Solve for x: 2sin^2(x) -sin(x)=1 over the interval 0 to 2pi

1 answer:

6 0

$2\sin^2x-\sin x=1\ \ \ \ |subtract\ 1\ from\ both\ sides\\\\2\sin^2x-\sin x-1=0\ \ \ \ |subtitute\ \sin x=t\ where\ t\in[-1;\ 1]\\\\2t^2-t-1=0\\\\2t^2-2t+t-1=0\\\\2t(t-1)+1(t-1)=0\\\\(t-1)(2t+1)=0\iff t-1=0\ or\ 2t+1=0\\\\t=1\ or\ 2t=-1\to t=-\dfrac{1}{2}\\\\therefore\\\sin x=1\ or\ \sin x=-\dfrac{1}{2}\\\\Answer:\boxed{t\in\left\{\dfrac{\pi}{2};\ \dfrac{7\pi}{6};\ \dfrac{11\pi}{6}\right\}}$

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A car dealership sells 0, 1, or 2 luxury cars on any day. When selling a car, the dealer also tries to persuade the customer to

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Answer:

Mean = 1.42

Variance = 0.58

Step-by-step explanation:

Given: X denote the number of luxury cars sold in a given day, and Y denote the number of extended warranties sold.

Also, joint probability function of X and Y are given.

To find:

mean and variance of X

Solution:

From the given joint probability function of X and Y,

$P(X=0)=\frac{1}{6}\\P(X=1)=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}=\frac{3}{12}\\P(X=2)=\frac{1}{12}+\frac{1}{3}+\frac{1}{6}=\frac{1+4+2}{12}=\frac{7}{12}$

Mean of X:

$E(X)=\sum XP(X)\\=0\left ( \frac{1}{6} \right )+1\left ( \frac{3}{12} \right )+2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{14}{12}\\=\frac{17}{12}=1.42$

Variance of X:

$E(X^2)=\sum X^2P(X)\\=0^2\left ( \frac{1}{6} \right )+1^2\left ( \frac{3}{12} \right )+2^2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{28}{12}\\=\frac{31}{12}$

$var(X)=E\left [ X^2 \right ]-\left ( E\left [ X \right ] \right )^2\\=\frac{31}{12}-\left ( \frac{17}{12} \right )^2\\=\frac{31}{12}-\frac{289}{144}\\=\frac{372-289}{144}\\=\frac{83}{144}\\=0.58$