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BabaBlast [244]
3 years ago
10

A set of data has a mean of 0.5 and a standard deviation of 0.01. A data point of the set has a z-score of 2.5. What does a z-sc

ore of 2.5 mean?
Mathematics
1 answer:
Paul [167]3 years ago
7 0
A z-score of 2.5 corresponds to a data point x such that

2.5=\dfrac{x-0.5}{0.01}\implies x=0.525
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(10 points) Consider the initial value problem y′+3y=9t,y(0)=7. Take the Laplace transform of both sides of the given differenti
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Answer:

The solution

Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3 t}

Step-by-step explanation:

<u><em>Explanation</em></u>:-

Consider the initial value problem y′+3 y=9 t,y(0)=7

<em>Step(i)</em>:-

Given differential problem

                           y′+3 y=9 t

<em>Take the Laplace transform of both sides of the differential equation</em>

                L( y′+3 y) = L(9 t)

 <em>Using Formula Transform of derivatives</em>

<em>                 L(y¹(t)) = s y⁻(s)-y(0)</em>

  <em>  By using Laplace transform formula</em>

<em>               </em>L(t) = \frac{1}{S^{2} }<em> </em>

<em>Step(ii):-</em>

Given

             L( y′(t)) + 3 L (y(t)) = 9 L( t)

            s y^{-} (s) - y(0) +  3y^{-}(s) = \frac{9}{s^{2} }

            s y^{-} (s) - 7 +  3y^{-}(s) = \frac{9}{s^{2} }

Taking common y⁻(s) and simplification, we get

             ( s +  3)y^{-}(s) = \frac{9}{s^{2} }+7

             y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}

<em>Step(iii</em>):-

<em>By using partial fractions , we get</em>

\frac{9}{s^{2} (s+3} = \frac{A}{s} + \frac{B}{s^{2} } + \frac{C}{s+3}

  \frac{9}{s^{2} (s+3} =  \frac{As(s+3)+B(s+3)+Cs^{2} }{s^{2} (s+3)}

 On simplification we get

  9 = A s(s+3) +B(s+3) +C(s²) ...(i)

 Put s =0 in equation(i)

   9 = B(0+3)

 <em>  B = 9/3 = 3</em>

  Put s = -3 in equation(i)

  9 = C(-3)²

  <em>C = 1</em>

 Given Equation  9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Comparing 'S²' coefficient on both sides, we get

  9 = A s²+3 A s +B(s)+3 B +C(s²)

 <em> 0 = A + C</em>

<em>put C=1 , becomes A = -1</em>

\frac{9}{s^{2} (s+3} = \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}

<u><em>Step(iv):-</em></u>

y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}

y^{-}(s)  =9( \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}) + \frac{7}{s+3}

Applying inverse Laplace transform on both sides

L^{-1} (y^{-}(s) ) =L^{-1} (9( \frac{-1}{s}) + L^{-1} (\frac{3}{s^{2} }) + L^{-1} (\frac{1}{s+3}) )+ L^{-1} (\frac{7}{s+3})

<em>By using inverse Laplace transform</em>

<em></em>L^{-1} (\frac{1}{s} ) =1<em></em>

L^{-1} (\frac{1}{s^{2} } ) = \frac{t}{1!}

L^{-1} (\frac{1}{s+a} ) =e^{-at}

<u><em>Final answer</em></u>:-

<em>Now the solution , we get</em>

Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3t}

           

           

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Answer:

w=4.9

Step-by-step explanation:

If you are solving for w, you want to isolate the variable. If it is easier to write it out, it would look something like this..

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To get w by itself, use the inverse of multiplication, which is division. Divide the expression to the left of the equal sign by -5. Remember, what you do to one side, you must do to the other. When you divide -24.5 by -5, you end up with

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