This is so provided that the velocity changes continuously in which case we can apply the mean value theorem.
<span>Velocity (v) is the derivative of displacement (x) : </span>
<span>v = dx/dt </span>
<span>Monk 1 arrives after a time t* and Monk 2 too. </span>
<span>Name v1(t) and v2(t) their respective velocities throughout the trajectory. </span>
<span>Then we know that both average velocities were equal : </span>
<span>avg1 = avg2 </span>
<span>and avg = integral ( v(t) , t:0->t*) / t* </span>
<span>so </span>
<span>integral (v1(t), t:0->t*) = integral (v2(t), t:0->t*) </span>
<span>which is the same of saying that the covered distances after t* seconds are the same </span>
<span>=> integral (v1(t) - v2(t) , t:0->t*) = 0 </span>
<span>Thus, name v#(t) = v1(t) - v2(t) , then we obtain </span>
<span>=> integral ( v#(t) , t:0->t*) = 0 </span>
<span>Name the analytical integral of v#(t) = V(t) , then we have </span>
<span>=> V(t*) - V(0) = 0 </span>
<span>=> V(t*) = V(0) </span>
<span>So there exist a c in [0, t*] so that </span>
<span>V'(c) = (V(t*) - V(0)) / (t* - 0) (mean value theorem) </span>
<span>We know that V(0) = V(t*) = 0 (covered distances equal at the start and finish), so we get </span>
<span>V'(c) = v#(c) = v1(c) - v2(c) = 0 </span>
<span>=> v1(c) = v2(c) </span>
<span>So there exist a point c in [0, t*] so that the velocity of monk 1 equals that of monk 2. </span>
Answer:
(0,-4)
Step-by-step explanation:
Answer:
Statistical significance relates to whether an effect exists.
Practical significance refers to the magnitude of the effect.
And you can have statistical significance but not practical.
Step-by-step explanation:
Let's analize it with an example.
Suppose that your new treatment involves hair recovery.
You divide the population of the test in two different groups.
And you apply the treatment to only one of them.
You can see that the treatment works and there is a 3% improvment.
You have statistical significance. The treatment worked.
Now, if the test was expensive, the 3% improvement might not be practical.
The fractions which cannbe formed are: 2/3, 3/2, 2/5, 5/2, 3/5 and 5/3.
Now, we find the products with the third number:
2/3*5= 10/3 = 3.33 > 2
3/2*5 = 15 >2= 7.5 > 2
2/5*3= 6/5 (rejected)
5/2*3 = 15/2 > 2
3/5*2= 6/5 (rejected)
5/3*2 = 15/2 = 7.5> 2
