Angle ABC measures 180 degrees find the value of X

The function F(x) = log2 x is increasing.<br> True or false

Ann [662]

Answer: i think true

Step-by-step explanation:

8 0

3 years ago

Please help me with my math

lianna [129]

C is the correct answer

3 0

3 years ago

Read 2 more answers

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

Genrish500 [490]

Answer:

a) 0.0537 = 5.37% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes

b) 0.9871 = 98.71% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mildly obese:

Mean 376 minutes and standard deviation 67 minutes, which means that $\mu = 376, \sigma = 67$

Sample of 6

This means that $n = 6, s = \frac{67}{\sqrt{6}} = 27.35$

Lean

Mean 520 minutes and standard deviation 110 minutes, which means that $\mu = 520, \sigma = 110$

Sample of 6

$n = 6, s = \frac{110}{\sqrt{6}} = 44.91$

A) What is the probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes?

This is 1 subtracted by the pvalue of Z when X = 420, using the mean and standard deviation for mildly obese people. So

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

By the Central Limit Theorem

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

$Z = \frac{420 - 376}{27.35}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463

1 - 0.9463 = 0.0537

0.0537 = 5.37% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes.

B) What is the probability that the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes?

This is 1 subtracted by the pvalue of Z when X = 420, using the mean and standard deviation for lean people. So

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

$Z = \frac{420 - 520}{44.91}$

$Z = -2.23$

$Z = -2.23$ has a pvalue of 0.0129

1 - 0.0129 = 0.9871

0.9871 = 98.71% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes

5 0

3 years ago

The lowest altitude of an altocumulus cloud is about 3^8 feet. The highest altitude of an altocumulus cloud is about 3 times the

Alona [7]

<span>3(3^8) equals 9^8 feet.</span>

5 0

4 years ago

Algebra 1 time and cost

Citrus2011 [14]

Answer:

What do you mean by time and cost

Step-by-step explanation:

8 0

3 years ago