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spin [16.1K]
3 years ago
8

Solve for x: 2/3(x − 2) = 4x. A. x = -2/5 B. x = -5/2 C. x = 2/5 D. x = −2

Mathematics
2 answers:
forsale [732]3 years ago
5 0

Answer:

a

Step-by-step explanation:

malfutka [58]3 years ago
3 0

Answer:

\large\boxed{A.\ x=-\dfrac{2}{5}}

Step-by-step explanation:

\dfrac{2}{3}(x-2)=4x\qquad\text{multiply both sides by 3}\\\\3\!\!\!\!\diagup^1\cdot\dfrac{2}{3\!\!\!\!\diagup_1}(x-2)=(3)(4x)\\\\2(x-2)=12x\qquad\text{divide both sides by 2}\\\\\dfrac{2\!\!\!\!\diagup^1(x-2)}{2\!\!\!\!\diagup_1}=\dfrac{12\!\!\!\!\!\diagup^6x}{2\!\!\!\!\diagup_1}\\\\x-2=6x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x-x-2=6x-x\\\\-2=5x\qquad\text{divide both sides by 5}\\\\\dfrac{-2}{5}=\dfrac{5\!\!\!\!\diagup^1x}{5\!\!\!\!\diagup_1}\\\\-\dfrac{2}{5}=x\to x=-\dfrac{2}{5}

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Answer:

the probability that the mean student loan debt for these people is between $31000 and $33000 is 0.1331

Step-by-step explanation:

Given that:

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The probability that the mean student loan debt for these people is between $31000 and $33000 can be computed as:

P(31000 < X < 33000) = P( X \leq 33000) - P (X \leq 31000)

P(31000 < X < 33000) = P( \dfrac{X - 30000}{\dfrac{\sigma}{\sqrt{n}}} \leq \dfrac{33000 - 30000}{\dfrac{9000}{\sqrt{100}}}    )- P( \dfrac{X - 30000}{\dfrac{\sigma}{\sqrt{n}}} \leq \dfrac{31000 - 30000}{\dfrac{9000}{\sqrt{100}}}    )

P(31000 < X < 33000) = P( Z \leq \dfrac{33000 - 30000}{\dfrac{9000}{\sqrt{100}}}    )- P(Z \leq \dfrac{31000 - 30000}{\dfrac{9000}{\sqrt{100}}}    )

P(31000 < X < 33000) = P( Z \leq \dfrac{3000}{\dfrac{9000}{10}}}) -P(Z \leq \dfrac{1000}{\dfrac{9000}{10}}})

P(31000 < X < 33000) = P( Z \leq 3.33)-P(Z \leq 1.11})

From Z tables:

P(31000 < X

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Therefore; the probability that the mean student loan debt for these people is between $31000 and $33000 is 0.1331

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