A simple random sample of 35 colleges and universities in the United States has a mean tuition of $17,700 with a standard deviation of $10,300. Construct a 95% confidence interval for the mean tuition for all colleges and universities in the United States. Round the answers to the nearest whole number.

The 95% confidence interval for the mean tuition for all colleges and universities in the United States is between $14,288 and $21,112.

Step-by-step explanation:

Sample size greater than 30, which means that we use the normal distribution to find the interval.

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find M as such

In which is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 17700 - 3412 = $14,288

The upper end of the interval is the sample mean added to M. So it is 17700 + 3412 = $21,112.

The 95% confidence interval for the mean tuition for all colleges and universities in the United States is between $14,288 and $21,112.

Value of x in given equation   is -52.

Option B

Step-by-step explanation:

We have, the following equation as One-third x minus two-thirds = negative 18 , i.e. .

In order to solve this, lets simplify the above equation:

⇒

⇒

⇒

⇒

⇒

⇒

∴ Value of x in the equation One-third x minus two-thirds = negative 18 i.e.   is -52. And so, correct option is B.