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alexdok [17]
3 years ago
6

Which two lines are parallel?

Mathematics
1 answer:
Masteriza [31]3 years ago
6 0

Answer:

  • 3x + y = -2
  • 2y = -6x – 8

Step-by-step explanation:

When the equations are written in the same form, the ratio of x coefficient to y coefficient is the same for parallel lines.

These equations in the same form are ...

  • 3x +y = -2
  • -6x +2y = 4
  • 6x +2y = -8

The first and last of these equations have a ratio of x-coefficient to y-coefficient of ...

  (x-coefficient)/(y-coefficient) = 3/1 = 6/2

so the lines are parallel.

The second equation has a ratio that is the opposite of this.

__

<em>Alternate approach</em>

You can put each equation into slope-intercept form (solve for y). The x-coefficient is then the slope of the line. Parallel lines** have equal slopes.

  • y = -3x -2 **
  • y = 3x +2
  • y = -3x -4 **

_____

I find a graphing calculator to be a big help for many problems involving equations and their characteristics.

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3 years ago
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Hal and Renee play the following game a bag has 11 tiles in it each with a letter from the word probably in it Hal and Renee tak
Masteriza [31]

Answer:

  • <em>D. The game is not fair because the probability of Hal drawing a winning letter is less then the probability of Renee drawing a winning letter </em>

Explanation:

The word is not probably, which has 8 letters, but probability, which has 11 letters

<u>1. Hal</u>

Probability of drawing a vowel

  • The vowels are:  o, a, i, i, and y: 5

  • The total number of letters in the bag is 11.

  • Probability of vowel = 5 / 11

<u>2. Renee</u>

Probability of drawing a consonant

  • The consonants are: p, r, b, b, l, t, and y: 7

  • Probability of a drawing a consonant: 7 /11

<h2>Conclusion</h2>

As you see, the probability of drawing a vowel (5/11), which is the winning letter for Hal, is less than the probability of drawing a consonant (7/11), which is the winning letter of Renee. Then,

  • The game is not fair because the probability of Hal drawing a winning letter is less then the probability of Renee drawing a winning letter.

8 0
3 years ago
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Step-by-step explanation:

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7/12 + 4/12 = 11/12

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LenaWriter [7]

Answer:

E[X^2]= \frac{2!}{2^1 1!}= 1

Var(X^2)= 3-(1)^2 =2

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

\phi(t) = E[e^{tX}]

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx

And we have that the moment generating function can be write like this:

\phi(t) = e^{\frac{t^2}{2}

And we can write this as an infinite series like this:

\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]

E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}

And then we have this:

E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...

And then we can find the E[X^2]

E[X^2]= \frac{2!}{2^1 1!}= 1

And we can find the variance like this :

Var(X^2) = E[X^4]-[E(X^2)]^2

And first we find:

E[X^4]= \frac{4!}{2^2 2!}= 3

And then the variance is given by:

Var(X^2)= 3-(1)^2 =2

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3 years ago
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