Answer:
To make a cheer dance routine in a solo performance I would still do the usual sections a group routine has, without the ones that are done with a partner. For example, I would have an opening to show my abilities, then I would do a standing tumbling sequence and finally different jumps. I would mix all these sequences with different dance moves.
hope that helped <3
Answer:
Puddles of water seen after a rainstorm that are clouded with sediment suggest erosion is occurring whereas puddles of clear water are a sign of minimal erosion. Lighter colored soil seen overtime is also a sign of erosion but can be more difficult to identify.
The use of imagery cannot be regarded as a cognitive technique for stress management in individuals and is denoted as option D from the list of available options below.
<h3>What is Cognitive technique?</h3>
These are the different types of ways and method in which our mind and thoughts can be influenced to help manage stress. It requires a mental method of ensuring that the stress doesn't build up and affect us in a negative manner.
Imagery on the other hand involves the creation of images which depicts more work being done instead of easing it which is therefore the reason why it isn't regarded as a cognitive technique for stress management in individuals.
The options include the following:
A - solving problems
b- modifying expectations.
c- maintaining positivity.
d- the use of imagery.
Read more about Imagery here brainly.com/question/25938417
#SPJ1
Using the normal distribution, it is found that 0.0764 = 7.64% of teenagers who will have waist sizes greater than 31 inches.
<h3>Normal Probability Distribution</h3>
In a normal distribution with mean 
and standard deviation 
, the z-score of a measure X is given by:
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem:
- The mean is of
.
- The standard deviation is of
.
The proportion of teenagers who will have waist sizes greater than 31 inches is <u>1 subtracted by the p-value of Z when X = 31</u>, hence:
has a p-value of 0.9236.
1 - 0.9236 = 0.0764.
0.0764 = 7.64% of teenagers who will have waist sizes greater than 31 inches.
More can be learned about the normal distribution at brainly.com/question/24663213
