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2 years ago

Give three examples of how scatterplots are used in the real world

Mathematics

2 answers:

Irina18 [472]2 years ago

5 0

Answer:

Scatter plots help visually illustrate relationships between two economic phenomena, such as employment and output, inflation and retail sales, and taxes and economic growth.

Step-by-step explanation:

I used a website for this, so answer this with your own words

Brut [27]2 years ago

3 0

Answer:

$\huge \text{\textbf{\textit{Hello! :)} }}$

<em>A scatterplot is a type of data chart that shows the relationship between two numerical variables. </em>

<em> Examples of how scatterplots are used in the real-life world include:↓</em>

<em />

<em>1. The correlation between the height of an individual and their weight.</em>

<em>2. The correlation number of hours spent by a student in watching movies and his exam scores.</em>

<em>3. In Economics, scatterplots are also applicable in the correlation between inflation and retail sales of commodities. </em>

Hope this helps!

Have a Wonderful Day!

$\bold{-InLoveWithNature-}$

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Ava and Kelly ran a road race, starting from the same place at the same time. Ava ran at an average speed of 6 miles per hour. K

Zepler [3.9K]

<h2>Hello!</h2>

The answer is:

Ava and Kelly will be 3/4 mile apart after 0.375 hours.

To calculate when will Ava and Kelly be 3/4 mile apart, we need to write the equations for both Ava's and Kelly's positions.

Writing the equations we have:

For Ava:

$x_A=x_o+v_ot\\\\x_A=0+v_o*t\\\\x_A=v_{o(ava)*t$

For Kelly:

We need to calculate when Kelly will be 3/4 mile apart becase is running faster than Ava, so, writing the equation we have:

$x_K=x_A+\frac{3}{4}mile=x_o+v_{o(Kelly)}*t$

Now, substituting the equation for Ava into the equation for Kelly, we have:

$x_A+\frac{3}{4}mile=x_o+v_{o(Kelly)}*t$

$v_{o(ava)*t+\frac{3}{4}mile}=x_o+v_{o(Kelly)}*t$

$v_{o(ava)*t+0.75mile}=x_{o}+v_{o(Kelly)}*t\\\\6mph*t+0.75mile=8mph*t\\\\0.75mile=2mph*t\\\\t=\frac{0.75miles}{2mph}=0.375hours$

To prove that the result is correct, we just need to substitute the obtained value for time into both equations, so, substutiting we have:

For Ava:

$x_A=v_{o(ava)*t=6mph*0.375hours=2.25miles$

For Kelly:

$x_K=v_{o(Kelly)}*t=8mph*0.375=3miles$

There is a difference of 0.75 miles or 3/4 mile between Ava and Kelly, so, the obtained value for time is correct.

Therefore, we have that Ava and Kelly will be 3/4 mile apart after 0.375 hours.

Have a nice day!

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