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marysya [2.9K]
3 years ago
14

Andrew needs a ladder to hang holiday lights. His house is 24 feet tall and he has a flower bed that extends 4 feet out from the

side of the house. How long of a ladder will he need to reach the top and be out of the flower bed?
Mathematics
1 answer:
blsea [12.9K]3 years ago
5 0

Answer:

24.33 feet

Step-by-step explanation:

The flower bed, the ladder and the house form a right angled triangle where the length of the ladder is the hypotenuse side.

Let L = length of ladder, H = height of house = 24 feet and l = length of flower bed = 4 feet.

Using Pythagoras' theorem,

L² = H² + l²

So, L = √(H² + l²)

substituting the values of the variables, we have

L = √[(24 ft)² + (4 ft)²]

L = √[576 ft² + 16 ft²]

L = √[592 ft²]

L = 24.33 ft

The ladder must be at least 24.33 feet long to reach the top and be out of the flower bed

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Josh travels frequently. For a particular airline, it takes 20 minutes for the first bag to arrive in baggage claim after a flig
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Using the uniform distribution, it is found that:

A,B) 0.3 of the time does it take longer than 27 minutes for Josh’s bag to arrive in baggage claim.

C) 20% of the time does Josh’s bag arrive in less than 22 minutes.

--------------------------

An uniform distribution has two bounds, a and b.  

The probability of finding a value of at lower than x is:

P(X < x) = \frac{x - a}{b - a}

The probability of finding a value between c and d is:

P(c \leq X \leq d) = \frac{d - c}{b - a}

The probability of finding a value above x is:

P(X > x) = \frac{b - x}{b - a}

--------------------------

  • In the graph, we have that the distribution is uniform between 20 and 30 minutes, thus a = 20, b = 30

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Itens a and b:

  • Above 27 minutes, thus:

P(X > 27) = \frac{30 - 27}{30 - 20} = 0.3

0.3 of the time does it take longer than 27 minutes for Josh’s bag to arrive in baggage claim.

--------------------------

Item c:

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P(X < 2) = \frac{22 - 20}{30 - 20} = 0.2

0.2*100 = 20%

20% of the time does Josh’s bag arrive in less than 22 minutes.

A similar problem is given at brainly.com/question/15855314

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A moon is about 420000 miles away from it orbits.
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A Ferris Wheel 22.0m in diameter rotates once every 12.5s. What is the ratio of a persons apperenet weight to her real weight (a
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a = omega^2*r 

omega and r in terms of given data: 
omega = 2*Pi/T 
r = d/2 

Thus: 
a = 2*Pi^2*d/T^2 

What forces cause this acceleration for the passenger, at either top or bottom? 

At top (acceleration is downward): 
Weight (m*g): downward 
Normal force (Ntop): upward 

Thus Newton's 2nd law reads: 
m*g - Ntop = m*a 

At top (acceleration is upward): 
Weight (m*g): downward 
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Thus Newton's 2nd law reads: 
Nbottom - m*g = m*a 

Solve for normal forces in both cases. Normal force is apparent weight, the weight that the passenger thinks is her weight when measuring by any method in the gondola reference frame: 
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Substitute a: 
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We are interested in the ratio of weight (gondola reference frame weight to weight when on the ground): 
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Nbottom/(m*g) = m*(g + 2*Pi^2*d/T^2)/(m*g) 

Simplify: 
Ntop/(m*g) = 1 - 2*Pi^2*d/(g*T^2) 
Nbottom/(m*g) = 1 + 2*Pi^2*d/(g*T^2) 

Data: 
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Results: 
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