Question

Can some please help me? this is my last question for homework and I can finaly be done for the day. Brainliest, rattings, thanks, etc.
Use the following table of a school bus during morning pickups to calculate its average speed between 0 h and 2.34 h.

Enter your answer in the space provided.

Answer 1

$\bf \begin{array}{|cc|ll} \cline{1-2} h&km\\ \cline{1-2} \underline{0.000}&0.0\\ 0.024&1.2\\ 0.051&2.8\\ 0.084&4.2\\ \underline{2.34}&16.34\\ \cline{1-2} \end{array}~\hspace{10em} (\stackrel{x_1}{0.000}~,~\stackrel{y_1}{0.0})\qquad (\stackrel{x_2}{2.34}~,~\stackrel{y_2}{16.34}) \\\\\\ \stackrel{\textit{average rate of change}}{slope = m\implies} \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{16.34-0.000}{2.34-0.0}\implies \cfrac{16.34}{2.34}~~\approx ~~6.98~\frac{km}{h}$

Answer 2

Alright lets get straight down to understanding the table.

Position(km)  Time(h)

0.0                  0.000

1.2                    0.024

2.8                   0.051

4.2                   0.084

16.34                2.34

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For every 1.2 km, it is 0.024th of a hour.

So for average, you and all numbers in position and divide it by the numbers they have, 5. So the equation would be, (0.0+1.2+2.8+4.2+16.34)/5. Then that is your answer, 4.908. Now we put that to the side and do the same thing for time. (0.000+0.024+0.051+0.084+2.34)/5. and that would be 0.4998.

Now to put it in a sentence, For every 4.908 km, there is 0.4998 hours.