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4 years ago

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!! Subract and simplify.

Mathematics

1 answer:

pantera1 [17]4 years ago

5 0

Answer: C

<u>Step-by-step explanation:</u>

$\dfrac{x+1}{x-5} - \dfrac{x-2}{x+3}$

= $\dfrac{x+1}{x-5}(\dfrac{x+3}{x+3}) - \dfrac{x-2}{x+3}(\dfrac{x-5}{x-5})$

= $\dfrac{x^{2}+4x+3}{(x-5)(x+3)} - \dfrac{x^{2}-7x+10}{(x-5)(x+3)}$

= $\dfrac{x^{2}+4x+3-(x^{2}-7x+10)}{(x-5)(x+3)}$

= $\dfrac{11x-7}{(x-5)(x+3)}$

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3 years ago

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(a) The percentage of gasoline stations had prices within 4 standard deviations of the mean is 93.75%.

(b) The percentage of gasoline stations had prices within 1.5 standard deviations of the mean is 55.56%. The prices for this stations goes from $3.325 to $3.535.

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The Chebyshev's inequality states that at least 1-(1/K^2) of data from a sample must fall within K standard deviations from the mean, being K any positive real number greater than one.

It can be expressed as

$P(|X-\mu| \geq k\sigma)\leq \frac{1}{k^2}$

In this problem, we have, for the gasoline price, a normal distribution with mean of 3.43 and standar deviation of 0.07.

(a) The percentage of gasoline stations had prices within 4 standard deviations of the mean is equal to <em>one less the percentage of gasoline stations that had prices out of 4 standard deviations of the mean:</em>

$P(|X-\mu| \leq k\sigma)=1-P(|X-\mu| \leq k\sigma) \\\\1-P(|X-\mu| \leq k\sigma) \geq 1-\frac{1}{k^2} \\\\1-P(|X-\mu| \leq 4\sigma) \geq 1-\frac{1}{4^2}\\\\1-P(|X-\mu| \leq 4\sigma) \geq 1-1/16\\\\1-P(|X-\mu| \leq 4\sigma) \geq 0.9375$

The percentage of gasoline stations had prices within 4 standard deviations of the mean is 93.75%.

(b) The percentage of gasoline stations had prices within 1.5 standard deviations of the mean is 55.56%.

$P(|X-\mu| \leq k\sigma)\geq 1-\frac{1}{k^2}\\\\P(|X-\mu| \leq 1.5\sigma)\geq 1-\frac{1}{1.5^2}\\\\P(|X-\mu| \leq 1.5\sigma)\geq 0.5556\\$

The prices for this stations goes from $3.325 to $3.535.

$X=\mu\pm 1.5\sigma=3.43\pm 1.5*0.07=3.43 \pm 0.105\\\\X_{upper} =3.43+0.105=3.535\\X_{lower}=3.43-0.105=3.325$

$x=\mu\pm k\sigma\\\\k=\frac{|x-\mu|}{\sigma} =\frac{|3.64-3.43|}{0.07}=\frac{0.21}{0.07}= 3$

For k=3, the Chebyshev's inequality states:

$P(|X-\mu| \leq 3\sigma)\geq 1-\frac{1}{3^2}\\\\P(|X-\mu| \leq 3\sigma)\geq 0.8889$

So, the minimum percentage of stations that had prices between $3.22 and $3.64 is 88.88%.

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