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3 years ago

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Which of the following statements concerning statistical probability is not true? A. Probability involves the level of certainty

a researcher has in a circumstance or event taking place. B. Probability influences how much a researcher is able to generalize research results back to the population. C. Probability influences whether or not a researcher can claim statistical significance. D. Probability levels for finding statistical significance increase as statistical power decreases.

2 answers:

Leto [7]3 years ago

8 0

The answer to this is D. Probability levels for finding statistical significance increase as statistical power decreases. The statement concerning statistical probability is not true as this is d.

I hope this helps!

Ne4ueva [31]3 years ago

3 0

Answer:

The answer is D

Explanation:

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The 100-m dash can be run by the best sprinters in 10.0 s. A 66-kg sprinter accelerates uniformly for the first 45 m to reach to

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(a) 154.5 N

Let's divide the motion of the sprinter in two parts:

- In the first part, he starts with velocity u = 0 and accelerates with constant acceleration $a_1$ for a total time $t_1$ During this part of the motion, he covers a distance equal to $s_1 = 45 m$ , until he finally reaches a velocity of $v_1 = u + a_1t_1$ . We can use the following suvat equation:

$s_1 = u t_1 + \frac{1}{2}a_1t_1^2$

which reduces to

$s_1 = \frac{1}{2}a_1 t_1^2$ (1)

since u = 0.

- In the second part, he continues with constant speed $v_1 = a_1 t_1$ , covering a distance of $d_2 = 55 m$ in a time $t_2$ . This part of the motion is a uniform motion, so we can use the equation

$s_2 = v_1 t_2 = a_1 t_1 t_2$ (2)

We also know that the total time is 10.0 s, so

$t_1 + t_2 = 10.0 s\\t_2 = (10.0-t_1)$

Therefore substituting into the 2nd equation

$s_2 = a_1 t_1 (10-t_1)$

From eq.(1) we find

$a_1 = \frac{2s_1}{t_1^2}$ (3)

And substituting into (2)

$s_2 = \frac{2s_1}{t_1^2}t_1 (10-t_1)=\frac{2s_1}{t_1}(10-t_1)=\frac{20 s_1}{t_1}-2s_1$

Solving for t,

$s_2+2s_1=\frac{20 s_1}{t_1}\\t_1 = \frac{20s_1}{s_2+2s_1}=\frac{20(45)}{55+2(45)}=6.2 s$

So from (3) we find the acceleration in the first phase:

$a_1 = \frac{2(45)}{(6.2)^2}=2.34 m/s^2$

And so the average force exerted on the sprinter is

$F=ma=(66 kg)(2.34 m/s^2)=154.5 N$

b) 14.5 m/s

The speed of the sprinter remains constant during the last 55 m of motion, so we can just use the suvat equation

$v_1 = u +a_1 t_1$

where we have

u = 0

$a_1 =2.34 m/s^2$ is the acceleration

$t_1 = 6.2 s$ is the time of the first part

Solving the equation,

$v_1 = 0 +(2.34)(6.2)=14.5 m/s$

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