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2 years ago

5

A cooler contains 5 bottle of lemonade, 4 bottles of water, and 3 bottles of juice. Each type is equally likely to be chosen. Ba

sed on your simulation, how many times must a drink be chosen in order to get each type?

1 answer:

lord [1]2 years ago

5 0

I think the answer is 4

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Ksenya-84 [330]

<span>Solve.

x² + 4x + 4 = 18

A x=−4±3√ 2

B x=2±3√ 2

C x=4±9√ 2

D x=−2±3√2

</span> $x^2 + 4x + 4 = 18 \\ \\ x^2+4x+4-18= 0 \\ \\ x^2+4x-14=0 \\ \\ x_1_y_2= \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} \qquad a= 1\qquad b= 4\qquad c= -14 \\ \\ \\ x_1_y_2= \dfrac{-4\pm \sqrt{4^2-4(1)(-14)} }{2(1)} \\ \\ \\ x_1_y_2= \dfrac{-4\pm \sqrt{16-(-56)} }{2} \\ \\ \\ x_1_y_2= \dfrac{-4\pm \sqrt{72} }{2} \\ \\ \\ x_1= \dfrac{-4+ \sqrt{72} }{2} \qquad\qquad x_2= \dfrac{-4+ \sqrt{72} }{2}\\ \\ \\ x_1= \dfrac{-4+ \sqrt{2^3*3^2} }{2} \qquad\qquad x_2= \dfrac{-4- \sqrt{2^3*3^2} }{2}$
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</span> $x_1= \dfrac{-4+2*3 \sqrt{2} }{2} \qquad\qquad x_2= \dfrac{-4- 2*3\sqrt{2} }{2} \\ \\ \\ x_1= \dfrac{-4+6 \sqrt{2} }{2} \qquad\qquad x_2= \dfrac{-4- 6\sqrt{2} }{2} \\ \\ \\ x_1= \dfrac{-4}{2} + \dfrac{6 \sqrt{2} }{2} \qquad\qquad x_2= \dfrac{-4}{2}- \dfrac{ 6\sqrt{2} }{2} \\ \\ \\ x_1= -2 + 3 \sqrt{2} \qquad\qquad\quad x_2= -2- 3\sqrt{2} \\ \\ \\ \boxed{x= -2\pm 3\sqrt{2} } \to D)$ <span>
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3 years ago

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kodGreya [7K]

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