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Dovator [93]
3 years ago
10

Which of the following is released by trees into the atmosphere? Check all that apply carbon dioxide nitrogen oxygen ozone

SAT
2 answers:
Zarrin [17]3 years ago
6 0

Answer:

Oxygen is released by oxygen from carbon dioxide through photosynthesis.

Explanation:

marusya05 [52]3 years ago
5 0

Answer:

oxygen

Explanation:

trees take in carbon dioxide and put out oxygen.

nitrogen doesn't have anything to do with trees.

the ozone protects us from the suns rays, again has little to do with trees

so therefore Oxygen is the correct answer

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B

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What is the volume of a cube that has a side length of 10 inches?
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Volume: 1000 cubic inches (a^3=V -----> 10^3=V ----> V=1000)

42 inches to mm= 1041.4 mm (multiply inch value (42) by 25.4 for mm)

231 cm to mm= 2310 mm (231*10)

5 ft to inches = 5*12 = 60 inches

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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
2 years ago
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