTEHA [7]

Answer:

Required equation of tangent plane is z=\frac{6}{5}(x-5y-11).

Step-by-step explanation:

Given surface function is,

f(x,y)=6-\frac{6}{5}(x-y)

To find tangent plane at the point (5,-1,1).

We know equation of tangent plane at the point $(x_0,y_0,z_0)[/tex] is,

z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)\hfill (1)

So that,

f(x_0,y_0)=6-\frac{6}{5}(5+1)=-\frac{6}{5}

f_x=-\frac{6}{5}y\implies f_x(5,-1,1)=\frac{6}{5}

f_y=-\frac{6}{5}x\implies f_y(5,-1,1)=-6

Substitute all these values in (1) we get,

z=\frac{6}{5}(x-5)-6(y+1)-\frac{6}{5}

\therefore z=\frac{6}{5}(x-5y-11)

Which is the required euation of tangent plane.

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I. Tell whether each statement is TRUE or FALSE. Refer to the figure below. Write your answer on the space provided.
Vika [28.1K]

Answer:

1. TRUE

2. TRUE

3. TRUE

4. TRUE

5. FALSE

6. TRUE

Step-by-step explanation:

1. From the diagram, we have that m∠2 and m∠7 are alternate interior angles

If m∠2 = 70° and m∠7 = 70°, then we have;

m∠2 = 70° = m∠7

m∠2 = m∠7

Therefore, the alternate interior angles of the lines l₁, and l₂ are equal, and therefore, the lines l₁ and l₂ are parallel, l₁ ║ l₂

TRUE

2. If m∠3 = 90° and m∠7 = 90°, therefore, the angle formed by the intersection of l₃ and l₂ = 90°

Therefore l₃ ⊥ l₂

TRUE

3. m∠5 and m∠7 are corresponding angles

If m∠5 = 85° and m∠7 = 85°, then, m∠5 = m∠7

Therefore, the corresponding angles formed by the lines l₁ and l₂ are equal, therefore;

l₁ ║ l₂

TRUE

4. Whereby we have, m∠1 = m∠5, we get;

m∠1 + m∠5 = 180° by sum of angles on a straight line

∴ m∠1 + m∠5 = m∠1 + m∠1 = 2·m∠1 = 180°

m∠1 = 180°/2 = 90°

∴ m∠1 = 90° = m∠5, and l₃ ⊥ l₁

TRUE

5. m∠1 and m∠8 are alternate exterior angles

If m∠1 = 98° and m∠8 = 82°

∴ m∠1 ≠ m∠8 and l₁ ∦ l₈

FALSE

6. Given that l₁║ l₂, then the angle formed between l₁ and l₃ will be equal to th angle formed between l₂ and l₃

Therefore;

If l₁║ l₂, and l₃ ⊥ l₁, then l₃ ⊥ l₂

(If l₁║ l₂, and l₃ is perpendicular to l₁, then l₃ is also perpendicular to l₂)

TRUE

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