Which of the following is not an advantage of decision making.

Pose to pose is a term used in animation to describe the process of creating key poses for characters and then inbetweening them in intermediate frames to give the illusion that the character is moving from one pose to the next.

The goal of Agnostic Pose Estimation is to develop a pose estimation model that can detect the pose of any class of object using only a few samples with keypoint definition.

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4 0

2 years ago

How do i delete my profilei have decided i want nothing to do with thid du to safety matters plsssssssssssss someone tell me?

Lilit [14]

Go to settings and delet it or long out

7 0

3 years ago

How can I be closer to god without having a bible?

dmitriy555 [2]

I'm a pastor's kid and I ask myself that question all the time, and what helps me the best is to go to church and listen to the pastors, listen to praise and worship music, and what I would recomend the most is downloading the bible app on your phone, because most often we are on our phones than reading paper, and personally I've read more of the bible on my phone than the actual book.

6 0

2 years ago

In a recent year, about 22% of Americans 18 years and older are single. What is the probability that in a random sample of 200 A

Pie

Using the normal approximation to the binomial, it is found that there is a 0.0107 = 1.07% probability that more than 30 are single.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

In this problem, the proportion and the sample size are, respectively, p = 0.22 and n = 200, hence:

$\mu = np = 200(0.22) = 44$

$\sigma = \sqrt{np(1 - p)} = \sqrt{200(0.22)(0.78)} = 5.8583$

The probability that more than 30 are single, using continuity correction, is P(X > 30.5), which is <u>1 subtracted by the p-value of Z when X = 30.5</u>, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{30.5 - 44}{5.8583}$

Z = -2.3

Z = -2.3 has a p-value of 0.0107.

0.0107 = 1.07% probability that more than 30 are single.

More can be learned about the normal distribution at brainly.com/question/24663213

6 0

1 year ago