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Tamiku [17]
3 years ago
14

Sunita creates a scale model of an aeroplane. The scale of the model is 5 cm to 13 m. Sunita's model has a wingspan of 24 cm. Wh

at is the wingspan of the real aeroplane?
Mathematics
1 answer:
ICE Princess25 [194]3 years ago
6 0

We have been given that Sunita creates a scale model of an aeroplane. The scale of the model is 5 cm to 13 m. Sunita's model has a wingspan of 24 cm. We are asked to find wingspan of the real aeroplane.

We will use proportion to solve our given problem.

\frac{\text{Wingspan of real plane}}{\text{Wingspan of model aeroplane}}=\frac{\text{Real length}}{\text{Scale length}}

\frac{\text{Wingspan of real plane}}{\text{24 cm}}=\frac{\text{13 m}}{\text{5 cm}}

\frac{\text{Wingspan of real plane}}{\text{24 cm}}\cdot\text{24 cm} =\frac{\text{13 m}}{\text{5 cm}}\cdot\text{24 cm}

\text{Wingspan of real plane}=\text{13 m}\cdot4.8

\text{Wingspan of real plane}=62.4\text{ m}

Therefore, the wingspan of real plane is 62.4 meters.

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Answer:

2.28%

Step-by-step explanation:

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The z score is used in probability to show how many standard deviation is a raw score below or above the mean. The formula for the z score (z) is given by:

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For a raw score (x) of 81 points, the z score can be calculated by:

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Therefore from the normal probability distribution table, the probability that a randomly selected score is greater than 81 can be given as:

P(x > 81) = P(z > 2) = 1 - P(z < 2) = 1 - 0.9772 = 0.0228 = 2.28%

7 0
3 years ago
Juniper wants to make a new rectangular garden in her backyard that has a length that is 2 more than twice the width. She needs
lisabon 2012 [21]

Answer:

2x+2y ≤ 64

x=2y+2

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she have 64ft of wire so the perimeter should be equal or less than 64 ft

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5 0
3 years ago
Eduardo and Paul are leaving the same airport in Florida. Eduardo’s flight to Jamaica is 273 miles long. Paul's flight to Hondur
k0ka [10]

Answer:

B. 41.8° is the correct answer

Step-by-step explanation:

We are given that,

Length of Eduardo's flight path = 273 miles

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Now, using the law of cosines, we get,

238^2=273^2+357^2-2\times 273\times 357\times \cos \theta

i.e. 56644=74529+127449-194922\times \cos \theta

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i.e. 194922\times \cos \theta=145334

i.e. \cos \theta=0.7456

i.e. \theta=\arccos 0.7456

i.e.  θ = 41.8°

Hence, the angle between their flight paths is 41.8°.

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Answer:

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is the answer

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