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4 years ago

2 3/5 x 2 1/3 nearest whole number

Mathematics

2 answers:

Alexeev081 [22]4 years ago

7 0

Answer:

Exact Form:

91 /15

Decimal Form:

6.0 6

Mixed Number Form:

6 1 /15

MARK ME BRAINLIEST PLEASE

slamgirl [31]4 years ago

4 0

Answer:

to the nearest whole number is 3

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20 quarts of peanut oil worth 19 cents a quart must be mixed.

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4 years ago

At one point the average price of regular unleaded gasoline was $3.43 per gallon. Assume that the standard deviation price per

disa [49]

Answer:

(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.

(c) The minimum percentage of stations that had prices between $3.22 and $3.64 is 88.88%.

Step-by-step explanation:

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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3 years ago

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riadik2000 [5.3K]

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4 years ago