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ELEN [110]
3 years ago
13

How do I do this what is it asking

Mathematics
1 answer:
padilas [110]3 years ago
3 0
\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}
\end{array}

now.. notice the template above

so hmm  \bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\end{array}

\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{  D}}\\
\qquad if\ {{  D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}
\end{array}

on a)  D is 2
on b)  D is -2
on c) A is 2
on d) A is 1/2


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Answer:

Probability to pick yellow or orange sweet = 3 / 8

Step-by-step explanation:

Given:

Number of yellow sweet = 5

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Number of green sweet = 8

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Number of white sweet = 3

Find:

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Computation:

Total number of sweets = 24 sweets

Probability to pick yellow = 5 / 24

Probability to pick orange = 4 / 24

Probability to pick yellow or orange sweet = 5/24 + 4 / 24

Probability to pick yellow or orange sweet = 9 / 24

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I hope this is good enough:

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What is the solution set of {x | x < 2} {x | x ≥ 2}?
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Step-by-step explanation:

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