<em>Acceleration (a) = 5.1 m/s</em>²
<em>Initial speed (u) = 16 km/h </em>⇒
<em>m/s </em>≈ <em>4.5m/s</em>
<em>
</em><em>Final speed (v) =118 km/h </em>⇒
<em>m/s</em> ≈ <em>32.8m/s</em>
<em>
</em>Distance(S) travel in that particular instant is carried out by 'Third equation of motion' i.e., v² = u² + 2aS
<em>So, When all quantities are in S.I. unit then,
</em>putting the values in the equation of motion,
<em /><em>As we have to carry out the distance covered,
</em><em>2</em>·<em>a</em>·<em>S = v</em>² <em>- u</em>²
<em>S = </em>
Putting values in derived equation,
⇒ <em>S = </em>
⇒ <em>S = </em>
⇒ <em>S = </em>
⇒ <em>S </em>≈ <em>103.489</em> <em>m
</em>
<em>
The total distance covered in that given condition is approx. 103.289 m.</em>
Answer:
45% of the people are in the water
Step-by-step explanation:
100%-29%=71%
71%-26%=45%
Answer:
The degrees of freedom are given by;
The significance level is 0.1 so then the critical value would be given by:
If the calculated value is higher than this value we can reject the null hypothesis that the arrivals are uniformly distributed over weekdays
Step-by-step explanation:
For this case we have the following observed values:
Mon 25 Tue 22 Wed 19 Thu 18 Fri 16 Total 100
For this case the expected values for each day are assumed:
The statsitic would be given by:
Where O represent the observed values and E the expected values
The degrees of freedom are given by;
The significance level is 0.1 so then the critical value would be given by:
If the calculated value is higher than this value we can reject the null hypothesis that the arrivals are uniformly distributed over weekdays
Answer:
Step-by-step explanation:
Equivalent expressions are expressions that are the same,even though they might look different. If you plug in the same variable value into equivalent expressions they will each give you the same value when you simplify.