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Montano1993 [528]
3 years ago
7

Simplify y-(-3)=5/3(x-4)

Mathematics
1 answer:
Anit [1.1K]3 years ago
7 0

Answer:

    x = 3 y + 29 / 5

Step-by-step explanation:

                       5  / 3  x +  −20 /  3  = y + 3

                        5 / 3 x + -20 / 3 +20 / 3 = y + 3 + 20 / 3 <em>( adding 20 / 3 at both the sides, LHS and RHS )</em>

                         5 / 3 x = y + 29 / 3

                         5 / 3 x / 5 / 3 = y + 29 / 3 / 5 / 3 (<em>dividing 5/3 at both the sides, LHS and RHS )</em>

                    x = 3 y + 29 / 5

Hope it helps

Wishing u a gr8 day ahead

:D

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Tell whether the angles are complementary or supplementary. Then find the value of x.
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Answer:

x = 33

Step-by-step explanation:

Two angles are complementary as the sum of two angles is 90

x + 24 + x = 90        {Combine like terms}

  2x + 24 = 90      {Subtract 24 from both sides}

          2x  = 90 - 24

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3 years ago
I need help I’m bad at math
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The width of each rect. is 1/2.

The heights of the 3 inscribed rect. are {-2.25+6, -1+6, -.25+6} = {3.75,5,5.75}.

The areas of these 3 inscribed rect. are (1/2)*{3.75,5,5.75}, which come out to:

{1.875, 2.5, 2.875}

Add these three areas together; you sum will represent the approx. area under the given curve on the given interval:  1.875+2.5+2.875 = ?
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3 years ago
the domain of a function f(x) is x &gt; 1, and the range is y &lt; -2. What are the domain and range of its inverse function, f^
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9514 1404 393

Answer:

  • domain: x < -2
  • range: y > 1

Step-by-step explanation:

Domain and range of the inverse function are swapped: the domain of the function is the range of its inverse, and vice versa.

__

The graph shows an example of a function and its inverse with the given domain and range. The function is red; its inverse is blue.

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Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????
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Answer:

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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 +....]

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And then we have this:

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

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

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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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Answer:

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Step-by-step explanation:

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