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3 years ago

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Flying against the wind, an airplane travels 6060 kilometers in 6 hours, Flying with the wind, the same plane travels 12,150 kil

ometers in 9 hours. What is the rate of the plane in still air and what is the rate of the wind?

1 answer:

geniusboy [140]3 years ago

7 0

Let <em>s</em> be the plane's speed in still air and <em>w</em> the wind speed. Then

<em>s</em> - <em>w</em> = (6060 km)/(6 h) = 1010 km/h

<em>s</em> + <em>w</em> = (12,150 km)/(9 h) = 1350 km/h

Adding these equations together eliminates <em>w</em> and lets you solve for <em>s</em> :

(<em>s</em> - <em>w</em>) + (<em>s</em> + <em>w</em>) = 1010 km/h + 1350 km/h

2<em>s</em> = 2360 km/h

<em>s</em> = 1180 km/h

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Answer:

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Step-by-step explanation:

Given

Low Temperature : 40−44 || 45−49 || 50−54 || 55−59 || 60−64

Frequency: --------------- 3 -----------6----------- 1-----------3--- -----7

Required

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The first step is to determine the midpoints of the given temperatures

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$Midpoint = \frac{40+44}{2}$

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$Midpoint = \frac{45+49}{2}$

$Midpoint = \frac{94}{2}$

$Midpoint = 47$

50 - 54:

$Midpoint = \frac{50+54}{2}$

$Midpoint = \frac{104}{2}$

$Midpoint = 52$

55- 59

$Midpoint = \frac{55+59}{2}$

$Midpoint = \frac{114}{2}$

$Midpoint = 57$

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$Midpoint = \frac{60+64}{2}$

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So, the new frequency table is as thus:

Low Temperature : 42 || 47 || 52 || 57 || 62

Frequency: ----------- 3 --||- -6-||- 1-||- --3- ||--7

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$Mean = \frac{\sum fx}{\sum x}$

$Mean = \frac{42 * 3 + 47 * 6 + 52 * 1 + 57 * 3 + 62 * 7}{3+6+1+3+7}$

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<em>The computed mean is greater than the actual mean </em>

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