From a random sample of 41 teens, it is found that on average they spend 43.1 hours each week online with a population standard deviation of 5.91 hours. what is the 90% confidence interval for the amount of time they spend online each week?

Step-by-step explanation:

We want to determine a 90% confidence interval for the mean amount of time that teens spend online each week.

Number of sample, n = 41

Mean, u = 43.1 hours

Standard deviation, s = 5.91 hours

For a confidence level of 90%, the corresponding z value is 1.645. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean +/- z ×standard deviation/√n

It becomes

43.1 ± 1.645 × 5.91/√41

= 43.1 ± 1.645 × 0.923

= 43.1 ± 1.52

The lower end of the confidence interval is 43.1 - 1.52 =41.58

The upper end of the confidence interval is 43.1 + 1.52 =44.62

Therefore, with 90% confidence interval, the mean amount of time that teens spend online each week is between 41.58 and 44.62

how many players are there?

step-by-step explanation:

c.

step-by-step explanation:

because first you have to add the 4x, 9x, and the 7x together and you get 20x and then you add the last numbers up and you get 5and then you put the subtraction sign in there and that's how you get the answer.