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2 years ago

What sort of commitment do you want or need to make to some larger goal?

SAT

1 answer:

KonstantinChe [14]2 years ago

4 0

Different people may do it different ways, but here are some ways:

Stay motivated. With commitment you need to stay persistent. No matter how big of a goal you have, you can reach it by being persistent. Actions speak wayyy louder than words. Know your goals. Learn to be ready to face your fears and to overcome obstacles. When it gets hard remember why you set those goals in the first place and focus on the big (or final) picture.

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What is the minimum value for g(x)=x2−10x+16? enter your answer in the box.

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To find the maximum and minimum (5,-9)

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Your credit limit is $1,500. You don’t pay your card off each month like you should. What is the max you can safely charge on th

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Can a puddle of water erode soil yes or no ASAP.

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A good letter of recommendation:

Vitek1552 [10]

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D) conveys unique qualities about a student

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2 years ago

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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

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1 year ago