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chubhunter [2.5K]
3 years ago
8

Do these 3 and show your work

Mathematics
1 answer:
Alinara [238K]3 years ago
6 0

Domain:\ x\neq-6\\\\\dfrac{5}{x+6}=10\to\dfrac{5}{x+6}=\dfrac{10}{1}\ \ \ \ |\text{cross multiply}\\\\10(x+6)=(5)(1)\ \ \ \ \ |\text{use distributive property}\\\\(10)(x)+(10)(6)=5\\\\10x+60=5\ \ \ \ |-60\\\\10x=-55\ \ \ \ |:10\\\\\boxed{x=-5.5}\in D

Domain:\\\dfrac{5}{x+6}\geq0\ \wedge\ x+6\neq0\to x+6 > 0\to x > -6\\\\\sqrt{\dfrac{5}{x+6}}=10\ \ \ \ \ |^2\\\\\left(\sqrt{\dfrac{5}{x+6}}\right)^2=10^2\\\\\dfrac{5}{x+6}=100\\\\\dfrac{5}{x+6}=\dfrac{100}{1}\ \ \ \ |\text{cross multiply}\\\\100(x+6)=5\ \ \ \ \ |:100\\\\x+6=0.05\ \ \ \ \ |-6\\\\\boxed{x=-5.95}\in D

Domain:x+3\geq0\to x\geq-3\\\\\sqrt{2x+3-x}=2\to\sqrt{x+3}=2\ \ \ \ |^2\\\\(\sqrt{x+3})^2=2^2\\\\x+3=4\ \ \ \ |-3\\\\\boxed{x=1}\in D

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A certain ball is dropped from a height of x feet. It always bounces up to 2/3 x feet. Suppose the ball is dropped from 10 feet
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Given that if a ball is dropped from x feet, it bounces up to 2/3 x feet.

And the ball is dropped from 10 feet, that is x=10 feet,

So,before the first bounce it travels 10 feet distance.

Between first and second bounce it travels  \frac{2}{3}* 10 + \frac{2}{3} * 10 = 20*(\frac{2}{3})

Between second and third bounce, it travels 20*(\frac{2}{3})^{2}

Between third and fourth bounce, it travels 20*(\frac{2}{3}) ^{3}

Like that between 29th and 30th bounce, it travels 20*(\frac{2}{3} )^{29}

Hence total distance traveled is

    10+20*(\frac{2}{3} )+20*(\frac{2}{3})^{2} + 20*(\frac{2}{3}) ^{3}+......+20*(\frac{2}{3})^{29}

=10+20[(\frac{2}{3}) +(\frac{2}{3}) ^{2} +(\frac{2}{3}) ^{3} +.....+(\frac{2}{3} )^{29} ]

= 10+20[\frac{\frac{2}{3}*(1-(\frac{2}{3})^{29})  }{1-\frac{2}{3} }]

= 10+20*2*(1-(\frac{2}{3}) ^{29})

= 49.9997 feet ≈ 50 feet approximately.

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3 years ago
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The mean student loan debt for college graduates in Illinois is $30000 with a standard deviation of $9000. Suppose a random samp
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Answer:

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Step-by-step explanation:

Given that:

Mean = 30000

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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}}})

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From Z tables:

P(31000 < X

P(31000 < X

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

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