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Paladinen [302]
4 years ago
14

A survey of 46 college athletes found that 24 played volleyball, while 22 played basketball. a) If we pick one athlete survey pa

rticipant at random, what is the probability they play basketbali? b) If we pick 2 athletes at random (without replacement), what is the probability we get one volleyball player and one basketball player?
Mathematics
1 answer:
PSYCHO15rus [73]4 years ago
5 0

Answer:

A) \dfrac{11}{23}

B) \dfrac{88}{345}

Step-by-step explanation:

A survey of 46 college athletes found that

  • 24 played volleyball,
  • 22 played basketball.

A) If we pick one athlete survey participant at random,  the probability they play basketball is

P_1=\dfrac{22}{46}=\dfrac{11}{23}

B) If we pick 2 athletes at random (without replacement),

  • the probability we get one volleyball player is \dfrac{24}{46}=\dfrac{12}{23};
  • the probability we get another basketball player is \dfrac{22}{45} (only 45 athletes left).

Thus, the probability we get one volleyball player and one basketball player is

P_2=\dfrac{12}{23}\cdot \dfrac{22}{45}=\dfrac{88}{345}

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3 1/6

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. A firm produces a commodity and receives $100 for each unit sold. The cost of producing and selling x units is 20x + 0.25x 2 d
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a)
now, what's the highest profit they can make?  check the picture below.

\bf p(x)=-0.25x^2+80x\\\\
-------------------------------\\\\
\textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lcclll}
p(x) = &{{ -0.25}}x^2&{{ +80}}x&{{ +0}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
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\\\\\\
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\bf p(x)=-0.25x^2+70x\\\\
-------------------------------\\\\
\textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lcclll}
p(x) = &{{ -0.25}}x^2&{{ +70}}x&{{ +0}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad  {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
\\\\\\
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3 years ago
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