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3 years ago

Is cutting an apple a chemical change or a physical change

SAT

2 answers:

sveticcg [70]3 years ago

6 0

It's a physical change because u can't reverse it

Alex777 [14]3 years ago

5 0

Cutting an apple is a physical change. the chemical make-up remains the same.

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If an organism is unicellular, a eukaryote, and autotrophic, into which domain would it be classified? question 10 options: doma

OverLord2011 [107]

Answer:

Domain Eukarya

Explanation:

For an organism to be in Domain Eukarya it needs to have at least one nucleus. The organism is a eukaryote (has organelles and a nucleus), unicellular and autotrophic.

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2 years ago

Of less than average height from top to bottom

ella [17]

I'm guessing it's "low". For Listia, right?

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2 years ago

Who is danny’s valentine actually from? explain your answer, using evidence from the story.

Inessa05 [86]

Can you provide more information? Who is Danny?

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1 year ago

Find the missing number

madreJ [45]

I think B

I am not sure tho

4 0

3 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