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Alchen [17]
3 years ago
11

An opinion poll asks a random sample of adults whether they favor banning ownership of handguns by private citizens. A commentat

or believes that more than half of all adults favor such a ban. The null and alternative hypotheses you would use to test this claim are (a)
H0:p^=0.5;Ha:p^>0.5
(b)
H0:p=0.5;Ha:p>0.5
(c)
H0:p=0.5;Ha:p<0.5
(d)
H0:p=0.5;Ha:p≠0.5
(e)
H0:p>0.5;Ha:p=0.5
Mathematics
1 answer:
andreev551 [17]3 years ago
3 0

Answer:

(b) H0: p = 0.5; Ha: p > 0.5

Step-by-step explanation:

p represents the population proportion.

A null hypothesis is a statement from the population parameter which is either rejected or accepted (fail to reject) upon testing. It is expressed using the equality sign.

An alternate hypothesis is also a statement from the population parameter which negates the null hypothesis and is accepted if the null hypothesis is rejected. It is expressed using any of the inequality signs.

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Consider the differential equation y'' − y' − 20y = 0. Verify that the functions e−4x and e5x form a fundamental set of solution
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Answer:

Therefore e^{-4x} and e^{5x} are fundamental solution of the given differential equation.

Therefore  e^{-4x} and e^{5x} are linearly independent, since W(e^{-4x},e^{5x})=9e^x\neq 0

The general solution of the differential equation is

y=c_1e^{-4x}+c_2e^{5x}

Step-by-step explanation:

Given differential equation is

y''-y'-20y =0

Here P(x)= -1, Q(x)= -20 and R(x)=0

Let trial solution be y=e^{mx}

Then, y'=me^{mx}   and   y''=m^2e^{mx}

\therefore m^2e^{mx}-m e^{mx}-20e^{mx}=0

\Rightarrow m^2-m-20=0

\Rightarrow m^2-5m+4m-20=0

\Rightarrow m(m-5)+4(m-5)=0

\Rightarrow (m-5)(m+4)=0

\Rightarrow m=-4,5

Therefore the complementary function is = c_1e^{-4x}+c_2e^{5x}

Therefore e^{-4x} and e^{5x} are fundamental solution of the given differential equation.

If y_1 and y_2 are the fundamental solution of differential equation, then

W(y_1,y_2)=\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|\neq 0

Then  y_1 and y_2 are linearly independent.

W(e^{-4x},e^{5x})=\left|\begin{array}{cc}e^{-4x}&e^{5x}\\-4e^{-4x}&5e^{5x}\end{array}\right|

                    =e^{-4x}.5e^{5x}-e^{5x}.(-4e^{-4x})

                    =5e^x+4e^x

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Therefore  e^{-4x} and e^{5x} are linearly independent, since W(e^{-4x},e^{5x})=9e^x\neq 0

Let the the particular solution of the differential equation is

y_p=v_1e^{-4x}+v_2e^{5x}

\therefore v_1=\int \frac{-y_2R(x)}{W(y_1,y_2)} dx

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\therefore v_2=\int \frac{y_1R(x)}{W(y_1,y_2)} dx

Here y_1= e^{-4x}, y_2=e^{5x},W(e^{-4x},e^{5x})=9e^x ,and  R(x)=0

\therefore v_1=\int \frac{-e^{5x}.0}{9e^x}dx

       =0

and

\therefore v_2=\int \frac{e^{5x}.0}{9e^x}dx

       =0

The the P.I = 0

The general solution of the differential equation is

y=c_1e^{-4x}+c_2e^{5x}

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