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ch4aika [34]
1 year ago
12

research using u.s. and canadian olympic athletes has shown that approximately what percent of olympic athletes use imagery? gro

up of answer choices 50% 75% 95% 85% 60%
Mathematics
1 answer:
IrinaVladis [17]1 year ago
5 0

The correct option is option (c) i.e 95%. Based on Shane Murphy's research, Doug Jowdy and Shirley Durtschi conducted a poll in U.S. about imagery use and find that approximately 95% of U.S. and Canadian Olympic athletes use imagery.

Imagery is a mental skill that many of us are familiar with. Imagery can be an effective mental skill that enhances performance when performed correctly and tailored to an athlete's needs. You can train your mind.

Physical repetition alone is not enough to develop a skill and take an athlete or team to the next level. Imagery is very important to be clear about what you want the result of your visualization to be, because the use of imagery can prepare the mind to ultimately direct the body. Asking athletes to take a "mental tour", visualizing their event and what they want to see, is a great tool for success and also helps them relax when playtime comes around increase. Shane Murphy, Doug Jowdy, and Shirley Durtschi conducted an imagery study by interviewing coaches at the US Olympic Training Center in Colorado Springs, Colorado, and found that 95% of coaches and 97% of athletes used imagery in their sport. It turns out that there is His 100% of coaches and his 97% of athletes surveyed agreed that imagery improve performance. When asked why an athlete uses an imagery, 80% used it to prepare for the contest and 48% used it to deal with technical errors , 44% learn new skills and 40% relax…

To learn more about Imagery , refer:

brainly.com/question/25938417

#SPJ4

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How do you do this question?
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0.001591

Step-by-step explanation:

The power series for arctan(x) is:

arctan(x) = ∑ (-1)ⁿ x²ⁿ⁺¹ / (2n + 1)

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arctan(5x) = ∑ (-1)ⁿ (5x)²ⁿ⁺¹ / (2n + 1)

Multiply both sides by x:

x arctan(5x) = ∑ (-1)ⁿ x (5x)²ⁿ⁺¹ / (2n + 1)

Simplify:

x arctan(5x) = ∑ (-1)ⁿ (5x) (5x)²ⁿ⁺¹ / (10n + 5)

x arctan(5x) = ∑ (-1)ⁿ (5x)²ⁿ⁺² / (10n + 5)

Multiply top and bottom by 5:

x arctan(5x) = ∑ (-1)ⁿ 5 (5x)²ⁿ⁺² / (50n + 25)

Integrate:

∫ x arctan(5x) = ∑ (-1)ⁿ (5x)²ⁿ⁺³ / ((50n + 25) (2n + 3))

Evaluate between x = 0.1 and x = 0:

∫₀⁰¹ x arctan(5x) = [∑ (-1)ⁿ (5x)²ⁿ⁺³ / ((50n + 25) (2n + 3))] |₀⁰¹

∫₀⁰¹ x arctan(5x) = ∑ (-1)ⁿ (0.5)²ⁿ⁺³ / ((50n + 25) (2n + 3))

This is an alternating series.  We can approximate it with Alternating Series Estimation.

bₙ₊₁ ≥ ε

(0.5)²⁽ⁿ⁺¹⁾⁺³ / ((50(n+1) + 25) (2(n+1) + 3)) ≥ 0.000001

(0.5)²ⁿ⁺⁵ / ((50n + 75) (2n + 5)) ≥ 0.000001

n ≥ 3

So the approximation is the sum of the terms from n=0 to n=3.

(-1)⁰ (0.5)²⁽⁰⁾⁺³ / ((50(0) + 25) (2(0) + 3))

+ (-1)¹ (0.5)²⁽¹⁾⁺³ / ((50(1) + 25) (2(1) + 3))

+ (-1)² (0.5)²⁽²⁾⁺³ / ((50(2) + 25) (2(2) + 3))

+ (-1)³ (0.5)²⁽³⁾⁺³ / ((50(3) + 25) (2(3) + 3))

= 0.0016667 − 0.0000833 + 0.0000089 − 0.0000012

= 0.001591

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Answer:

Step-by-step explanation:

There is no solution, because the lines do not intersect.

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