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leva [86]
3 years ago
6

Convert the Cartesian equation (x 2 + y 2)2 = 4(x 2 - y 2) to a polar equation.

Mathematics
1 answer:
SashulF [63]3 years ago
7 0

<u>ANSWER</u>

{r}^{2}  = 4  \cos2\theta

<u>EXPLANATION</u>

The Cartesian equation is

{( {x}^{2}  +  {y}^{2} )}^{2}  = 4( {x}^{2} -  {y}^{2}  )

We substitute

x = r \cos( \theta)

y = r \sin( \theta)

and

{x}^{2}  +  {y}^{2}  =  {r}^{2}

This implies that

{( {r}^{2} )}^{2}  = 4(( { r \cos\theta)  }^{2} -  {(r \sin\theta) }^{2}  )

Let us evaluate the exponents to get:

{r}^{4}  = 4({  {r}^{2} \cos^{2}\theta } -   {r}^{2}  \sin^{2}\theta)

Factor the RHS to get:

{r}^{4}  = 4{r}^{2} ({   \cos^{2}\theta } -   \sin^{2}\theta)

Divide through by r²

{r}^{2}  = 4 ({   \cos^{2}\theta } -   \sin^{2}\theta)

Apply the double angle identity

\cos^{2}\theta -\sin^{2}\theta=  \cos(2 \theta)

The polar equation then becomes:

{r}^{2}  = 4  \cos2\theta

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3 years ago
The random variable x has the following probability distribution: x f(x) 0 .25 1 .20 2 .15 3 .30 4 .10 a. Is this probability di
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Given : The random variable x has the following probability distribution.

To find :

a. Is this probability distribution valid? Explain and list the requirements for a valid probability distribution.

b. Calculate the expected value of x.

c. Calculate the variance of x.

d. Calculate the standard deviation of x.

Solution :

First we create the table as per requirements,

x      P(x)         xP(x)           x²            x²P(x)

0    0.25           0               0                0

1     0.20        0.20             1              0.20

2    0.15          0.3               4             0.6

3    0.30         0.9               9             2.7

4    0.10          0.4               16             1.6

   ∑P(x)=1     ∑xP(x)=1.8               ∑x²P(x)=5.1

a) To determine that table shows a probability distribution we add up all five probabilities if the sum is 1 then it is a valid distribution.

\sum P(X)=0.25+0.20+0.15+0.30+0.10

\sum P(X)=1

Yes it is a probability distribution.

b) The expected value of x is defined as

E(x)=\sum xP(x)=1.8

c) The variance of x is defined as

V=\sum x^2P(x)-(\sum xP(x))^2\\V=5.1-(1.8)^2\\V=5.1-3.24\\V=1.86

d) The standard deviation of x is  defined as

\sigma=\sqrt{V}

\sigma=\sqrt{1.86}

\sigma=1.136

5 0
3 years ago
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