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3 years ago

Can somebody please explain how to do this please? (5 points)

Mathematics

1 answer:

galina1969 [7]3 years ago

8 0

Look at the picture.

∠2 and ∠3 are Corresponding Angles. j || k, thereofre m∠2 = m∠3.

∠1 and ∠3 are supplementary angles, therefore m∠1 + m∠3 = 108°.

m∠1 = 10x + 4, m∠2 = m∠3 = 4x - 6. Therefore we have the equation:

(10x + 4) + (4x - 6) = 180

(10x + 4x) + (4 - 6) = 180

14x - 2 = 180 <em>add 2 to both sides</em>

14x = 182 <em>divide both sides by 14</em>

x = 13

10x + 4 = 10(13) + 4 = 130 + 4 = 134

4x - 6 = 4(13) - 6 = 52 - 6 = 46

<h3>Answer x = 13, m∠1 = 134°, m∠2 = 46°</h3>

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Option B: 8x+1 has the steepest slope

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2 years ago

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It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

NemiM [27]

Answer:

a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 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 373 minutes and standard deviation 67 minutes. So $\mu = 373, \sigma = 67$

A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?

So $n = 5, s = \frac{67}{\sqrt{5}} = 29.96$

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

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

$Z = \frac{410 - 373}{29.96}$

$Z = 1.23$

$Z = 1.23$ has a pvalue of 0.8907.

So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

Lean

Normally distributed with mean 526 minutes and standard deviation 107 minutes. So $\mu = 526, \sigma = 107$