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mel-nik [20]
2 years ago
10

Based on a survey conducted by Greenfield Online, 25 - 34-year-olds spend the most each week on fast food. The average weekly am

ount of $44 was reported in a May 2009 USA Today Snapshot. Assuming that weekly fast food expenditures are normally distributed with a standard deviation of $14.50, what is a probability that a person in this age group will spend more than $60 on fast food in a week?
Mathematics
1 answer:
DIA [1.3K]2 years ago
7 0

The correct answer is P = 0.1357.

In solving for the probability (P) that the 34-year olds will spend more than $60 on fastfood, the mean and the standard deviation is to be taken into account. The mean (μ) being the $44 average weekly expenditure; while the standard deviation (σ) is $14.50. The formula goes this way P(X > 60) = P( X−μ > 60−44 ) = P ( X−μ/σ > 60−44/14.50). 

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Answer:

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Step-by-step explanation:

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The total mixture would then have 8 gallons and 82 %.

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Then you would do 2.4x + 5.6(1) = 6.56

2.4x + 5.6 = 6.56

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0.96/2.4 = 0.4 = 40%

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3 years ago
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Answer:

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Step-by-step explanation:

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3 years ago
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julsineya [31]

Answer:

Her second lap was <u>8%</u> better when compared to her first lap.

Step-by-step explanation:

Given:

Emily ran her first lap in 75 seconds she ran her second lap in 69 seconds.

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Emily ran her first lap in 75 seconds.

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So, we get the difference in seconds:

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Now, to get the percentage of her first lap better than second:

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=0.08\times 100

=8\%.

Therefore, her second lap was 8% better when compared to her first lap.

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Step-by-step explanation:

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is the right answer

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2 years ago
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Step-by-step explanation:

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3 years ago
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