Each aluminum can is a right cylinder. Find the amount of paper needed for the cans label and the total amount of aluminum needed to make a can. Round to the nearest tenth.

Fig .A=>L=169.6 cm² and S= 226.2 cm²

Fig B=>L=31.4in² and S=70.7in²

Step-by-step explanation:

See attachment for the figure

For figure A:

Given:

Radius 'r'= 3cm and height 'h'=9cm

->Lateral area formula is given by:

L = 2πrh => 2π(3)(9)

L= 54π cm² ≈169.6 cm²

-> next is to find surface area 'S'

S= L + 2πr² =>= 54π + 2π(3)²

S= 72π cm²≈ 226.2 cm²

Therefore,

The amount of paper needed for the label is the lateral area, 169.6 cm²

The amount of aluminum needed for the can is the surface area, 226.2 cm²

For figure B:

Given:

Radius 'r'= 2.5in and height 'h'=2in

->Lateral area formula is given by:

L = 2πrh => 2π(2.5)(2)

L= 10π in² ≈31.4in²

-> next is to find surface area 'S'

S= L + 2πr² =>= 10π + 2π(2.5)²

S= 22.5π in²≈ 70.7in²

The amount of paper needed for the label is the lateral area, 31.4in²

The amount of aluminum needed for the can is the surface area,70.7in²

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: