Find the zeros of the function y = -4x^2 + 16

2 answers:

7 0

$-4x^2+16 & =0\\\\\implies -4(x^2-4) & =0 \\\\\implies x^2-4 =0\\ \\\implies x^2-2^2 =0\\\\\implies (x+2)(x-2)=0\\\\\implies x=2,~~x = -2$

azamat1 year ago

3 0

Step-by-step explanation:

\begin{gathered}-4x^2+16 & =0\\\\\implies -4(x^2-4) & =0 \\\\\implies x^2-4 =0\\ \\\implies x^2-2^2 =0\\\\\implies (x+2)(x-2)=0\\\\\implies x=2,~~x = -2\end{gathered}

−4x

+16

⟹−4(x

−4)

⟹x

−4=0

⟹x

−2

⟹(x+2)(x−2)=0

⟹x=2, x=−2

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Using the definition of expected value, it is found that Ayo can be expected to make a profit of £55.8.

The <em>expected value</em> is given by the <u>sum of each outcome multiplied by it's respective probability.</u>

In this problem:

The player wins $6, that is, Ayo loses £6, if he rolls a 6 and spins a 1, hence the probability is $\frac{1}{6} \times \frac{1}{12} = \frac{1}{72}$ .

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In the other cases, Ayo wins £1.40, with $1 - \frac{18}{72} = \frac{54}{72}$ probability.

Hence, his expected profit for a single game is:

$E(X) = -6\frac{1}{72} - 3\frac{17}{72} + 1.4\frac{54}{72} = \frac{-6 - 3(17) + 54(1.4)}{72} = 0.2583$

For 216 games, the expected value is:

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To learn more about expected value, you can take a look at brainly.com/question/24855677

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