2. hospital records show that a certain surgical procedure takes on the average of 120 minutes with a standard deviation of 10 minutes. between how many minutes must be the lengths of at least 93.75% of these surgical procedures?

93.75% of these surgical procedures takes at-least 80 to 160 minutes.

Step-by-step explanation:

Given : Hospital records show that a certain surgical procedure takes on the average of 120 minutes with a standard deviation of 10 minutes.

To find : Between how many minutes must be the lengths of at least 93.75% of these surgical procedures?

Solution :

The average mean is minutes.

The standard deviation is minutes.

At least 93.75% of these surgical procedures,

Applying Chebyshev's theorem,

For any constant k>1, no more than of teh data set lie outside the k standard deviations away from the mean i.e. at-least of the distribution value fall within 'k' standard deviations.

So,

Length must be given by,

Therefore, 93.75% of these surgical procedures takes at-least 80 to 160 minutes.

3< x< 5

step-by-step explanation:

|x - 4| < 1

there are two solutions, one positive and one negative.   remember to flip the inequality when taking the negative solution

x-4 < 1       x-4 > -1

add 4 to each side

x-4+4 < 1+4       x-4+4 > -1+4

x< 5                 x> 3

3< x< 5

the answer is 8/21

hope this