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ExtremeBDS [4]
3 years ago
12

Which solution set describes the solutions to the inequality below?

Mathematics
1 answer:
Nuetrik [128]3 years ago
7 0
<h2>Hello!</h2>

The answer is:

The sixth option,

[-6,6]

<h2>Why?</h2>

To solve the problem, we need to find which of the given options match with the solution to the inequality. As we know, solving inequalities is almost the same that solving equalities, we need to isolate the variable and then, find where it satisfies the given condition (inequality)

So, solving the inequality we have:

-3x^{2} +1\leq -107\\-3x^{2}\leq -107-1\\-3x^{2} \leq -108\\3x^{2} \leq 108\\x^{2} \leq \frac{108}{3}\\\sqrt{x^{2} } \leq \sqrt{36}\\ x^{2\leq } +-36\\(x-6)(x+6)\leq 0

Therefore, we have that the solutions to the inequality are:

First solution:

x-6\leq 0\\x\leq 6

Second solution:

x+6\leq 0\\x\leq-6

Hence, the correct option is the sixth option, the solution is described by the following solution set :

[-6,6]

Have a nice day!

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Thus, the expected value of points for Derek and Mia are \dfrac{-1}{9} and \dfrac{1}{9} respectively.

Step-by-step explanation:

Number of green marbles = 2 and Number of Yellow marbles = 1

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A person selects two marbles one after another after replacing them.

So, the probabilities of selecting different combinations of colors are,

1.\ P(GG)=P(G)\times P(G)\\\\P(GG)=\dfrac{2}{3}\times \dfrac{2}{3}\\\\P(GG)=\dfrac{4}{9}

2.\ P(GY)=P(G)\times P(Y)\\\\P(GY)=\dfrac{2}{3}\times \dfrac{1}{3}\\\\P(GY)=\dfrac{2}{9}

3.\ P(YG)=P(Y)\times P(G)\\\\P(YG)=\dfrac{1}{3}\times \dfrac{2}{3}\\\\P(YG)=\dfrac{2}{9}

4.\ P(YY)=P(Y)\times P(Y)\\\\P(YY)=\dfrac{1}{3}\times \dfrac{1}{3}\\\\P(YY)=\dfrac{1}{9}

Now, we have that,

If two marbles are of same color, then Mia gains 1 point and Derek loses 1 point.

If two marbles are of different color, then Derek gains 1 point and Mia loses 1 point.

<h3>Also, the expected value of a random variable X is E(X)=\sum_{i=1}^{n} x_i\times P(x_i).</h3>

Then, the expected value of points for Derek is,

E(D)= (-1)\times \dfrac{4}{9}+1\times \dfrac{2}{9}+1\times \dfrac{2}{9}+(-1)\times \dfrac{1}{9}\\\\E(D)= \dfrac{-5}{9}+\dfrac{4}{9}\\\\E(D)=\dfrac{-1}{9}

And the expected value of points for Mia is,

E(M)= 1\times \dfrac{4}{9}+(-1)\times \dfrac{2}{9}+(-1)\times \dfrac{2}{9}+1\times \dfrac{1}{9}\\\\E(M)= \dfrac{5}{9}-\dfrac{4}{9}\\\\E(M)=\dfrac{1}{9}.

Thus, the expected value of points for Derek and Mia are \dfrac{-1}{9} and \dfrac{1}{9} respectively.

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