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Art [367]
4 years ago
12

A restaurant offers a "Create Your Own Pizza" menu option. There are 3 types of meat, 6 types of vegetables, and 3 kinds of chee

se. Customers may choose one meat, one vegetable, and one cheese. You decide to have the chef choose the toppings randomly. What is the probability that the chef will make a pizza with chicken, banana peppers, and ricotta cheese?
Mathematics
1 answer:
zheka24 [161]4 years ago
6 0
Multiply 3 times 6 times 3 to get the total number of choices.
Since there can only be one meat, one vegetable, and one cheese, the chances are 1/54.
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Step-by-step explanation:

If x_1 and x_2 are the solutions to the quadratic equation, then this equation can be written as

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In your case,

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Then the equation is

\left(x-\left(-9+\dfrac{\sqrt{65}}{2}\right)\right) \left(x-\left(-9-\dfrac{\sqrt{65}}{2}\right)\right)=0\\ \\\left(x+9-\dfrac{\sqrt{65}}{2}\right)\left(x+9+\dfrac{\sqrt{65}}{2}\right)=0\\ \\x^2+9x+\dfrac{\sqrt{65}}{2}x+9x+81+\dfrac{9\sqrt{65}}{2}-\dfrac{\sqrt{65}}{2}x-\dfrac{9\sqrt{65}}{2}-\dfrac{65}{4}=0\\ \\x^2+18x+\dfrac{259}{4}=0\\ \\4x^2+72x+259=0

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