21.6 Computer use. A random sample of 197 12th-grade students from across the United States was surveyed, and it was observed that these students spent an average of 23.5 hours on the computer per week, with a standard deviation of 8.7 hours. If we plan to use these data to construct a 99% confident interval, the margin of error will be approximately (a) 0.07. (b) 0.62. (c) 1.6. (d) 8.7.

(c) 1.6

Step-by-step explanation:

Calculation for what the margin of error will be approximately

First step is to calculate Total hours spent

Total hours spent = (197 * 23.5 hours)

Total hours spent =4,625 hours

Second step is to find the square root of the sample size

Sample size=√197

Sample size=14.04

Third step is to divide the Standard deviation of 8.7 hours by square root of sample size of 14.04

= 8.7 hours /14.04

= 0.6196

Fourth step is to use the Z table to find Z value at 99% confidence

Z value at 99% confidence = 2.58

Last step is to calculate the Margin of error

Margin of error = 0.6196 * 2.58

Margin of error = 1.59857

Margin of error= 1.6 (Approximately)

Therefore what the margin of error will be approximately is 1.6

reduce the friction between its moving parts

step-by-step explanation:

in fact, by reducing the friction between the moving parts of the machine, it is possible to reduce the energy wasted due to this friction; therefore, more input energy is converted into useful work, and this will improve the efficiency of the machine.

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.