Answer:
Height of the fighter plane =1.5km=1500 m
Speed of the fighter plane, v=720km/h=200 m/s
Let be the angle with the vertical so that the shell hits the plane. The situation is shown in the given figure.
Muzzle velocity of the gun, u=600 m/s
Time taken by the shell to hit the plane =t
Horizontal distance travelled by the shell =u
x
t
Distance travelled by the plane =vt
The shell hits the plane. Hence, these two distances must be equal.
u
x
t=vt
u Sin θ=v
Sin θ=v/u
=200/600=1/3=0.33
θ=Sin
−1
(0.33)=19.50
In order to avoid being hit by the shell, the pilot must fly the plane at an altitude (H) higher than the maximum height achieved by the shell for any angle of launch.
H
max
=u
2
sin
2
(90−θ)/2g=600
2
/(2×10)=16km
Answer:
0.9792
Step-by-step explanation:
Data provided in the question:
Average gross sales = $1,240
Standard deviation = $180
sample size = 40
Now,
standard deviation of sample average
= 
= 
= 28.46
Now,
z value for 1200 = 
= -1.4,
and,
p value for (z = -1.4) = 0.0808
therefore,
P(average < $1200) = 0.0808
Thus,
probability that the average over the next 40 weekdays will exceed $1,200
= 1 - 0.808
= 0.9792
Step-by-step explanation:
by having two equations that have the same point, you can set the two equations equal to each other
y = 12x + 3, y = x + 1
12x + 3 = x + 1
11x = -2
x = -2/11
when you find the the x value, plug it back into either one of the equations
y = x + 1
y = (-2/11) + 1
y = 11/11 - 2/11
y = 9/11
the point (solution):
(-2/11, 9/11)
Answer:
I believe the answer is -176º
