Answer
given,
SAT is 500 with a standard deviation of 100.
a sample of 400 students whose family income was between $70,000 and $80,000 had an average verbal SAT score of 511.
sample mean = ![\dfrac{standard\ deviation}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Cdfrac%7Bstandard%5C%20deviation%7D%7B%5Csqrt%7Bn%7D%7D)
= ![\dfrac{100}{\sqrt{400}}](https://tex.z-dn.net/?f=%5Cdfrac%7B100%7D%7B%5Csqrt%7B400%7D%7D)
= 5
95% confidence level is achieved within +/- 1.960 standard deviations.
1.960 standard deviations x 5 is equal to +/- 9.8
confidence interval = 511 - 9.8 --- 511 + 9.8
= 501.2-----520.8
Answer:
Explanation:
a ) y = A sin(B) ; here B is the phase of the wave which moves so that it remains constant
ωt - kx = constant
differentiating on both sides
ωdt - kdx =0
ωdt = kdx
dx / dt = ω / k
wave velocity = ω / k
b ) ω = 14.5 rad / s ,
k = 18 rad / m
wave velocity = ω / k
= 14.5 / 18
= .805 m /s
80.5 cm / s
c )
Amplitude = A
= 9.5 m
Gas stations or sewage treatment facility
a) 6.25 rad/s
The law of conservation of angular momentum states that the angular momentum must be conserved.
The angular momentum is given by:
![L=I\omega](https://tex.z-dn.net/?f=L%3DI%5Comega)
where
I is the moment of inertia
is the angular speed
Since the angular momentum must be conserved, we can write
![L_1 = L_2\\I_1 \omega_1 = I_2 \omega_2](https://tex.z-dn.net/?f=L_1%20%3D%20L_2%5C%5CI_1%20%5Comega_1%20%3D%20I_2%20%5Comega_2)
where we have
is the initial moment of inertia
is the initial angular speed
is the final moment of inertia
is the final angular speed
Solving for
, we find
![\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(2.25 kg m^2)(5.00 rad/s)}{1.80 kg m^2}=6.25 rad/s](https://tex.z-dn.net/?f=%5Comega_2%20%3D%20%5Cfrac%7BI_1%20%5Comega_1%7D%7BI_2%7D%3D%5Cfrac%7B%282.25%20kg%20m%5E2%29%285.00%20rad%2Fs%29%7D%7B1.80%20kg%20m%5E2%7D%3D6.25%20rad%2Fs)
b) 28.1 J and 35.2 J
The rotational kinetic energy is given by
![K=\frac{1}{2}I\omega^2](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7DI%5Comega%5E2)
where
I is the moment of inertia
is the angular speed
Applying the formula, we have:
- Initial kinetic energy:
![K=\frac{1}{2}(2.25 kg m^2)(5.00 rad/s)^2=28.1 J](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7D%282.25%20kg%20m%5E2%29%285.00%20rad%2Fs%29%5E2%3D28.1%20J)
- Final kinetic energy:
![K=\frac{1}{2}(1.80 kg m^2)(6.25 rad/s)^2=35.2 J](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7D%281.80%20kg%20m%5E2%29%286.25%20rad%2Fs%29%5E2%3D35.2%20J)