The claim of the study is that the average attendance at games is over 523.
Under the claim the null and alternative hypotheses are,
After the performing the test, the test yields the results that, fail to reject the null hypothesis.
As null hypothesis is fail to be rejected, then it can be concluded that there is no sufficient evidence to support the claim that the average attendance at games is over 523.
Therefore the correct answer is C: <em>There is not sufficient evidence to support the claim that the mean attendance is greater than 523.</em>
Samburu's weight is (mass x gravity) = (1650kg x 9.8 m/s²) = 16,170 Newtons.
Samburu's contact area with the ground is (0.25 m²/hoof) x (4 hoofs) = 1 m² .
Pressure = (force) / (area) = (16,170 Newtons) / (1 m²) = 16,170 Pascal .
Answer:
dR/dt = 10.2 ft / s
Explanation:
Let's work this problem by finding the distance between the balloon and the motorcycle and then drift for the speed change of the distance
Balloon
y = y₀ + t
Motorcycle
x = v₀ₓ t
Distance, let's use Pythagoras' theorem
R² = x² + y²
R² = (v₀ₓ t)² + (y₀ + t)²
v₀ₓ = 88 ft / s
= 8 ft / s
y₀ = 150 ft
R² = (8 t)² + (150 + 8 t )²
R² = 64 t² + (150 + 8t )²
This is the expression for the distance between the two bodies, the rate of change is the derivative with respect to time (d / dt)
2RdR / dt = 64 2 t + 2 (150 + 8t) 8
dR / dt = [64 t + (1200 + 64t )] / R
dR/dt = (1200 +128 t)/R
Let's calculate for the time of 10 s
dR / dt = (1200 + 128 10) / R = 2480 /R
R = √ [64 10² + (150 + 8 10)²
R = √ [6400 + 52900]
R = 243.5 ft
dR / dt = (2480) / 243.5
dR / dt = 10.2 ft / s
Answer:
11.5 m/s²
Explanation:
The centripetal acceleration, a = rω² where r = radius of cylinder = 10.7 cm = 0.107 m and ω = angular speed = 2πN where N = number of revolutions per second = 1.65 rev/s
So, a = rω²
a = r(2πN)²
a = 4π²rN²
substituting the values of the variables into the equation, we have
a = 4π²rN²
a = 4π²(0.107 m)(1.65 rev/s)²
a = 4π²(0.107 m)(2.7225 rev²/s)²
a = 4π² × 0.2913075 mrev²/s)²
a = 11.5 m/s²