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Illusion [34]
3 years ago
12

3.74 rounded to the nearest hundredth

Mathematics
2 answers:
Anvisha [2.4K]3 years ago
4 0
3.75 is 3.74 rounded to the nearest hundredth
rewona [7]3 years ago
4 0
The answer is 3.74. Hope this helps :)
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Alexandra [31]
Parallel lines<span> are coplanar </span>lines<span> that do not intersect</span>
6 0
3 years ago
It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So \mu = 375, \sigma = 68

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So n = 6, s = \frac{68}{\sqrt{6}} = 27.76

This probability is 1 subtracted by the pvalue of Z when X = 410.

Z = \frac{X - \mu}{s}

Z = \frac{410 - 375}{27.76}

Z = 1.26

Z = 1.26 has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So \mu = 522, \sigma = 106

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?

So n = 6, s = \frac{106}{\sqrt{6}} = 43.27

This probability is 1 subtracted by the pvalue of Z when X = 410.

Z = \frac{X - \mu}{s}

Z = \frac{410 - 523}{43.27}

Z = -2.61

Z = -2.61 has a pvalue of 0.0045.

So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

6 0
3 years ago
The angles of a quadrilateral are represented by 9x, (20x + 16) (12x + 10), and (7x- 2)°. What is the value of x?
Pepsi [2]

Answer:

x=7

Step-by-step explanation:

9x+20x+16+12x+10+7x-2=360

48x+24=360

48x=360-24

48 x=336

x=336/48=7

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3 years ago
There are 1,200 students at Roosevelt middle school. There are 700 boys at the school. What percentage of the students at Roosev
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To convert this into a percentage, divide 700 by 1,200.

\frac{700}{1200} = 0.583333 (or just 0.583 if you round)

Multiply 0.583 by 100 to get a percentage.

0.583 · 100 = 58.3

So, the percent of boys at the school is 58.3%.
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Yuki888 [10]
2(x + 3) = 18
2x + 6 = 18
      - 6 = 
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Divide 2
x = 6

7 0
3 years ago
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