Answer: The point estimate for the mean weight of all spawning Chinook salmon in the Columbia River is the mean weight from your sample
= 31.2 pounds .
Step-by-step explanation:
The weights of spawning Chinook salmon in the Columbia river are normally distributed.
Given : The mean weight from your sample is 31.2 pounds with a standard deviation of 4.4 pounds.
The point estimate for the mean
weight of all spawning Chinook salmon in the Columbia River is the mean weight from your sample
i.e. is 31.2 pounds because the best point estimate for population mean is the sample mean. [in normal distribution]
Answer with Step-by-step explanation:
Let a mass weighing 16 pounds stretches a spring
feet.
Mass=![m=\frac{W}{g}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7BW%7D%7Bg%7D)
Mass=![m=\frac{16}{32}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B16%7D%7B32%7D)
![g=32 ft/s^2](https://tex.z-dn.net/?f=g%3D32%20ft%2Fs%5E2)
Mass,m=
Slug
By hook's law
![w=kx](https://tex.z-dn.net/?f=w%3Dkx)
![16=\frac{8}{3} k](https://tex.z-dn.net/?f=16%3D%5Cfrac%7B8%7D%7B3%7D%20k)
![k=\frac{16\times 3}{8}=6 lb/ft](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B16%5Ctimes%203%7D%7B8%7D%3D6%20lb%2Fft)
![f(t)=10cos(3t)](https://tex.z-dn.net/?f=f%28t%29%3D10cos%283t%29)
A damping force is numerically equal to 1/2 the instantaneous velocity
![\beta=\frac{1}{2}](https://tex.z-dn.net/?f=%5Cbeta%3D%5Cfrac%7B1%7D%7B2%7D)
Equation of motion :
![m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}+f(t)](https://tex.z-dn.net/?f=m%5Cfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%3D-kx-%5Cbeta%20%5Cfrac%7Bdx%7D%7Bdt%7D%2Bf%28t%29)
Using this equation
![\frac{1}{2}\frac{d^2x}{dt^2}=-6x-\frac{1}{2}\frac{dx}{dt}+10cos(3t)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%3D-6x-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bdx%7D%7Bdt%7D%2B10cos%283t%29)
![\frac{1}{2}\frac{d^2x}{dt^2}+\frac{1}{2}\frac{dx}{dt}+6x=10cos(3t)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%2B%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bdx%7D%7Bdt%7D%2B6x%3D10cos%283t%29)
![\frac{d^2x}{dt^2}+\frac{dx}{dt}+12x=20cos(3t)](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%2B%5Cfrac%7Bdx%7D%7Bdt%7D%2B12x%3D20cos%283t%29)
Auxillary equation
![m^2+m+12=0](https://tex.z-dn.net/?f=m%5E2%2Bm%2B12%3D0)
![m=\frac{-1\pm\sqrt{1-4(1)(12)}}{2}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B-1%5Cpm%5Csqrt%7B1-4%281%29%2812%29%7D%7D%7B2%7D)
![m=\frac{-1\pmi\sqrt{47}}{2}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B-1%5Cpmi%5Csqrt%7B47%7D%7D%7B2%7D)
![m_1=\frac{-1+i\sqrt{47}}{2}](https://tex.z-dn.net/?f=m_1%3D%5Cfrac%7B-1%2Bi%5Csqrt%7B47%7D%7D%7B2%7D)
![m_2=\frac{-1-i\sqrt{47}}{2}](https://tex.z-dn.net/?f=m_2%3D%5Cfrac%7B-1-i%5Csqrt%7B47%7D%7D%7B2%7D)
Complementary function
![e^{\frac{-t}{2}}(c_1cos\frac{\sqrt{47}}{2}+c_2sin\frac{\sqrt{47}}{2})](https://tex.z-dn.net/?f=e%5E%7B%5Cfrac%7B-t%7D%7B2%7D%7D%28c_1cos%5Cfrac%7B%5Csqrt%7B47%7D%7D%7B2%7D%2Bc_2sin%5Cfrac%7B%5Csqrt%7B47%7D%7D%7B2%7D%29)
To find the particular solution using undetermined coefficient method
![x_p(t)=Acos(3t)+Bsin(3t)](https://tex.z-dn.net/?f=x_p%28t%29%3DAcos%283t%29%2BBsin%283t%29)
![x'_p(t)=-3Asin(3t)+3Bcos(3t)](https://tex.z-dn.net/?f=x%27_p%28t%29%3D-3Asin%283t%29%2B3Bcos%283t%29)
![x''_p(t)=-9Acos(3t)-9sin(3t)](https://tex.z-dn.net/?f=x%27%27_p%28t%29%3D-9Acos%283t%29-9sin%283t%29)
This solution satisfied the equation therefore, substitute the values in the differential equation
![-9Acos(3t)-9Bsin(3t)-3Asin(3t)+3Bcos(3t)+12(Acos(3t)+Bsin(3t))=20cos(3t)](https://tex.z-dn.net/?f=-9Acos%283t%29-9Bsin%283t%29-3Asin%283t%29%2B3Bcos%283t%29%2B12%28Acos%283t%29%2BBsin%283t%29%29%3D20cos%283t%29)
![(3B+3A)cos(3t)+(3B-3A)sin(3t)=20cso(3t)](https://tex.z-dn.net/?f=%283B%2B3A%29cos%283t%29%2B%283B-3A%29sin%283t%29%3D20cso%283t%29)
Comparing on both sides
![3B+3A=20](https://tex.z-dn.net/?f=3B%2B3A%3D20)
![3B-3A=0](https://tex.z-dn.net/?f=3B-3A%3D0)
Adding both equation then, we get
![6B=20](https://tex.z-dn.net/?f=6B%3D20)
![B=\frac{20}{6}=\frac{10}{3}](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B20%7D%7B6%7D%3D%5Cfrac%7B10%7D%7B3%7D)
Substitute the value of B in any equation
![3A+10=20](https://tex.z-dn.net/?f=3A%2B10%3D20)
![3A=20-10=10](https://tex.z-dn.net/?f=3A%3D20-10%3D10)
![A=\frac{10}{3}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B10%7D%7B3%7D)
Particular solution, ![x_p(t)=\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)](https://tex.z-dn.net/?f=x_p%28t%29%3D%5Cfrac%7B10%7D%7B3%7Dcos%283t%29%2B%5Cfrac%7B10%7D%7B3%7Dsin%283t%29)
Now, the general solution
![x(t)=e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)](https://tex.z-dn.net/?f=x%28t%29%3De%5E%7B-%5Cfrac%7Bt%7D%7B2%7D%7D%28c_1cos%28%5Cfrac%7B%5Csqrt%7B47%7Dt%7D%7B2%7D%29%2Bc_2sin%28%5Cfrac%7B%5Csqrt%7B47%7Dt%7D%7B2%7D%29%2B%5Cfrac%7B10%7D%7B3%7Dcos%283t%29%2B%5Cfrac%7B10%7D%7B3%7Dsin%283t%29)
From initial condition
x(0)=2 ft
x'(0)=0
Substitute the values t=0 and x(0)=2
![2=c_1+\frac{10}{3}](https://tex.z-dn.net/?f=2%3Dc_1%2B%5Cfrac%7B10%7D%7B3%7D)
![2-\frac{10}{3}=c_1](https://tex.z-dn.net/?f=2-%5Cfrac%7B10%7D%7B3%7D%3Dc_1)
![c_1=\frac{-4}{3}](https://tex.z-dn.net/?f=c_1%3D%5Cfrac%7B-4%7D%7B3%7D)
![x'(t)=-\frac{1}{2}e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+e^{-\frac{t}{2}}(-c_1\frac{\sqrt{47}}{2}sin(\frac{\sqrt{47}t}{2})+\frac{\sqrt{47}}{2}c_2cos(\frac{\sqrt{47}t}{2})-10sin(3t)+10cos(3t)](https://tex.z-dn.net/?f=x%27%28t%29%3D-%5Cfrac%7B1%7D%7B2%7De%5E%7B-%5Cfrac%7Bt%7D%7B2%7D%7D%28c_1cos%28%5Cfrac%7B%5Csqrt%7B47%7Dt%7D%7B2%7D%29%2Bc_2sin%28%5Cfrac%7B%5Csqrt%7B47%7Dt%7D%7B2%7D%29%2Be%5E%7B-%5Cfrac%7Bt%7D%7B2%7D%7D%28-c_1%5Cfrac%7B%5Csqrt%7B47%7D%7D%7B2%7Dsin%28%5Cfrac%7B%5Csqrt%7B47%7Dt%7D%7B2%7D%29%2B%5Cfrac%7B%5Csqrt%7B47%7D%7D%7B2%7Dc_2cos%28%5Cfrac%7B%5Csqrt%7B47%7Dt%7D%7B2%7D%29-10sin%283t%29%2B10cos%283t%29)
Substitute x'(0)=0
![0=-\frac{1}{2}\times c_1+10+\frac{\sqrt{47}}{2}c_2](https://tex.z-dn.net/?f=0%3D-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20c_1%2B10%2B%5Cfrac%7B%5Csqrt%7B47%7D%7D%7B2%7Dc_2)
![\frac{\sqrt{47}}{2}c_2-\frac{1}{2}\times \frac{-4}{3}+10=0](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B47%7D%7D%7B2%7Dc_2-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7B-4%7D%7B3%7D%2B10%3D0)
![\frac{\sqrt{47}}{2}c_2=-\frac{2}{3}-10=-\frac{32}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B47%7D%7D%7B2%7Dc_2%3D-%5Cfrac%7B2%7D%7B3%7D-10%3D-%5Cfrac%7B32%7D%7B3%7D)
![c_2==-\frac{64}{3\sqrt{47}}](https://tex.z-dn.net/?f=c_2%3D%3D-%5Cfrac%7B64%7D%7B3%5Csqrt%7B47%7D%7D)
Substitute the values then we get
![x(t)=e^{-\frac{t}{2}}(-\frac{4}{3}cos(\frac{\sqrt{47}t}{2})-\frac{64}{3\sqrt{47}}sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)](https://tex.z-dn.net/?f=x%28t%29%3De%5E%7B-%5Cfrac%7Bt%7D%7B2%7D%7D%28-%5Cfrac%7B4%7D%7B3%7Dcos%28%5Cfrac%7B%5Csqrt%7B47%7Dt%7D%7B2%7D%29-%5Cfrac%7B64%7D%7B3%5Csqrt%7B47%7D%7Dsin%28%5Cfrac%7B%5Csqrt%7B47%7Dt%7D%7B2%7D%29%2B%5Cfrac%7B10%7D%7B3%7Dcos%283t%29%2B%5Cfrac%7B10%7D%7B3%7Dsin%283t%29)
10.5 is the answer to your question
Guess - early childhood education?
A way a teacher can help the student by using preferential seating, so that the student can see the teacher's mouth when he/she is speaking. Reduce background noise in the classroom. Use preferential seating, so that the teacher can more privately model correct articulation to the student. Give additional time for the student to speak. This way the child will have an easier time learning.