Answer:
The height of the pole is 167 m
Step-by-step explanation:
The given parameters are;
Increase in the length of the shadow = 90 m
Initial angle of elevation of the Sun = 58°
Final angle of elevation of the Sun = 36°
We have a triangle formed by the change in the length of the shadow and the rays from the two angle of elevation to the top of the pole giving an angle 22° opposite to the increase in the length of the shadow
We have by sin rule;
90/(sin (22°) = (Initial ray from the top of the pole to the end of the shadow's length)/(sin(122°)
Let the initial ray from the top of the pole to the end of the shadow's length = l₁
90/(sin (22°) = l₁/(sin(122°)
l₁ = 90/(sin (22°) ×(sin(122°) = 283.3 m
Therefore;
The height of the pole = 283.3 m × sin(36°) = 166.52 m
The height of the pole= 167 m to three significant figures.
Answer:
f(-3) = 21
g(3) = 69
Step-by-step explanation:
We have been given the following functions;
f(x) = –6x + 3
g(x) = 3x + 21x–3.
To find f(-3) , we simply substitute x = -3 in the function of f(x);
f(-3) = -6(-3) + 3
= 21
To find g(3) , substitute x = 3 in the function of g(x);
g(3) = 3(3) +21(3) - 3
= 69
Answer:
travel from 5 to 10 cm/yr.
Step-by-step explanation:
The correct one is multiplication.
-- Take the number of square miles.
-- Multiply it by the fraction (5,280 feet / 1 mile)² .
-- After canceling things where possible, simplifying, and
cleaning up the product, what you have is the same area
expressed in square feet.
Answer:
P = 0,0012 or P = 0.12 %
Step-by-step explanation:
We know for normal distribution that:
μ ± σ in that range we find 68.3 % of all values
μ ± 2σ ⇒ 95.5 % and
μ ± 3σ ⇒ 99.7 %
Fom problem statement
We have to find (approximately) % of cars that reamain in service between 71 and 83 months
65 + 6 = 71 ( μ + σ ) therefore 95.5 % of values are from 59 and up to 71 then by symmetry 95.5/2 = 49.75 of values will be above mean
Probability between 65 and 71 is 49.75 %
On the other hand 74 is a value for mean plus 1, 5 σ and
74 is the value limit for mean plus 1,5 σ and correspond to 49,85 (from z=0 or mean 65).
Then the pobabilty for 83 have to be bigger than 49.85 and smaller than 0,5 assume is 49.87
Finally the probability approximately for cars that remain in service between 71 and 83 months is : 0,4987 - 0.49.75
P = 0,0012 or P = 0.12 %
