The rectangular front windshield of the bus has the length of 1.5√3 m and the width of 1.5 m . The wipers are attached to the bottom corners. If the length of each wiper is equal to the width of the windshield, find the percentage of the windshield that wipers can reach. (Solve this task without using Trigonometry)

See the attached sketch. From their respective corners, each wiper sweeps out a portion of the total area of the windshield equal to the area of a quarter-circle with radius 1.5 m (shaded green), and they overlap in a half-lens-shaped region (shaded red). The area that the wipers together cover is equal to the sum of the green areas minus the area of the red overlap. It's easy to see that this red area is equal to the area of the highlighted circular segment (bordered in blue).

Not sure how exactly you can avoid finding the area of the segment without resorting to trigonometry, though... The radii AC and AD form an angle θ such that the segment has area

1/2 (1.5 m)² (θ - sin(θ))

but you can avoid mentioning the sine function by writing it as

1/2 (1.5 m)² (θ - √3/2)

which follows from

sin(θ) = 2 sin(θ/2) cos(θ/2) = 2 ((BD/2)/AB) ((AC/2)/AB)

and the fact that BD has length equal to the radii, since by the Pythagorean theorem

(AC/2)² + (BD/2)² = AB²

(whether this theorem falls under trigonometry is also up for debate)

You would need trig to find the measure of the angle θ as well, unless you just take for granted the fact that a triangle with side lengths 1.5√3/2, 1.5/2, and 1.5 (i.e. AC/2, BD/2, and AB), since they occur in a ratio of √3 : 1 : 2 is a 30°-60°-90° triangle, so that θ = 60° = π/3 rad.

At any rate, the percentage is about (2×(quarter-circle area) - (segment area))/(rectangle area) = 85.46%, since

• area of rectangle = (1.5 √3 m) • (1.5 m) ≈ 3.897 m²

• area of quarter-circle = π/4 (1.5 m)² ≈ 1.767 m²

• area of segment = 1/2 (1.5 m)² (π/3 - √3/2) ≈ 0.204 m²

Step-by-step explanation:

Refer to attached

The area of coverage is the sum of two sectors of 60° and a triangle.

Find each area and calculate the percentage of covered region

Rectangle area:

Triangle height

Triangle area

Sectors' area

Area covered by wipers

Percentage of covered region:

it is a. 4.2 because 7.6-3.4 gets the remaining length of 4.2. i might be wrong but i am pretty sure this is the answer.

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: