All categories

3 years ago

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

Mathematics

1 answer:

Vinvika [58]3 years ago

5 0

Answer:

The data in shelter A show more spread than the data in shelter B.

Step-by-step Explanation:

Judging from the data set given for shelter A and shelter B in the diagram of the box plots attached below, shelter A has a data set that ranges from 8 pounds to 30 pounds, while shelter B ranges from e can infer that the data in shelter A has more spread than the data in shelter B ranges from 10 to 28.

Taking into consideration the range of the data set of each shelter, shelter A obviously has a larger range compared to shelter B. This implies that, the data in shelter A show more spread than the data in shelter B.

You might be interested in

Find the slope of the line running<br> through these two points (-4,5) and (8,-3)

Artemon [7]

Answer:

The slope is

$\frac{-3-5}{8-(-4)} =\frac{-8}{8+4} =\frac{-8}{12} =-\frac{2}{3}$

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

the product of two consecutive integers n and n+1 is 42. what is the positive integer that satisfies the situation?

MrMuchimi

$n(n+1)=42 \\
n^2+n=42 \\
n^2+n-42=0 \\ \\
a=1 \\ b=1 \\ c=-42 \\ b^2-4ac=1^2-4 \times 1 \times (-42)=1+168=169 \\ \\
n=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-1 \pm \sqrt{169}}{2 \times 1}=\frac{-1 \pm 13}{2} \\
n=\frac{-1 -13}{2} \ \lor \ n=\frac{-1+13}{2} \\
n=\frac{-14}{2} \ \lor \ n=\frac{12}{2} \\
n=-7 \ \lor \ n=6$

The positive integer is 6.
The consecutive integers are 6 and 7.

8 0

3 years ago

A. ASA<br> B. CPCTC<br> C. AAS<br> D. SAS

Ede4ka [16]

Answer:

It is ASA congruence rule as the 2 angles and the included side of the triangle are equal

6 0

3 years ago

Read 2 more answers

Please help me! Thank you

Yuki888 [10]

Answer:

Anne

Step-by-step explanation:

The expression cannot be simplified anymore, but it can be laid out correctly.

The greatest number for the exponents should come first, so Anne is the one who laid it out correctly.

Hope it helps!!

Let me know if I'm wrong...

3 0

3 years ago

How do I solve

Maslowich

$Use:\\\\(ab)^n=a^nb^n\\\\(a^n)^m=a^{nm}\\\\a^n\cdot a^m=a^{n+m}\\\\\sqrt[n]{a}=a^\frac{1}{n}\\------------------------\\\\\left(16n^4\right)^{\frac{5}{4}}=16^\frac{5}{4}\left(n^4\right)^\frac{5}{4}=16^{1\frac{1}{4}}n^{4\cdot\frac{5}{4}}=16^{1+\frac{1}{4}}n^5=16^1\cdot16^\frac{1}{4}\cdot n^5\\\\=16\cdot\sqrt[4]{16}\cdot n^5=16\cdot2\cdot n^5=\boxed{32n^5}\\\\\sqrt[4]{16}=2\ because\ 2^4=16\\\\\text{Answer}\ \boxed{\left(16n^4\right)^\frac{5}{4}=32n^5}$

4 0

3 years ago