Americans spend an average of 4 hours per day online. If the standard deviation is 26 minutes, find the range in whitch at least 88.89% of the data will lie. Use Chebyshev's theorem. Round your K to the nearest whole number.

The range in which at least 88.89% of the data will lie is (162,318).

Step-by-step explanation:

Given:

Average time spend online, = hrs = mins

Standard deviation, = mins

We have to find the range in which at least 88.89% of the data will lie by using Chebyshev's theorem :

Formula :

Lower range =

Higher range =

Now we have to find k values.

We know that the percentage of the data that lies within 'k' standard deviation is at least,1-1/k^2 and k > 1.

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Plugging the value of k=3 in the formula above.

The range is .

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So,

The range in which at least 88.89% of the data will lie is (162,318).

7 weeks

step-by-step explanation:

convert each fraction to the least common multiple of 20, so 3/4 = 15/20 & 1/10 = 2/20, then divide 15 by 2 and you get a little over 7, so the food will last 7 weeks.

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation: