Answer:
y=2x*6
Step-by-step explanation:
Answer:
0.25 rad to the nearest hundredth radian
Step-by-step explanation:
Here is the complete question
Suppose a projectile is fired from a cannon with velocity vo and angle of elevation (theta). The horizontal distance R(θ) it travels (in feet) is given by the following.
R(θ) = v₀²sin2θ/32
If vo=80ft/s what angel (theta) (in radians) should be used to hit a target on the ground 95 feet in front of the cannon?
Do not round any intermediate computations, and round your answer(s) to the nearest hundredth of a radian.
(θ)= ?rad
Solution
R(θ) = v₀²sin2θ/32
If v₀ = 80 ft/s and R(θ) = 95 ft
θ = [sin⁻¹(32R(θ)/v₀²)]/2
= [sin⁻¹(32 × 95/80²)]/2
= [sin⁻¹(3040/6400)]/2
= [sin⁻¹(0.475)]/2
= 28.36°/2
= 14.18°
Converting 14.18° to radians, we have 14.18° × π/180° = 0.2475 rad
= 0.25 rad to the nearest hundredth radian
Answer:
a) 54 mph to 144 mph
Step-by-step explanation:
We don't know the shape of the distribution, so we use Chebyshev's Theorem to solve this question. It states that:
At least 75% of the measures are within 2 standard deviations of the mean.
At least 89% of the measures are within 3 standard deviations of the mean.
At least eight-ninths of the player's serves.
8/9 is approximately 89%
So
Mean: 99, standard deviation: 15
99 - 3*15 = 54
99 + 3*15 = 144
So the correct answer is:
a) 54 mph to 144 mph
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