Find the higest common division of x² - 4 , x²

WILL GIVE YOU BRAINLIEST IF IT IS CORRECT!! PLS HELP!!

777dan777 [17]

Answer:

AFH, BFG, CDG, EFG

Step-by-step explanation:

Please check the point position with your question, if there is any difference... the way to find out a right angle triangle is just like the attached picture.

4 0

3 years ago

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I need help fast I appreciate the help

Neko [114]

Answer:

3/5

Step-by-step explanation:

Sin is opp/hyp so it would be 30/50 in lowest form it would become 3/5

8 0

3 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

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

If you were to use the substitution method to solve the following system, choose the new equation after the expression equivalen

MArishka [77]

Answer:

all work is shown and pictured

8 0

3 years ago

Solve for all values of x by factoring.<br> x^2 + 4x – 30 = -3x

vichka [17]

Answer:

${x}^{2} + 4x - 30 = - 3x \\{x}^{2} + 7x - 30 = 0 \\ {x}^{2} - 3x + 10x - 30 = 0 \\ x(x - 3) + 10(x - 3) = 0 \\ (x - 3)(x + 10) = 0 \\ \boxed{x = 3} \\ \boxed{x = - 10}$

<h3><em><u>x=3 or x=-10</u></em> is the right answer.</h3>

7 0

3 years ago

Find the higest common division of x² - 4 , x² - 6x and x}^{2-8​

Find the higest common division of x² - 4 , x² - 6x and x}^{2-8