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>
The answer to this question is 650m.
Answer:
<E=<G=115°
since opposite angle of rhombus is equal
Answer:
The standard error of your estimate of the average height in the city is 0.5 inches.
Step-by-step explanation:
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
You begin by collecting the heights of a random sample of 196 people from the city.
This means that
The standard deviation of the heights in your sample is 7 inches.
This means that
The standard error of your estimate of the average height in the city is
The standard error of your estimate of the average height in the city is 0.5 inches.
The answer is 804.25 ft squared