7/2x + 1/2x = 9 + 9/2x<br> the value of x is

How do you find the area.

serious [3.7K]

For a square and rectangle it is Area=length*width

6 0

3 years ago

If f(x)=X-3/x and g(x) = 5x – 4, what is the domain of (fog)(x)?<br><br><br> NEED HELP FAST!!

dedylja [7]

Answer:

Step-by-step explanation:

This is function composition. We will take the g function and plug it into the f function wherever we see an x. That will look like this when it's done but before it's simplified:

$f(g(x))=\frac{5x-4-3}{5x-4}$ and that can be simplified down to

$f(g(x))=\frac{5x-7}{5x-4}$

The domain of this function will be all real numbers EXCEPT for the one(s) that cause the denominator to become undefined. Set the denominator equal to 0 and solve for x:

5x - 4 = 0 so

5x = 4 and

x = 4/5

In other words, D = {x | x ≠ 4/5}

6 0

3 years ago

The Office of Student Services at a large western state university maintains information on the study habits of its full-time st

Vera_Pavlovna [14]

Answer:

0.8254 = 82.54% probability that the mean of this sample is between 19.25 hours and 21.0 hours

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 $s = \frac{\sigma}{\sqrt{n}}$ .