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aniked [119]
3 years ago
15

Alice and her friend Emma leave on different flights from the same airport. Alice's flight flies 100 miles due south, then turns

70° toward west and flies 50 miles. Emma's flight flies 100 miles due north, then turns 50° toward east and flies 50 miles. Which fight among you is farther from the airport? Explan your rreasoning
don't spam!
Mathematics
1 answer:
DaniilM [7]3 years ago
3 0

9514 1404 393

Answer:

  Emma's flight is farther from the airport

Step-by-step explanation:

Even without doing any detailed calculation, you can see on a graph that the larger turn Alice's flight makes keeps it closer to the airport.

Emma's flight is farther from the airport.

__

The distance can be calculated using the Law of Cosines. The angle of interest is the supplement of the turn angle. Since the leg distances are the same in each case, the only variable affecting the distance from the airport is the angle measure. Again, the larger the turn, the shorter the distance from the airport. If T is the turn angle, the distance is ...

  d² = 100² +50² -2·100·50·cos(180° -T)

  d² = 12500 +10000·cos(T) . . . . . . . using cos(180°-T) = -cos(T)

  d = 100√(1.25+cos(T))

For Alice's 70° turn, the distance is ...

  d = 100√(1.25+0.342) ≈ 126.18 mi

For Emma's 50° turn, the distance is ...

  d = 100√(1.25 +0.643) ≈ 137.58 mi

Emma's flight is farther from the airport.

_____

The law of cosines formula is ...

  c² = a² +b² -2ab·cos(C)

where triangle sides are lengths a, b, c and angle C is opposite side c.

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GarryVolchara [31]

Answer:

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

Step-by-step explanation:

Check for conditions

For this case in order to use the normal distribution for this case or the 68-95-99.7% rule we need to satisfy 3 conditions:

a) Independence : we assume that the random sample of 400 students each student is independent from the other.

b) 10% condition: We assume that the sample size on this case 400 is less than 10% of the real population size.

c) np= 400*0.74= 296>10

n(1-p) = 400*(1-0.74)=104>10

So then we have all the conditions satisfied.

Solution to the problem

For this case we know that the distribution for the population proportion is given by:

p \sim N(p, \sqrt{\frac{p(1-p)}{n}})

So then:

\mu_p = 0.74

\sigma_p =\sqrt{\frac{0.74(1-0.74)}{400}}=0.0219

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

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I cant give you the answer without the picture but I can tell you how to solve it.

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Step-by-step explanation:

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