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Kay [80]
1 year ago
15

in a study by giacobbi et al. (2014), it was determined that about 60% of the participants found imagery very helpful in enhanci

ng their physical activity, but 40% said they would no longer use imagery in their exercise. the implication from these results is that
Mathematics
1 answer:
Vanyuwa [196]1 year ago
3 0

Before offering imaging programs, practitioners need to be aware of the reasons why exercise participants are doing it.

<h3>Define imagery in sports.</h3>

When we use imagery, we simulate an actual situation in our minds rather than actually going through it. It differs significantly from daydreaming or simply thinking about anything because it is a cognitive activity that is consciously used by an athlete or exerciser to accomplish a certain task.

In this study, an analysis of secondary data from a recently published randomized controlled trial. In a community-based, group-mediated physical activity intervention for sedentary people 50 and older, the Active Adult Mentoring Program (AAMP) tested the effectiveness of peer volunteers as delivery agents. The AAMP was built on the social-cognitive and self-determination theories, and mentors were trained to lead discussions in groups that would help reinforce key ideas from both theories.

The adaptability of images makes it useful at different times and in varied settings. Athletes employ imagery most frequently right before a competition or during practice, but they do so during the entire season, including the off-season. Similar to how it's reported by athletes, visualization is frequently used before an activity session. For example, it would be more effective for a swimmer to mentally practice her race start by adopting the proper position on the starting block at the swimming pool, as opposed to sitting on a chair at home. Both types of people will typically imagine within the sport and exercise environment where the benefits of this technique are maximized.

To know more about imagery in sports visit:

brainly.com/question/14319340

#SPJ4

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470,925.8 in expanded form is ...

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Answer:

The statistic for this case would be:

z=\frac{\hat p -p_o}{\sqrt{\frac{\hat p(1-\hat p)}{n}}}

And replacing we got:

z= \frac{0.436-0.4}{\sqrt{\frac{0.436*(1-0.436)}{700}}}= 1.92

Step-by-step explanation:

For this case we have the following info:

n =700 represent the sample size

X= 305 represent the number of employees that earn more than 50000

\hat p=\frac{305}{700}= 0.436

We want to test the following hypothesis:

Nul hyp. p \leq 0.4

Alternative hyp : p>0.4

The statistic for this case would be:

z=\frac{\hat p -p_o}{\sqrt{\frac{\hat p(1-\hat p)}{n}}}

And replacing we got:

z= \frac{0.436-0.4}{\sqrt{\frac{0.436*(1-0.436)}{700}}}= 1.92

And the p value would be given by:

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