Mathematics, 02.03.2020 21:21 dimontsho4717

At one point the average price of regular unleaded gasoline was $3.52 per gallon. Assume that the standard deviation price per gallon is $0.06 per gallon and use Chebyshev�s inequality to answer the following.

a. At least % of gasoline stations had prices within 4 standard deviations of the mean. (round to the nearest hundredth as needed)
b. At least % of gasoline stations had prices within 1.5 standard deviations of the mean.
c. The gasoline prices that are within 1.5 standard deviations of the mean are $ to $ (Use ascending order)
d. % is the minimum percentage of gasoline stations that had prices between $3.40 and $3.64. (round to the nearest hundredth as needed)

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Answer from: rue61

a) $1-\frac{1}{4^2} = 1- \frac{1}{16} = \frac{15}{16} =0.9375 = 93.75\%$

At least 93.75% of gasoline stations had prices within 4 standard deviations of the mean. (round to the nearest hundredth as needed)

b) $1-\frac{1}{1.5^2} =0.5556= 55.56\%$

At least 55.56% of gasoline stations had prices within 1.5 standard deviations of the mean. (round to the nearest hundredth as needed)

c) Lower = 3.52 -1.5*0.06 = 3.43

Upper = 3.52 +1.5*0.06 = 3.61

d) $k = \frac{3.52-3.40}{0.06} = 2$

And since we have the value of k we have:

$1- \frac{1}{2^2}= 1-\frac{1}{4} = \frac{3}{4} = 75\%$

75% is the minimum percentage of gasoline stations that had prices between $3.40 and $3.64. (round to the nearest hundredth as needed)

Step-by-step explanation:

Data given

$\bar X =3.52$ reprsent the sample mean

s=0.06 represent the sample standard deviation

The Chebyshev's Theorem states that for any dataset

•We have at least 75% of all the data within two deviations from the mean.

•We have at least 88.9% of all the data within three deviations from the mean.

•We have at least 93.8% of all the data within four deviations from the mean.

Or in general words "For any set of data (either population or sample) and for any constant k greater than 1, the proportion of the data that must lie within k standard deviations on either side of the mean is at least: $1-\frac{1}{k^2}"$

Part a

$1-\frac{1}{4^2} = 1- \frac{1}{16} = \frac{15}{16} =0.9375 = 93.75\%$

At least 93.75% of gasoline stations had prices within 4 standard deviations of the mean. (round to the nearest hundredth as needed)

Part b

$1-\frac{1}{1.5^2} =0.5556= 55.56\%$

At least 55.56% of gasoline stations had prices within 1.5 standard deviations of the mean. (round to the nearest hundredth as needed)

Part c

For this case we have:

Lower = 3.52 -1.5*0.06 = 3.43

Upper = 3.52 +1.5*0.06 = 3.61

Part d

For this case we can find the number of k like this:

$k = \frac{3.52-3.40}{0.06} = 2$

And since we have the value of k we have:

$1- \frac{1}{2^2}= 1-\frac{1}{4} = \frac{3}{4} = 75\%$

75% is the minimum percentage of gasoline stations that had prices between $3.40 and $3.64. (round to the nearest hundredth as needed)

Answer from: Quest

step-by-step explanation:

Answer from: Quest

x=17

step-by-step explanation:

solve 3x-24=x+10

then that makes it 2x=34 because you subtract an x from both sides to get rid of the one on the right, and then you add 24 to both sides to get rid of the 24 on the left

divide by both sides by 2 to get your answer so that you can get your answer

In circle a, bae dae. what is the value of x?

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