1. D
2. A
3. 79
I'm not confident with 3 but 1 and 2 are for sure correct
9514 1404 393
Answer:
9.5°, yes
Step-by-step explanation:
The relevant trig relation is ...
Tan = Opposite/Adjacent
The distance opposite the angle of elevation is the plane's height, 500 m. The distance adjacent to the angle of elevation is the horizontal distance to the plane, 3 km = 3000 m. Then the angle is found from ...
tan(α) = 500/3000 = 1/6
α = arctan(1/6) ≈ 9.46°
The plane is approaching at an angle of 9.46°. It is safe to land, since that angle is less than 15°.
_____
<em>Additional comment</em>
The usual descent angle for most commercial air traffic is 3°. Some airport geography demands it be different (steeper). A higher descent angle can put undue stress on the landing gear.
Step-by-step explanation:
the solution is found in the images above
here is the data set for the complete question
x: 18 21 19 21 20 21
y; 2 14 5 6 18 18
Answer:
B. 0.652
Step-by-step explanation:
x y rank of x rank of y d d²
18 2 1 1 0 0
21 14 4 4 0 0
19 5 2 2 0 0
21 6 4 3 1 1
20 18 3 5.5 -2.5 6.25
21 18 4 5.5 -1.5 2.25
∑d² = 8.5
rs = 1 - 6[∑di² + ∑m(m²-1)]/n(n²-1)
= 1 - 6[8.5 +{3(3²-10/12 + 2(2² - 1)/12}]/6(6²-1)
= 1 - 0.348
= 0.652
therefore option b is the right answer.
The answer is -2-m- plz make brainliest
