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Mashutka [201]
3 years ago
5

A model airplane is built at a scale of 1 inch to 6 feet. If the model plane is 8 inches long, how many feet long is the actual

airplane?
Type your answer as a whole number in the box below.

_____ feet
Mathematics
2 answers:
AnnyKZ [126]3 years ago
7 0

Answer:

48 feet

Step-by-step explanation:

Let's create a proportion using the following setup.

inches/feet=inches/feet

We know the scale is 1 inch to 6 feet. We also know that the plane is 8 inches long, but we don't know how many feet long it is, so we can say <em>x </em>feet.

1 inch/6 feet= 8 inches/ x feet

1/6=8/x

Cross multiply. Multiply the numerator of the first fraction by the denominator of the second. Multiply the denominator of the first by the numerator of the second.

1*x=6*8

x=6*8

x=48

Add appropriate units, in this case: feet.

x= 48 feet

The actual airplane is 48 feet long.

noname [10]3 years ago
5 0

The airplane is 48 feet long in real dimensions.

To find out how long the airplane actually is, we can use the scale and input our amount of inches that it is measured by.

So, 1 inch equals to 6 feet.

Our model plane is 8 inches.

Now, we plug it in.

8 inches = 6(8)

8 inches = 48 feet

Therefore, the length of the real airplane is 48 feet.

48 feet

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Answer:

A. I only

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Step-by-step explanation:

Proper labelling of the axis would help graph less misleading. When plotting graph, both x and y axis must be properly scaled by representing each unit with a reasonable measurement e.g taking 2cm as 1unit, 5cm as 1unit etc.

Note that rescaling the axis will only make the graph look ambiguous and confusing, therefore we must desist from scale resizing.

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3 years ago
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Step-by-step explanation:

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What is the probability of drawing the compliment of a king or a
inna [77]

Answer:

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Step-by-step explanation:

<u><em>Step(i):-</em></u>

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n (S) = 52C₁ = 52

Let E₁ be the event of the card drawn being a king

n( E_{1} ) = 4 _{C_{1} }  = 4

Let E₂ be the event of the card drawn being a queen

n( E_{2} ) = 4 _{C_{1} }  = 4

But E₁ and E₂ are mutually exclusive events

since E₁ U E₂ is the event of drawing a king or a queen

<u><em>step(ii):-</em></u>

The probability  of drawing of a king or a  queen from a standard deck of playing cards

P( E₁ U E₂ ) = P(E₁) +P(E₂)

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P( E₁ U E₂ ) = \frac{8}{52}

<u><em>step(iii):-</em></u>

The probability of drawing the compliment of a king or a  queen from a standard deck of playing cards

P(E_{1}UE_{2})  ^{-} = 1- P(E_{1} U E_{2} )

P(E_{1}UE_{2})  ^{-} = 1- \frac{8}{52}

P(E_{1}UE_{2})  ^{-} = \frac{52-8}{52} = \frac{44}{52} = 0.846

<u><em>Conclusion</em></u>:-

The probability of drawing the compliment of a king or a  queen from a standard deck of playing cards = 0.846

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