一平面简谐波沿 OX轴负方向传播。已知 x = b处质点的振动方程为 y = Acos (w t +j0),波速为 u , 则波动方程为
A.
B.
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D.
- · 有3位网友选择 C,占比33.33%
- · 有3位网友选择 B,占比33.33%
- · 有3位网友选择 D,占比33.33%
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A、y =Acos{ω[t-(x0-x)/u]+φ0}
B、y =Acos{ω[t-(x-x0)/u]+φ0}
C、y =Acos{ω[t+(x0-x)/u]+φ0}
D、y =Acos{ω[t+(x-x0)/u]+φ0}
一平面简谐波沿X轴正向传播,已知x=L(L<λ)处质点的振动方程为y=Acos(∞t+φ0),波速为u,那么x=0处质点的振动方程为:()
A. y=Acos[ω(t+L/u)+φ0]
B. y=Acos[ω(t-L/u)+φ0]
C. y=Acos[ωt+L/u+φ0]
D. y=Acos[ωt-L/u+φ0]
A、y=Acos{w[t-(x0-x)/u]+f0}.
B、y=Acos{w[t-(x-x0)/u]+f0}.
C、y=Acos{w t-[(x0-x)/u]+f0}.
D、y=Acos{w t+[(x0-x)/u]+f0}.
A.y=Acosω[t-(x-L)/u]
B.y=Acosω[t-(x+L)/u]
C.y=Acosω[t+(x+L)/u]
D.y=Acosω[t+(x-L)/u]
一平面简谐波沿X轴正向传播,已知x=1(1<λ)处质点的振动方程为y=Acoswt+φ0),波速为u,那么x=0处质点的振动方程为:()
A. y=Acos[w(t+1/u)+φ0]
B. y=ACOS[w(t-1/u)+φ0]
C. y=Acos[wt+1/u+φ0]
D. y=Acos[wt-1/u+φ0]
A.y=Acos[w(t+1/u)+φ0]
B.y=ACOS[w(t-1/u)+φ0]
C.y=Acos[wt+1/u+φ0]
D.y=Acos[wt-1/u+φ0]
A、y=0.3cos[2π(t-x/20)+π/2](m);
B、y=0.3cos[2π(t+x/20)-π/2](m);
C、y=0.3cos[2π(t-x/20)-π/2](m);
D、y=0.3cos[2π(t+x/20)+π/2](m);
To collect data in an introductory statistics course, recently I gave the students a questionnaire. One question asked whether the student was a vegetarian. Of 25 students, 0 answered “yes.”They were not a random sample, but let us use these data to illustrate inference for a proportion. (You may wish to refer to Section1.4.1onmethodsofinference.) Let π denote the population proportion who would say “yes.” Consider H0:π =0.50 and Ha:π =0.50. a. What happens when you try to conduct the “Wald test,” for whichuses the estimated standard error? b. Find the 95% “Wald confidence interval” for π. Is it believable? (When the observation falls at the boundary of the sample space, often Wald methods do not provide sensible answers.) c. Conduct the “score test,” for whichuses the null standard error. Report the P-value. d. Verify that the 95% score confidence interval (i.e., the set of π0 for which |z|< 1.96 in the score test) equals (0.0,0.133). (Hint: What do the z test statistic and P-value equal when you test H0:π =0.133 against Ha:π ≠ 0.133.)
Data posted at the FBI website (www.fbi.gov) stated that of all blacks slain in 2005, 91% were slain by blacks, and of all whites slain in 2005, 83% were slain by whites. Let Y denote race of victim and X denote race of murderer. a. Which conditional distribution do these statistics refer to, Y given X, orX given Y? b. Calculate and interpret the odds ratio between X and Y. c. Given that a murderer was white, can you estimate the probability that the victim was white?What additional information would you need to do this? (Hint: How could you use Bayes’s Theorem?)
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