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2 years ago

11

A cooler contains 7 cans of soda; 2 colas, 4 orange, and 1 cherry. Two cans are selected at random without replacement. Find

the probability that at least one can is cherry

1 answer:

Mazyrski [523]2 years ago

7 0

The probability is 2/14

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$\bf ~\hspace{10em}\textit{function transformations} \\\\\\ \begin{array}{llll} f(x)= A( Bx+ C)^2+ D \\\\ f(x)= A\sqrt{ Bx+ C}+ D \\\\ f(x)= A(\mathbb{R})^{ Bx+ C}+ D \end{array}\qquad \qquad \begin{array}{llll} f(x)=\cfrac{1}{A(Bx+C)}+D \\\\\\ f(x)= A sin\left( B x+ C \right)+ D \end{array} \\\\[-0.35em] ~\dotfill\\\\ \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\ \bullet \textit{ flips it upside-down if } A\textit{ is negative}$

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

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

now with that template in mind

$\bf h(x)=\stackrel{A}{2}|\stackrel{B}{1}x+\stackrel{C}{3}|\stackrel{D}{-1}\qquad \qquad \stackrel{\textit{C=C-2 a translation to the right}}{h(x)=2|x\boxed{+3-2}|-1} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill h(x)=2|x+1|-1~\hfill$

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Because first you do 36/4 which gets you to 9 and then subtract it by 5 to get 4!

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