When being burned in a writable CD-R drive, the angular speed of a CD (diameter of 120 mm) is often much faster than when playing audio, but the angular speed still varies to keep the linear speed constant where the disc is being written. When wirting along the outer edge of the disc, the angular speed of one drive is about 4800 RPM. Find the linear speed.

30.1 m/s

Step-by-step explanation:

For a point in uniform circular motion, the relationship between angular speed and linear speed is given by

where

v is the linear speed

is the angular speed

r is the distance of the point from the axis of rotation

In this problem, we have:

is the angular speed, in revolutions/minute

Converting into radians per second,

The diameter of the CD is

d = 120 mm

So the radius is

r = 60 mm = 0.06 m

Therefore, a point at the outer edge of the disc has a distance of

r = 0.06 m from the axis.

Therefore, its linear speed is:

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation:

yes

step-by-step explanation:

so of course you want to start by plugging the 5 in which goes to 7(5)+3(5)=50 because 7*5=35 and 3*5=15 and then add them together, then plug in the 5 to the other equation as well 9(5)+5=50 because 9*5= 45 and then add 5 and so yes they are equivalent.