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16. Make a Conjecture
Complete the table. Then find the rates.

distance
———
time


time
———
distance

a. Are the rates equivalent? Explain.

b. Suppose you graph the points (time, distance) and your friend graphs (distance, time). How will your graphs be different?

17. To graph a rate or ratio from a table, how do you determine the scales to use on each axis?

16 Make A Conjecture Complete The Table Then Find The Rates Distance Time Time Distance A Are The Rates Equivalent Explain B Suppose You Graph The Points Time D class=

Sagot :

The table can be completed by assuming a constant rate, such that the

graph of the values form a straight line.

  • a. The rates are different [tex]\dfrac{time}{distance}[/tex] < [tex]\dfrac{distance}{time}[/tex]
  • b. The graphs will have different slopes due to them having different rates
  • 17. The scales to use on each axis are determined by the range of values of the data and the data points in the table.

Reasons:

The given table is presented as follows;

[tex]\begin{tabular}{|l|c|c|c|c|}\underline{Time (min)}&1&2&5&\\Distance (m)&&&25&100\end{array}\right][/tex]

Whereby the distance and time have a proportional relationship, we have;

Distance = Constant of proportionality × Time

From the table, we have;

At 5 minutes, the distance is 25 meters

Which gives;

25 = Constant of proportionality × 5

[tex]\displaystyle Constant \ of \ proportionality = \frac{25}{5} = \mathbf{ 5}[/tex]

Therefore;

Each time value is multiplied by 5 to get the distance, while the distance is

divided by 5 to give the time, which gives the completed table as follows;

[tex]\begin{tabular}{|l|c|c|c|c|}\underline{Time (min)}&1&2&5&100 \div 5 = 20\\Distance (m)&1 \times 5 = 5&2 \times 5 = 10&25&100\end{array}\right][/tex]

The completed table is therefore

[tex]\begin{tabular}{|l|c|c|c|c|}\underline{Time (min)}&1&2&5& 20\\Distance (m)&5&10&25&100\end{array}\right][/tex]

a. The rates are;

[tex]\begin{array}{|l|c|c|c|c|}\underline{Time \ (min)}&1&2&5& 20\\Distance \ (m)&5&10&25&100\\&&&&\\Rate= \dfrac{distance}{time} &\dfrac{5}{1} = 5 &\dfrac{10}{2} = 5&\dfrac{25}{5} = 5& \dfrac{100}{20} = 5 \end{array}\right][/tex]

The rate [tex]\dfrac{distance}{time}[/tex] is constant and equals 5, therefore;

The rate [tex]\dfrac{time}{distance}[/tex] equals [tex]\dfrac{1}{5}[/tex] = 0.2

Therefore;

The rates are different (not equivalent), given that the distance is changing at a rate of 5 times the time while the time is changing at a rate of 0.2 times the distance.

b. The graphs will be different given that the rates are different.

The graph of [tex]\dfrac{time}{distance}[/tex] (time, distance) will have a gentler slope (rise to run) than the graph of (distance, time) [tex]\dfrac{distance}{time}[/tex]

17. The given table is a ratio table. To graph a ratio table, determine the

range of values, in the table. The range is divided into amounts that will

preferable show points on the table, this is known as the scale of the

graph.

With the acceptable scale that shows points on the graph, the axes of the

graph are marked, and the values on the table are marked on the graph.

The rate or ratio is given by the slope of the graph.

Learn more proportional relationship tables here:

https://brainly.com/question/13203560

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