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slavikrds [6]
2 years ago
13

Explain how the graph of y=|x+1|-2 will differ from the graph of y=|x|.

Mathematics
1 answer:
Natali5045456 [20]2 years ago
3 0

\bf ~~~~~~~~~~~~\textit{function transformations}\\\\\\f(x)=  A(  Bx+  C)+  D\\\\~~~~y=  A(  Bx+  C)+  D\\\\f(x)=  A\sqrt{  Bx+  C}+  D\\\\f(x)=  A(\mathbb{R})^{  Bx+  C}+  D\\\\f(x)=  A sin\left( B x+  C  \right)+  D\\\\--------------------\\\\\bullet \textit{ stretches or shrinks horizontally by  }   A\cdot   B\\\\\bullet \textit{ flips it upside-down if }  A\textit{ is negative}

\bf ~~~~~~\textit{reflection over the x-axis}\\\\\bullet \textit{ flips it sideways if }  B\textit{ is negative}\\~~~~~~\textit{reflection over the y-axis}\\\\\bullet \textit{ horizontal shift by }\frac{  C}{  B}\\~~~~~~if\ \frac{  C}{  B}\textit{ is negative, to the right}

\bf ~~~~~~if\ \frac{  C}{  B}\textit{ is positive, to the left}\\\\\bullet \textit{ vertical shift by }  D\\~~~~~~if\   D\textit{ is negative, downwards}\\\\~~~~~~if\   D\textit{ is positive, upwards}\\\\\bullet \textit{ period of }\frac{2\pi }{  B}

with that template in mind,

\bf y=\stackrel{A}{1}|\stackrel{B}{1}x\stackrel{C}{+1}|\stackrel{D}{-2}

so is really just |x| but shifted some,

C/B = 1/1 = +1, horizontally to the right by 1 unit.

D = -2, vertically downwards by 2 units.

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Step-by-step explanation:

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what's the probability that at most two of them are defective

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- We will define random variable X : The number of defective bulbs picked.

Such that,               P ( X ≤ 2 ) is required!

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        X = 0 ,       Number of choices = 15 C r = 15C2 = 105 ways

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      P ( X = 0 ) = number of choices with no defective / Total choices

                       = 105 / 20C2 = 105 / 190

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- The number of ways we choose 2 bulbs such that one of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 1 and out of defective n = 5 choose r = 1 defective bulb:

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      P ( X = 1 ) = number of choices with 1 defective / Total choices

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                       = 0.3947

- The number of ways we choose 2 bulbs such that both of them are defective, out of 5 available defective bulbs choose r = 2 defective.

        X = 2 ,       Number of choices = 5 C 2 = 10 ways

- The probability of selecting 2 defective bulbs:

      P ( X = 2 ) = number of choices with 2 defective / Total choices

                       = 10 / 20C2 = 10 / 190

                       = 0.05263

- Hence,

    P ( X ≤ 2 ) = P ( X =0 ) + P ( X = 1 ) + P (X =2)

                     = 0.5526 + 0.3947 + 0.05263

                     = 1

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