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3 years ago

13

What is the best way to display data if there is a wide range and we want to see individual data?

1 answer:

Vikki [24]3 years ago

7 0

Answer:

The correct option is;

b. Stem and leaf plot

Step-by-step explanation:

A stem-and-leaf plot shows the data shape and the densities of the different classes in a graphical form such that the information in the data such as the data characteristics are self displayed by the data numbers arranged in the stem-and-leaf plot. The stem-and-leaf plot keeps the raw numerical data, in the form they were obtained and by their arrangement, outliers are easily spotted.

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Here is a map of the Missouri Botanical Garden. Clare walked all the way around the garden

Karo-lina-s [1.5K]

Answer:

yk you could try searching it up

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3 years ago

The strip below is divided into 10 equal parts. Fill in the missing fractions and percentages.Note that your fractions don't nee

boyakko [2]

Answer:

1. First box, the missing fraction = 1/10

2. For the 2nd box

a. The missing fraction = 3/10

b. Percentage = 30%

3. For 7/10

Percentage = 70%

Step-by-step explanation:

From the question given above, the following data were obtained:

The strip is divided into 10 equal parts.

1. For the first box:

The missing fraction = 1/10

2. For the 2nd box

a. The missing fraction = 3/10

b. Percentage = 3/10 × 100 = 30%

3. For 7/10

Percentage = 7/10 × 100 = 70%

7 0

3 years ago

Simplify each expression.<br> -4.1 + 2n +1 +0.3n<br> What is the answer

Gnom [1K]

Answer: -4/5, in decimal form -0.8

Step-by-step explanation: −4.1+2n+1+0.3n

Combine Like Terms:

=−4.1+2n+1+0.3n

=(2n+0.3n)+(−4.1+1)

=2.3n+−3.1

=2.3n−3.1

7 0

2 years ago

(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

1.

$m^{-1}=\dfrac{1}{m}$

2.

$q^0=1$

3.

$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

4.

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

5.

$m^{-1}=\dfrac{1}{m}$

6.

$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

7.

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

8.

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

8 0

3 years ago

The set of ordered pairs below represents a relation that is a function.

Allushta [10]

Answer:

B (4,2)

Step-by-step explanation:

This coordinate pair would make the relation not a function because there would be two X values that are the same.

3 0

3 years ago

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